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AP®︎/College Chemistry

Course: ap®︎/college chemistry   >   unit 4, oxidation–reduction (redox) reactions.

  • Worked example: Using oxidation numbers to identify oxidation and reduction
  • Balancing redox equations
  • Worked example: Balancing a simple redox equation
  • Worked example: Balancing a redox equation in acidic solution
  • Worked example: Balancing a redox equation in basic solution

What is an oxidation–reduction reaction?

Oxidation numbers.

  • An atom of a free element has an oxidation number of 0 ‍   . For example, each Cl ‍   atom in Cl A 2 ‍   has an oxidation number of 0 ‍   . The same is true for each H ‍   atom in H A 2 ‍   , each S ‍   atom in S A 8 ‍   , and so on.
  • A monatomic ion has an oxidation number equal to its charge. For example, the oxidation number of Cu A 2 + ‍   is + 2 ‍   , and the oxidation number of Br A − ‍   is − 1 ‍   .
  • When combined with other elements, alkali metals (Group 1 A ‍   ) always have an oxidation number of + 1 ‍   , while alkaline earth metals (Group 2 A ‍   ) always have an oxidation number of + 2 ‍   .
  • Fluorine has an oxidation number of − 1 ‍   in all compounds.
  • Hydrogen has an oxidation number of + 1 ‍   in most compounds. The major exception is when hydrogen is combined with metals, as in NaH ‍   or LiAlH A 4 ‍   . In these cases, the oxidation number of hydrogen is − 1 ‍   .
  • Oxygen has an oxidation number of − 2 ‍   in most compounds. The major exception is in peroxides (compounds containing O A 2 A 2 − ‍   ), where oxygen has an oxidation number of − 1 ‍   . Examples of common peroxides include H A 2 O A 2 ‍   and Na A 2 O A 2 ‍   .
  • The other halogens ( Cl ‍   , Br ‍   , and I ‍   ) have an oxidation number of − 1 ‍   in compounds, unless combined with oxygen or fluorine. For example, the oxidation number of Cl ‍   in the ion ClO A 4 A − ‍   is + 7 ‍   (since O ‍   has an oxidation number of − 2 ‍   and the overall charge on the ion is − 1 ‍   ).
  • The sum of the oxidation numbers for all atoms in a neutral compound is equal to zero, while the sum for all atoms in a polyatomic ion is equal to the charge on the ion. Consider the polyatomic ion NO A 3 A − ‍   . Each O ‍   atom has an oxidation number of − 2 ‍   (for a total of − 2 × 3 = − 6 ‍   ). Since the overall charge on the ion is − 1 ‍   , the oxidation number of the N ‍   atom must be + 5 ‍   .

Example 1: Assigning oxidation numbers

Recognizing redox reactions, example 2: using oxidation numbers to identify oxidation and reduction, want to join the conversation.

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What Are the Rules for Assigning Oxidation Numbers?

Redox Reactions and Electrochemistry

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Electrochemical reactions involve the transfer of electrons . Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules.

Rules for Assigning Oxidation Numbers

  • The convention is that the cation is written first in a formula, followed by the anion . For example, in NaH, the H is H-; in HCl, the H is H+.
  • The oxidation number of a free element is always 0. The atoms in He and N 2 , for example, have oxidation numbers of 0.
  • The oxidation number of a monatomic ion equals the charge of the ion. For example, the oxidation number of Na + is +1; the oxidation number of N 3- is -3.
  • The usual oxidation number of hydrogen is +1. The oxidation number of hydrogen is -1 in compounds containing elements that are less ​ electronegative than hydrogen, as in CaH 2 .
  • The oxidation number of oxygen in compounds is usually -2. Exceptions include OF 2 because F is more electronegative than O, and BaO 2 , due to the structure of the peroxide ion, which is [O-O] 2- .
  • The oxidation number of a Group IA element in a compound is +1.
  • The oxidation number of a Group IIA element in a compound is +2.
  • The oxidation number of a Group VIIA element in a compound is -1, except when that element is combined with one having a higher electronegativity. The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. For example, the sum of the oxidation numbers for SO 4 2- is -2.
  • Assigning Oxidation States Example Problem
  • How to Balance Net Ionic Equations
  • Balance Redox Reaction Example Problem
  • The Difference Between Oxidation State and Oxidation Number
  • 5 Steps for Balancing Chemical Equations
  • Chemistry Vocabulary Terms You Should Know
  • Reduction Definition in Chemistry
  • Learn About Redox Problems (Oxidation and Reduction)
  • Oxidation Reduction Reactions—Redox Reactions
  • How to Neutralize a Base With an Acid
  • Net Ionic Equation Definition
  • Oxidation Definition and Example in Chemistry
  • Types of Chemical Reactions
  • Reactions in Water or Aqueous Solution
  • How Many Protons, Neutrons, and Electrons in an Atom?
  • Valence Definition in Chemistry

CHEM101: General Chemistry I

Oxidation numbers and redox reactions.

In the example in the previous text, we could tell, which species were gaining and losing electrons because the species were single elements and elemental ions. However, most redox chemistry is much more complex and cannot be determined by quickly looking at the reaction.

To determine if a substance is being oxidized or reduced in a reaction, we use oxidation numbers. Oxidation numbers are simply a record-keeping tool. We can calculate the oxidation numbers for all elements in a reaction. If an element's oxidation number increases from reactant to product, it is being oxidized. If an element's oxidation number decreases from reactant to product, it is being reduced.

Read this text, which lists the rules we use to determine oxidation numbers. You need to know these rules and be able to apply them to determine the oxidation numbers of elements in compounds. After you study the rules, try the example problems.

Redox reactions  may involve proton transfers and other bond-breaking and bond-making processes, as well as electron transfers, and therefore the equations involved are much more difficult to deal with than those describing  acid-base reactions . In order to be able to recognize redox reactions, we need a method for keeping a careful account of all the electrons. This is done by assigning  oxidation numbers  to each atom before and after the reaction.

For example, in NO 3 –  the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. In NO 2 , on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO 3 – .

This arbitrarily assigned gain of one electron corresponds to reduction of the nitrogen atom on going from NO 3 –  to NO 2 . As a general rule, reduction corresponds to a lowering of the oxidation number of some atom. Oxidation corresponds to increasing the oxidation number of some atom. Applying the oxidation number rules to the following equation, we have:

Since the oxidation number of copper increased from 0 to +2, we say that copper was oxidized and lost two negatively charged electrons. The oxidation number of nitrogen went down from 5 to 4, and so the nitrogen (or nitrate ion) was reduced. Each nitrogen gained one electron, so 2 e –  were needed for the 2 NO 3 – . The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agent. Copper was oxidized because its electrons were accepted by an oxidizing agent, nitrogen (or nitrate ion).

Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO 3 –  does not really have a +5 charge which can be reduced to +4 in NO 2 . Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atoms in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid.

The general rules for oxidation numbers are seen below, taken from the following page in the Analytical Chemistry Core Textbook: Oxidation States

Determining Oxidation States

Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states.These rules provide a simpler method:

  • The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2 ,  S 8 , and large structures of carbon or silicon each have an oxidation state of zero.
  • The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
  • The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
  • The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
  • Some elements almost always have the same oxidation states in their compounds:

Exceptions:

Hydrogen in the metal hydrides : Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H - . The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.

Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).

Oxygen in peroxides : Peroxides include hydrogen peroxide, H 2 O 2 . This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F 2 O : The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen : Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is  not  -1, and work the correct state out using fluorine or oxygen as a reference.

Example: Redox Reactions

Identify the redox reactions and the reducing and oxidizing agents from the following:

a)  The appropriate oxidation numbers are

The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO 2  has been oxidized by MnO 4 – , and so MnO 4 – is the oxidizing agent. MnO 4 –  has been reduced by SO 2 , and so SO 2  is the reducing agent.

b)  The oxidation numbers

show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred.

H 2 S has been oxidized, losing two electrons to form elemental S. Since H 2 S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl – . Since it accepts electrons, HClO is the oxidizing agent.

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Chapter 4. Chemical Reactions and Equations

Oxidation-reduction reactions, learning objectives.

  • Define oxidation and reduction .
  • Assign oxidation numbers to atoms in simple compounds.
  • Recognize a reaction as an oxidation-reduction reaction.

Consider this chemical reaction:

2 Na(s) + Cl 2 (g) → 2 NaCl

The reactants are elements, and it is assumed that they are electrically neutral; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Na + and Cl − ions. Somehow, the individual sodium atoms as reactants had to lose an electron to make the Na + ion, while the chlorine atoms as reactants had to each gain an electron to become the Cl − ion. This reaction involves the transfer of electrons between atoms.

In reality, electrons are lost by some atoms and gained by other atoms simultaneously. However, mentally we can separate the two processes. Oxidation  is defined as the loss of one or more electrons by an atom. Reduction  is defined as the gain of one or more electrons by an atom. In reality, oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactions .

Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use an artificial count called the oxidation number  to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on a series of rules. Oxidation numbers are not necessarily equal to the charge on the atom; we must keep the concepts of charge and oxidation numbers separate.

The rules for assigning oxidation numbers to atoms are as follows:

  • Atoms in their elemental state are assigned an oxidation number of 0.
  • Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, which differentiates them from charges.
  • In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion, H − , in which case rule 2 prevails).
  • In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).

Let us work through a few examples for practice. In H 2 , both hydrogen atoms have an oxidation number of 0, by rule 1. In NaCl, sodium has an oxidation number of +1, while chlorine has an oxidation number of −1, by rule 2. In H 2 O, the hydrogen atoms each have an oxidation number of +1, while the oxygen has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound as per rule 3. By contrast, by rule 3 in hydrogen peroxide (H 2 O 2 ), each hydrogen atom has an oxidation number of +1, while each oxygen atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO 2 . Each oxygen atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the sulfur atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it only means that the sulfur atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.

Assign oxidation numbers to the atoms in each substance.

  • Ba(NO 3 ) 2
  • Br 2 is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.
  • By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.

The compound barium nitrate can be separated into two parts: the Ba 2+ ion and the nitrate ion. Considering these separately, the Ba 2+ ion has an oxidation number of +2 by rule 2. Now consider the NO 3 − ion. Oxygen is assigned an oxidation number of −2, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation

x + 3(−2) = −1

where x is the oxidation number of the nitrogen atom and −1 represents the charge on the species. Evaluating,

x + (−6) = −1 x = +5

Thus, the oxidation number on the N atom in the nitrate ion is +5.

Test Yourself

Assign oxidation numbers to the atoms in H 3 PO 4 .

H = +1, O = −2, P = +5

All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized . When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced . Oxidation and reduction are thus also defined in terms of increasing or decreasing oxidation numbers, respectively.

Identify what is being oxidized and reduced in this redox equation.

2 Na + Cl 2 → 2 NaCl

Consider the reactants. Because both reactants are the elemental forms of their atoms, the Na and Cl atoms as reactants have oxidation numbers of 0. In the ionic product, the sodium ions have an oxidation number of +1, while the chloride ions have an oxidation number of −1:

NaCl-1

We note that the sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; chlorine is decreasing its oxidation number from 0 to −1, so it is being reduced:

NaCl-2

Because oxidation numbers are changing, this is a redox reaction. Note that the total number of electrons lost by the sodium (two, one lost from each atom) is gained by the chlorine atoms (two, one gained for each atom).

C + O 2 → CO 2

C is being oxidized from 0 to +4; O is being reduced from 0 to −2.

In this introduction to oxidation-reduction reactions, we are using straightforward examples. However, oxidation reactions can become quite complex; the following equation represents a redox reaction:

Redox Reaction

To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation.

Food and Drink App: Acids in Foods

Many foods and beverages contain acids. Acids impart a sour note to the taste of foods, which may add some pleasantness to the food. For example, orange juice contains citric acid, H 3 C 6 H 5 O 7 . Note how this formula shows hydrogen atoms in two places; the first hydrogen atoms written are the hydrogen atoms that can form H + ions, while the second hydrogen atoms written are part of the citrate ion, C 6 H 5 O 7 3− . Lemons and limes contain much more citric acid—about 60 times as much—which accounts for these citrus fruits being more sour than most oranges. Vinegar is essentially a ~5% solution of acetic acid (HC 2 H 3 O 2 ) in water. Apples contain malic acid (H 2 C 4 H 4 O 5 ; the name malic acid comes from the apple’s botanical genus name, malus ), while lactic acid (HC 3 H 5 O 3 ) is found in wine and sour milk products, such as yogurt and some cottage cheeses.

Table 4.2 “Various Acids Found in Food and Beverages” lists some acids found in foods, either naturally or as an additive. Frequently, the salts of acid anions are used as additives, such as monosodium glutamate (MSG), which is the sodium salt derived from glutamic acid. As you read the list, you should come to the inescapable conclusion that it is impossible to avoid acids in food and beverages.

Table 4.2 Various Acids Found in Food and Beverages

Key Takeaways

  • Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another.
  • Oxidation numbers are used to keep track of electrons in atoms.
  • There are rules for assigning oxidation numbers to atoms.
  • Oxidation is an increase of oxidation number (a loss of electrons); reduction is a decrease in oxidation number (a gain of electrons).

Is the reaction

2 K(s) + Br 2 (ℓ) → 2 KBr(s)

an oxidation-reduction reaction? Explain your answer.

NaCl(aq) + AgNO 3 (aq) → NaNO 3 (aq) + AgCl(s)

In the reaction

2 Ca(s) + O 2 (g) → 2 CaO

indicate what has lost electrons and what has gained electrons.

16 Fe(s) + 3 S 8 (s) → 8 Fe 2 S 3 (s)

2 Li(s) + O 2 (g) → Li 2 O 2 (s)

indicate what has been oxidized and what has been reduced.

2 Ni(s) + 3 I 2 (s) → 2 NiI 3 (s)

What are two different definitions of oxidation?

What are two different definitions of reduction?

Assign oxidation numbers to each atom in each substance.

c)  SO 2 2−

d)  Ca(NO 3 ) 2

10.  Assign oxidation numbers to each atom in each substance.

b)  (NH 4 ) 2 S

d)  Li 2 O 2 (lithium peroxide)

11.  Assign oxidation numbers to each atom in each substance.

12.  Assign oxidation numbers to each atom in each substance.

a)  NaH (sodium hydride)

13.  Assign oxidation numbers to each atom in each substance.

c)  Rb 2 SO 4

14.  Assign oxidation numbers to each atom in each substance.

a)  C 6 H 6

b)  B(OH) 3

15.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 NO + Cl 2 → 2 NOCl

16.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

Fe + SO 3 → FeSO 3

17.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 KrF 2 + 2 H 2 O → 2 Kr + 4 HF + O 2

18.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

SO 3 + SCl 2 → SOCl 2 + SO 2

19.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 K + MgCl 2 → 2 KCl + Mg

20.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

C 7 H 16 + 11 O 2 → 7 CO 2 + 8 H 2 O

Yes; both K and Br are changing oxidation numbers.

Ca has lost electrons, and O has gained electrons.

Li has been oxidized, and O has been reduced.

loss of electrons; increase in oxidation number

b)  S: +4; O: −2

c)  S: +2; O: −2

d)  Ca: 2+; N: +5; O: −2

a)  C: +2; O: −2

b)  C: +4; O: −2

c)  Ni: +2; Cl: −1

d)  Ni: +3; Cl: −1

a)  C: 0; H: +1; O: −2

b)  N: −3; H: +1

c)  Rb: +1; S: +6; O: −2

d)  Zn: +2; C: 0; H: +1; O: −2

N is being oxidized, and Cl is being reduced.

O is being oxidized, and Kr is being reduced.

K is being oxidized, and Mg is being reduced.

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Oxidation - Reduction Reactions

Oxidation - Reduction Equations

Oxidizing and Reducing Agents

Oxidation-Reduction Equations

Balancing Oxidation-Reduction Equations

There are two situations in which relying on trial and error can get you into trouble. Sometimes the equation is too complex to be solved by trial and error within a reasonable amount of time. Consider the following reaction, for example.

3 Cu( s ) + 8 HNO 3 ( aq ) 3 Cu 2+ ( aq ) + 2 NO( g ) + 6 NO 3 - ( aq ) + 4 H 2 O( l )

Other times, more than one equation can be written that seems to be balanced. The following are just a few of the balanced equations that can be written for the reaction between the permanganate ion and hydrogen peroxide, for example.

Equations such as these have to be balanced by a more systematic approach than trial and error.

The Half-Reaction Method of Balancing Redox Equations

A powerful technique for balancing oxidation-reduction equations involves dividing these reactions into separate oxidation and reduction half-reactions. We then balance the half-reactions, one at a time, and combine them so that electrons are neither created nor destroyed in the reaction.

The steps involved in the half-reaction method for balancing equations can be illustrated by considering the reaction used to determine the amount of the triiodide ion (I 3 - ) in a solution by titration with the thiosulfate (S 2 O 3 2- ) ion.

STEP 1 : Write a skeleton equation for the reaction . The skeleton equation for the reaction on which this titration is based can be written as follows.

I 3 - + S 2 O 3 2- I - + S 4 O 6 2-

STEP 2 : Assign oxidation numbers to atoms on both sides of the equation . The negative charge in the I 3 - ion is formally distributed over the three iodine atoms, which means that the average oxidation state of the iodine atoms in this ion is - 1 / 3 . In the S 4 O 6 2- ion, the total oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is therefore +2 1 / 2 .

STEP 3 : Determine which atoms are oxidized and which are reduced .

STEP 4 : Divide the reaction into oxidation and reduction half-reactions and balance these half-reactions one at a time . This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes what happens during oxidation.

The other describes the reduction half of the reaction.

It doesn't matter which half-reaction we balance first, so let's start with the reduction half-reaction. Our goal is to balance this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the number of iodine atoms on both sides of the equation.

We then balance the charge by noting that two electrons must be added to an I 3 - ion to produce 3 I - ions,

as can be seen from the Lewis structures of these ions shown in the figure below.

We now turn to the oxidation half-reaction. The Lewis structures of the starting material and the product of this half-reaction suggest that we can get an S 4 O 6 2- ion by removing two electrons from a pair of S 2 O 3 2- ions, as shown in the figure below.

STEP 5 : Combine these half-reactions so that electrons are neither created nor destroyed . Two electrons are given off in the oxidation half-reaction and two electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced chemical equation by simply combining these half-reactions.

STEP 6 : Balance the remainder of the equation by inspection, if necessary . Since the overall equation is already balanced in terms of both charge and mass, we simply introduce the symbols describing the states of the reactants and products.

Redox Reactions In Acidic Solutions

Some might argue that we don't need to use half-reactions to balance equations because they can be balanced by trial and error. The half-reaction technique becomes indispensable, however, in balancing reactions such as the oxidation of sulfur dioxide by the dichromate ion in acidic solution.

The reason why this equation is inherently more difficult to balance has nothing to do with the ratio of moles of SO 2 to moles of Cr 2 O 7 2- ; it results from the fact that the solvent takes an active role in both half-reactions.

  • Samples of KMnO 4 are usually contaminated by MnO 2 .
  • Some of the KMnO 4 reacts with trace contaminants when it dissolves in water, even when distilled water is used as the solvent.
  • The presence of traces of MnO 2 in this system catalyzes the decomposition of MnO 4 - ion on standing.

Solutions of this ion therefore have to be standardized by titration just before they are used. A sample of reagent grade sodium oxalate (Na 2 C 2 O 4 ) is weighed out, dissolved in distilled water, acidified with sulfuric acid, and then stirred until the oxalate dissolves. The resulting oxalic acid solution is then used to titrate MnO 4 - to the endpoint of the titration, which is the point at which the last drop of MnO 4 - ion is decolorized and a faint pink color persists for 30 seconds.

Solutions of the MnO 4 - ion that have been standardized against oxalic acid, using the equation balanced in the previous practice problem, can be used to determine the concentration of aqueous solutions of hydrogen peroxide, using the equation balanced in the following practice problem.

Use the half-reaction method to determine the correct stoichiometry for this reaction.

Click here to check your answer to Practice Problem 5.

Redox Reactions in Basic Solutions

Half-reactions are also valuable for balancing equations in basic solutions. The key to success with these reactions is recognizing that basic solutions contain H 2 O molecules and OH - ions. We can therefore add water molecules or hydroxide ions to either side of the equation, as needed.

The following equation describes the reaction between the permanganate ion and hydrogen peroxide in an acidic solution.

It might be interesting to see what happens when this reaction occurs in a basic solution.

Click here to check your answer to Practice Problem 6.

Reactions in which a single reagent undergoes both oxidation and reduction are called disproportionation reactions . Bromine, for example, disproportionates to form bromide and bromate ions when a strong base is added to an aqueous bromine solution.

Molecular Redox Reactions

Lewis structures can play a vital role in understanding oxidation-reduction reactions with complex molecules. Consider the following reaction, for example, which is used in the Breathalyzer to determine the amount of ethyl alcohol or ethanol on the breath of individuals who are suspected of driving while under the influence.

We could balance the oxidation half-reaction in terms of the molecular formulas of the starting material and the product of this half-reaction.

It is easier to understand what happens in this reaction, however, if we assign oxidation numbers to each of the carbon atoms in the Lewis structures of the components of this reaction, as shown in the figure below.

The carbon atom in the CH 3 group in ethanol is assigned an oxidation state of -3 so that it can balance the oxidation states of the three H atoms it carries. Applying the same technique to the CH 2 OH group in the starting material gives an oxidation state of -1.

The carbon in the CH 3 group in the acetic acid formed in this reaction has the same oxidation state as it did in the starting material: -3. There is a change in the oxidation number of the other carbon atom, however, from -1 to +3. The oxidation half-reaction therefore formally corresponds to the loss of four electrons by one of the carbon atoms.

Because this reaction is run in acidic solution, we can add H + and H 2 O molecules as needed to balance the equation.

The other half of this reaction involves a six-electron reduction of the Cr 2 O 7 2- ion in acidic solution to form a pair of Cr 3+ ions.

Adding H + ions and H 2 O molecules as needed gives the following balanced equation for this half-reaction.

We are now ready to combine the two half-reactions by assuming that electrons are neither created nor destroyed in this reaction.

Simplifying this equation by removing 3 H 2 O molecules and 12 H + ions from both sides of the equation gives the balanced equation for this reaction.

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  • Balancing Redox Equations Using the Oxidation Number Method

Key Questions

You follow a series of steps in order.

For example, balance the equation

HNO₃ + H₃AsO₃(aq) → NO(g) + H₃AsO₄(aq) + H₂O(l)

Identify the oxidation number of every atom . Left hand side: H= +1; N= +5; O = -2; As = +3 Right hand side: N = +2; O = -2; H = +1; As = +5

Determine the change in oxidation number for each atom that changes . N: +5 → +2; Change = -3 As: +3 → +5; Change = +2

Make the total increase in oxidation number equal to the total decrease in oxidation number . We need 2 atoms of N for every 3 atoms of As. This gives us total changes of -6 and +6.

Place these numbers as coefficients in front of the formulas containing those atoms . 2HNO₃ + 3H₃AsO₃(aq) → 2NO(g) + 3H₃AsO₄(aq) + H₂O(l)

Balance all remaining atoms other than H and O . Done.

Balance O . Done.

Balance H . Done.

Now try to balance the equations in the link below (answers included).

http://bowvalleycollege.ca/Documents/Learning%20Resource%20Services/Library%20Learning%20Commons/E-Resources/Study%20guides/chemistry30%20ox_num_method.pdf

assigning oxidation numbers in equations

Step 1. Identify the atoms that change oxidation number

Left hand side: #"Zn"# = 0; #"H"# = +1; #"Cl"# = -1 Right hand side: #"Zn"# = +2; #"Cl"# = -1; #"H"# = +1

The changes in oxidation number are: #"Zn"# : 0 → +2; Change = +2 #"H"# : +1 → 0; Change = -1

Step 2. Equalize the changes in oxidation number

Each #"Zn"# atom has lost two electrons, and each #"H"# atom has gained one electron.

You need 2 atoms of #"H"# for every 1 atom of #"Zn"# . This gives us total changes of +2 and -2.

Step 3. Insert coefficients to get these numbers

#color(red)(1)"Zn" + color(red)(2)"HCl" → color(red)(1)"ZnCl"_2 + color(red)(1)"H"_2#

The balanced equation is

#color(red)("Zn" + 2"HCl" → "ZnCl"_2 + "H"_2)#

  • How do you balance redox reactions in basic solution?
  • How do you balance redox equations by oxidation number method?
  • How do you balance redox equations in acidic solutions?
  • What is the oxidation number method?
  • What are some examples of balancing redox equations using the oxidation number method?
  • Why is the oxidation number method useful?
  • Is it necessary to break the equation into half reactions in the oxidation number method?
  • What is the difference between the oxidation number method and the ion-electron method?
  • Can you balance the equation using the oxidation states method MnO2+Al--->Mn+Al2O3?
  • How do you balance this redox reaction using the oxidation number method? Fe2+(aq) + MnO4–(aq) --> Fe3+(aq) + Mn2+(aq)
  • How do you balance this redox reaction using the oxidation number method? HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l)
  • What is the difference between the oxidation number method and the half-reaction method?
  • How do you balance this redox reaction using the oxidation number method? HNO3(aq) + C2H6O(l) + K2Cr2O7(aq) → KNO3(aq) + C2H4O(l) + H2O(l) + Cr(NO3)3(aq)
  • How do you balance this redox reaction using the oxidation number method? Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g)
  • How do you balance this redox reaction using the oxidation number method? KClO3(s) → KCl(s)+O2(g)
  • Question #91e2a
  • Question #51215
  • How do you balance this chemical equation? #"KMnO"_4 + "Na"_2"C"_2"O"_4 + "H"_2"SO"_4 → "K"_2"SO"_4 + "Na"_2"SO"_4 + "MnSO"_4 + "CO"_2 + "H"_2"O"#
  • How do you balance the acid equation #"MnO"_4^"-" + "H"^"+" + "HSO"_3^"-" → "Mn"^"2+" + "SO"_4^"2-" + "H"_2"O"#?
  • Question #33342
  • Question #4e8ae
  • What are the coefficients in front of NO3-(aq) and Cu(s) when the following redox equation is balanced in an acidic solution: ___ NO3-(aq) + ___ Cu(s) → ___ NO(g) + ___ Cu2+(aq)? A) 2, 6 B) 3, 6 C) 3, 4 D) 2, 3
  • Question #79236
  • Question #ebe14
  • How do you balance the following redox reaction? #"MnO"_4^(-) + "I"^(-) -> "MnO"_2 + "IO"_3^(-)# ?
  • Question #de191
  • Question #01c4d
  • Question #84f03
  • Question #6f441
  • Question #4b8db
  • How to I write half reactions and balance this oxidation-reduction reaction? #S_2O_(8(aq))^(2-) + Cr_((aq))^(3+) -> SO_(4(aq))^(2-) + Cr_2O_(7(aq))^(2-)#
  • Question #8d261
  • How to balance #Zn + HNO_3 -> Zn(NO_3)_2 + NH_4NO_3 + H_2O#?
  • How do you balance the redox reaction? #"KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O"# ?
  • Question #92fde
  • Question #e32d1
  • How does lead sulfide react with oxygen gas to form #PbO#, and #SO_2#?
  • Question #2e551
  • Question #3d7dc
  • Question #1d773
  • Question #21c09
  • Question #15bad
  • How do we represent the oxidation of copper metal to #"cupric ion"#, with accompanying reduction of nitrate ion to #NO(g)#?
  • How would you balance the following equation: #"S" + "HNO"_3 -> "H"_2"SO"_4 + "NO"_2 + "H"_2"O"# ?
  • Question #c9eb6
  • Question #755f2
  • Question #86362
  • Question #92bc1
  • Question #7eb16
  • How do we solve a redox equation when BASIC conditions are specified?
  • How do you balance #Cu + H_2SO_4 -> CuSO_4 + SO_2 + H_2O# and identify the element oxidized and the oxidizing agent?
  • How is #"thiosulfate anion"#, #S_2O_3^(2-)#, oxidized by #"permanganate anion"#, #MnO_4^(-)#?
  • How could sulfur be oxidized to #SO_3# by the action of #NO_3^(-)#?
  • How do you balance #Al(s) + S(s) -> Al_2S_3(s)#?
  • How do you balance #K+B_2O_3 -> K_2O + B#?
  • What is the oxidation half reaction for #Fe(s) + 2HCl(aq) -> FeCl_2(aq) + H_2(g)#?
  • Question #319b1
  • What is the difference between oxidation number and oxidation state?
  • How do you balance #CO(g) + I_2O_5(s) -> I_2(s) + CO_2(g)#?
  • Can you represent the reduction of dinitrogen pentoxide by dihydrogen gas?
  • Balance this redox reaction in acidic media? #"As"_2"O"_3(s) + "NO"_3^(-)(aq) -> "H"_3"AsO"_4(aq) + "N"_2"O"_3(aq)#
  • How do you use the half-equation to represent oxidation-reduction reactions?
  • How do you balance #SO_3(g) -> SO_2(g) + O_2(g)#?
  • What do we call the process when oxygen gas reacts with a SINGLE electron?
  • Will methylated spirit dissolve iodine? Will it dissolve sodium chloride or potassium permanganate?
  • Upon treatment of a #14.75*g# mass of #MnO_2# with excess #HCl(aq)#, what VOLUME of chlorine gas is generated under standard conditions...?
  • For a certain reaction #2A + B rightleftharpoons C + 3D#, #K_(eq) = 4.2 xx 10^3#, which of the following is true?
  • Question #cc55e
  • Question #f3a1e
  • Question #59ff2
  • How do represent the oxidation of chloride anion to chlorine in acidic solution?
  • In the oxidation of oxalate ion, #C_2O_4^(2-)# to give carbon dioxide by potassium permanganate, #K^(+)MnO_4^(-)#, how do we vizualize the endpoint?
  • What is the first step to balance a redox equation using the half-reaction method?
  • What is the reduction half-reaction for the unbalanced redox equation #Cr_2O_7^(2-) + NH_4^+ -> Cr_2O_3 + N_2#?
  • Question #2fc11
  • How is zinc metal oxidized, and nitrate ion reduced to give ammonium ion in aqueous by the action of nitric acid on zinc?
  • How do you balance disproportionation redox reactions?
  • How do you balance this reaction? #"K"_2"Cr"_2"O"_7(aq) + "H"_2"SO"_4(aq) + "H"_2"S"(aq) -> "Cr"_2("SO"_4)_3(aq) + "H"_2"O"(l) + "S"(s) + "K"_2"SO"_4(aq)#
  • Question #d0498
  • How do we represent #(i)# represent the reduction of #"chlorate ion"#, #ClO_3^(-)#, to give #ClO_2#?
  • When using the half reaction method of balancing redox reactions, what should you do first?
  • When balancing a redox reaction, what are you balancing?
  • Question #ebe51
  • How do you balance #Al+Cl_2->AlCl_3#?
  • The change in oxidation state of Mn element in the reaction , 3MnO4²- + 4H+→ MnO2+2MnO4- + 2H2O is? Please help me, thanks.
  • Is the reaction of #"lead sulfide"# with #"dioxygen gas"# to give #"lead oxide"#, and #"sulfur dioxide"#, a redox reaction?
  • How is pyrophosphate anion, #P_2O_7^(4-)# reduced to elemental phosphorus, with oxidation of hydrogen sulfide to sulphur?
  • Question #ddcdf
  • Question #a8904
  • How do you represent the redox reaction of chlorate ion, #ClO_3^(-)# with #SO_2(g)# to give #Cl^(-)# and #SO_4^(2-)#?
  • Question #bfe80
  • Question #bb687
  • What is the redox equation for the oxidation of potassium chloride to chlorine gas by potassium permanganate in the presence of sulfuric acid?
  • How do we represent the oxidation of hydrogen sulfide by nitric acid to give sulfur and #NO(g)#?
  • What is the coefficient for #H^+# when the redox equation below is balanced?
  • Given that the sulfate(IV) ion, #SO_2^(-2)#, is converted to the sulfate(VI) ion, #SO_4^(-2)#, in the presence of water, deduce the balanced equation for the redox reaction between #Cr_2O_7^(-2)# (aq) and #SO_3^(-2)#?
  • Question #ec000
  • Question #de72a
  • Can you help me balance this equation? As2S3 + K2Cr2O7 + H2SO4 = H3AsO4 + K2SO4 + Cr2(SO4)3 + H2O + SO2
  • How do you represent the oxidation of #SO_2(g)# by #"nitrous acid"# to give #"sulfuric acid"#?
  • How do you balance the following redox equation in acidic solution: #P_4 + HOCl -> H_3PO_4 + Cl^-#?
  • Question #f0628
  • Question #d0c67
  • Question #6f8ee
  • Question #3a2a3
  • How to balance the following redox problems using both methods?
  • How do you solve #I^- + ClO^- rarr I_3^- +Cl# using the redox reaction method in a base solution?
  • How would we represent the oxidation of sucrose to give oxalic acid with nitric acid oxidant?
  • How does #"iodic acid"#, #HIO_3#, react with iodide anion, #I^-#, to give elemental iodine?
  • Question #75947
  • How do you balance redox equations?
  • How do we represent the oxidation of #Cr^(3+)# ion to #CrO_4^(2-)# by hydrogen peroxide, using the method of half-equations?
  • Question #e3210
  • Question #64b84
  • What is the Balanced redox reaction for the following equation: I2 + Ca(ClO)2➡ Ca(IO3)2 +CaCl2 ?
  • How do I balance this equation using the "half reaction" method: #CH_4+O_2+H_2O=CO+CO_2+H_2# ?
  • Balance this reaction using ion electron method in shortest way possible . but must be clear .. #Fe^(2+)+MnO_4^(-)+H^(+) -> Fe^(3+)+Mn^(2+)+H_2O# ?
  • Question #78b35
  • Question #62a46
  • Question #d49b8
  • Question #c3e76
  • How is #NO_2# oxidized by #Cr_2O_7^(2-)# to give nitrate ion?
  • Question #06291
  • Can you represent the reduction of tellurite ion, #TeO_3^(2-)#, to tellurium metal by the oxidation of iodide ion?
  • How would you balance the following reactions using the oxidation number change method?
  • Question #2bf15
  • Question #6b532
  • How to balance this redox reaction?
  • How do we represent the oxidation of ethanol to acetic acid by potassium permanganate using the oxidation number method?
  • Question #5e7c6
  • Question #3ef9d
  • Question #8470b
  • What redox reaction occurs between #MnO_2# and #HCl#?
  • How does aluminum metal react with nitrate ion under basic conditions to give ammonia, and aluminum ion?
  • Question #be447
  • Question #53cf1
  • How do you write a balanced equation for this redox reaction using the smallest whole number coefficients?
  • Question #e7fd5
  • Question #065fa
  • Question #412b9
  • Balance each of the following half-reactions, assuming that they occur in basic solution?
  • What is the balanced redox reaction between lead and ammonia?
  • Balance this reaction? #"H"_ ((aq))^(+) + "MnO"_ (4(aq))^(2-) -> "MnO"_ ((aq))^(-) + "MnO"_ (2(s)) + "H"_ 2"O"_ ((l))#
  • Balance the below in acidic solution (redox reaction)?
  • Write a balanced redox equation for the following in acidic solution?
  • Write a balanced oxidation reduction equation, in acidic solution, for the below reaction?

Electrochemistry

  • Oxidation and Reduction Reactions
  • Oxidation Numbers
  • Metal Activity Series
  • Galvanic Cells
  • Electrolysis
  • Calculating Energy in Electrochemical Processes

Chapter 2: Atoms and Elements

Chapter 3: molecules, compounds, and chemical equations, chapter 4: chemical quantities and aqueous reactions, chapter 5: gases, chapter 6: thermochemistry, chapter 7: electronic structure of atoms, chapter 8: periodic properties of the elements, chapter 9: chemical bonding: basic concepts, chapter 10: chemical bonding: molecular geometry and bonding theories, chapter 11: liquids, solids, and intermolecular forces, chapter 12: solutions and colloids, chapter 13: chemical kinetics, chapter 14: chemical equilibrium, chapter 15: acids and bases, chapter 16: acid-base and solubility equilibria, chapter 17: thermodynamics, chapter 18: electrochemistry, chapter 19: radioactivity and nuclear chemistry, chapter 20: transition metals and coordination complexes, chapter 21: biochemistry.

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assigning oxidation numbers in equations

Redox reactions between metals and nonmetals typically involve a complete transfer of electrons to form ionic compounds; hence, they are easy to identify. However, redox reactions involving only nonmetals with a partial transfer of electrons are not as easily identifiable. 

Redox reactions are characterized by changes in the oxidation states of the atoms, which indicates electron movement between the atoms.

The oxidation state, or oxidation number, of an atom in a compound is the charge it would have if the shared electrons in each heteronuclear bond were completely transferred to the more electronegative atom. Homonuclear bonds are divided equally.

For instance, in gaseous hydrogen chloride, chlorine is more electronegative. If hydrogen’s electron is transferred completely to chlorine, chlorine gets a 1− charge, corresponding to the −1 oxidation state, and hydrogen gets a 1+ charge, corresponding to the +1 oxidation state.

Oxidation states can be assigned to atoms in elemental form and in most ions and compounds using specific rules. The first three rules are always followed. The remaining rules are applied one by one until the first three rules are satisfied. 

These rules will now be applied to identify whether the formation reactions of sulfur dioxide and calcium carbonate are redox reactions.

According to rule number 1, elements in the free state have an oxidation number of zero, so elemental sulfur and oxygen are both assigned the oxidation number zero.

According to rule number 3, the sum of the oxidation numbers in a neutral compound is zero so the oxidation numbers of sulfur and oxygen in SO 2 must sum to zero.

In accordance with rule number 6, the oxidation number of each oxygen is −2 in SO 2 . Two oxygen atoms sum to −4. The oxidation number of sulfur is, therefore, +4.

The oxidation number of sulfur increases from zero to +4, so it is oxidized, while the oxidation number of oxygen decreases from zero to −2, so it is reduced. Thus, this is a redox reaction.

In the case of calcium carbonate, the oxidation number of oxygen is −2 in all three compounds, and calcium is +2 in calcium oxide and calcium carbonate. According to rule 3, carbon must be +4 in carbon dioxide and calcium carbonate.

Since there is no change in the oxidation numbers of the atoms during the reaction, this is not a redox reaction.

4.11: Oxidation Numbers

In redox reactions, the transfer of electrons occurs between reacting species. Electron transfer is described by a hypothetical number called the oxidation number (or oxidation state). It represents the effective charge of an atom or element, which is assigned using a set of rules.

Oxidation Number (Oxidation State)

In the case of an ionic compound, oxidation numbers are assigned based on the number of electrons transferred between reacting species. For example, in the formation of calcium chloride (CaCl 2 ), calcium loses two valence electrons, and the two chlorine atoms gain one electron each. In CaCl 2 , calcium’s oxidation state is +2, and each chlorine’s oxidation state is −1. 

In the case of covalent compounds, electrons are not gained or lost but instead are shared between the atoms. The atom with a greater attraction for electrons pulls the shared pair more strongly. Reactions involving covalent compounds are identified as redox by applying the concept of oxidation number to track electron movements. Oxidation states help us easily identify the species being oxidized and reduced in redox reactions.  

The Rules for Assigning Oxidation Number

Oxidation numbers can be positive, negative, or zero. They are assigned based on the following rules:

  • All free elements have an oxidation number zero. The elements could be monoatomic, diatomic, or polyatomic. 
  • In a compound, group 1A elements (all alkali metals) have an oxidation number of +1, while group 2A elements (all alkaline earth metals) have an oxidation number of +2. 
  • Halogens usually have an oxidation number of −1, except in their compounds with oxygen, where they have a positive oxidation state.  Fluorine is the most electronegative element. It has a −1 oxidation state in all its compounds.
  • For monoatomic ions, the oxidation number is the same as the charge on the ion. 
  • Oxygen always has an oxidation number of −2, except in peroxides, where its oxidation number is −1.
  • Hydrogen has an oxidation state of +1 with nonmetals and −1 with metals.
  • The sum of the oxidation number for a neutral compound is zero, while for a polyatomic ion, it is equal to the charge on the ion.

This text is adapted from Openstax, Chemistry 2e, Section 4.2: Classifying Chemical Reactions.

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Chemistry LibreTexts

7.3: Oxidation-Reduction Reactions

  • Last updated
  • Save as PDF
  • Page ID 453139

  • Lisa Sharpe Elles
  • University of Kansas

Learning Objectives

  • To identify oxidation–reduction reactions in solution.

The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O 2− ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.

Any oxidation must ALWAYS be accompanied by a reduction and vice versa.

Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows:

\[ \ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1} \]

Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or "oil rig" . The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O 2 or H 2 , are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is

\[\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2} \]

Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al 2 O 3 , electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements):

\[ 4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3} \]

Equation \(\ref{4.4.1}\) and Equation \(\ref{4.4.2}\) are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation \(\ref{4.4.3}\) , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:

\[ \begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \\[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*} \]

\[ \begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \\[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*} \]

The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in Figure \(\PageIndex{1}\) .

In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.

Assigning Oxidation States

Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure \(\PageIndex{1}\) ), magnesium oxide (MgO), and calcium chloride (CaCl 2 ). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.

Diagram of electron transfer in sodium chloride. The reductant, sodium, loses the electron and becomes oxidized. The oxidant, chlorine, gains and electron and becomes reduced.

A set of rules for assigning oxidation states to atoms in chemical compounds follows.

Rules for Assigning Oxidation States

  • The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
  • The oxidation state of a monatomic ion is the same as its charge—for example, Na + = +1, Cl − = −1.
  • The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
  • Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
  • Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
  • The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.

In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.

Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6 . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H − ion, whereas HCl forms H + and Cl − ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF 2 but −½ in KO 2 . Note that an oxidation state of −½ for O in KO 2 is perfectly acceptable.

The reduction of copper(I) oxide shown in Equation \(\ref{4.4.5}\) demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H 2 and Cu. From rule 4, hydrogen in H 2 O has an oxidation state of +1, and from rule 5, oxygen in both Cu 2 O and H 2 O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu 2 O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:

\[ \overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5} \]

Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:

\[ \text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a} \]

\[ \text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b} \]

Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.

Example \(\PageIndex{1}\): Oxidation States

Assign oxidation states to all atoms in each compound.

  • sulfur hexafluoride (SF 6 )
  • methanol (CH 3 OH)
  • ammonium sulfate [(NH 4 )2SO 4 ]
  • magnetite (Fe 3 O 4 )
  • ethanoic (acetic) acid (CH 3 CO 2 H)

Given : molecular or empirical formula

Asked for : oxidation states

Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.

a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF 6 ), the oxidation state of sulfur must be +6:

[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0

b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:

[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0

c. Note that (NH 4 ) 2 SO 4 is an ionic compound that consists of both a polyatomic cation (NH 4 + ) and a polyatomic anion (SO 4 2 − ) (see Table 2.4 ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH 4 + , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:

[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH 4 + ion

For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:

[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion

d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:

Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe 3 O 4 can be viewed as having two Fe 3 + ions and one Fe 2 + ion per formula unit, giving a net positive charge of +8 per formula unit. Fe 3 O 4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”

e. Initially, we assign oxidation states to the components of CH 3 CO 2 H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of

[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0

So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH 3 ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO 2 H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of

[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3

To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus

\[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber \]

Thus the sum of the oxidation states of the two carbon atoms is indeed zero.

Exercise \(\PageIndex{1}\): Oxidation States

  • barium fluoride (BaF 2 )
  • formaldehyde (CH 2 O)
  • potassium dichromate (K 2 Cr 2 O 7 )
  • cesium oxide (CsO 2 )
  • ethanol (CH 3 CH 2 OH)

Ba, +2; F, −1

C, 0; H, +1; O, −2

K, +1; Cr, +6; O, −2

Cs, +1; O, −½

C, −3; H, +1; C, −1; H, +1; O, −2; H, +1

Types of Redox Reactions

Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include:

  • Synthesis reactions: The formation of any compound directly from the elements is a redox reaction, for example, the formation of water from hydrogen and oxygen: \[\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber \]
  • Decomposition reactions: Conversely, the decomposition of a compound to its elements is also a redox reaction, as in the electrolysis of water: \[\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \nonumber \]
  • Combustion reactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water as the products: \[\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \nonumber \]

The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution.

Redox Reactions of Solid Metals in Aqueous Solution

\[\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81} \]

In subsequent steps, \(\ce{FeCl2}\) undergoes oxidation to form a reddish-brown precipitate of \(\ce{Fe(OH)3}\).

Tree growing through hood of rusted abandoned car in a forest.

Many metals dissolve through reactions of this type, which have the general form

\[\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82} \]

Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

\[ \ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83} \]

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!

Single-Displacement Reactions

Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation \(\ref{4.4.84}\)) and the reduction of silver salts by copper (Equation \(\ref{4.4.85}\) and Figure \(\PageIndex{3}\)):

\[ \ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84} \]

\[ \ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85} \]

The reaction in Equation \(\ref{4.4.84}\) is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.

Top-down view of beaker with copper wire placed in silver nitrate solution. Solid silver nitrate can be seen forming around the wire.

The Activity Series

By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing \(\ce{Zn2+}\). Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:

\[ \ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10} \]

\[ \ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11} \]

Magnesium has a greater tendency to be oxidized than zinc does.

Pairwise reactions of this sort are the basis of the activity series (Figure \(\PageIndex{4}\)), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).

imageedit_13_2662634503.jpg

When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series . Because magnesium is above zinc in Figure \(\PageIndex{4}\), magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H 2 . Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example \(\PageIndex{2}\) demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.

Example \(\PageIndex{2}\): Activity

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

  • A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
  • A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
  • Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.

Given: reactants

Asked for: overall reaction and net ionic equation

  • Locate the reactants in the activity series in Figure \(\PageIndex{4}\) and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
  • Write the net ionic equation for the redox reaction.

\[ \ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber \]

Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.

  • A Mercury lies below lead in the activity series, so no reaction will occur.

\[ \ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber \]

Lead(II) sulfate is the white solid that forms on corroded battery terminals.

Closeup view of rusted battery terminal with white solid forming around screw head.

Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.

Exercise \(\PageIndex{2}\)

  • A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
  • A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
  • A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).

\(no\: reaction\)

\(3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)\)

\(2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)\

Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table \(\PageIndex{1}\)), in which the overall reaction is separated into an oxidation equation and a reduction equation. There are many types of redox reactions. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure \(\PageIndex{4}\)), which arranges metals and H 2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.

  • Chemistry Concept Questions and Answers

Oxidation Number Questions

The oxidation number is a theoretical concept. It tells about the hypothetical charge of an atom if all of its bonds are ionic. It tells us about the extent of oxidation of an atom in a molecule. It can be either positive, negative or neutral.

Oxidation Number Chemistry Questions with Solutions

Q1. What is the oxidation number of sulfur in sulphuric acid (H 2 SO 4 )?

  • None of the above

Answer: (b), The oxidation number of sulphur in sulphuric acid (H 2 SO 4 ) is 6.

Calculation:

1 X 2 + x + 4 X – 2 = 0

2 + x – 8 = 0

So, option (b) is correct.

Q2. What is the oxidation number of chromium in calcium dichromate (CaCr 2 O 7 )?

Answer: (b), The oxidation number of chromium in calcium dichromate (CaCr 2 O 7 ) is 6.

2 + 2x + 7 X (- 2) = 0

2 + 2x – 14 = 0

Q3. What is the oxidation number of nitrogen in nitric acid (HNO 3 )?

Answer: (b), The oxidation number of nitrogen in nitric acid (HNO 3 ) is 5.

1 + x + 3 X ( -2) = 0

1 + x – 6 = 0

x – 5 = 0

Q4. Which of the following statements between HClO 4 and HClO 3 is true?

  • The oxidation number of chlorine in HClO 4 has been decreased in HClO 3
  • The oxidation numbers for all atoms are the same in both molecules
  • The oxidation number for chlorine in HClO 4 has increased in HClO 3
  • The oxidation number of oxygen in HClO 4 has been decreased in HClO 3

Answer: (a), The oxidation number of chlorine in HClO 4 has been decreased in HClO 3 is true.

Explanation: To understand it better, foremost we will calculate the oxidation number of HClO 4 and HClO 3.

The oxidation number of chlorine in HClO 4 will be:

1 + x + 4 X ( -2) = 0

1 + x – 8 = 0

x – 7 = 0

The oxidation number of chlorine in HClO 3 will be:

Thus, we can clearly see that oxidation number in HClO 4 has been decreased in HClO 3. So option (a) is correct.

Q5. What is the oxidation number of carbon in CH 2 Cl 2 ?

  • None of the above.

Answer: (d), The oxidation number of carbon in CH 2 Cl 2 is zero.

x + 2 X 1 + 2 X ( -1) = 0

x + 2 – 2 = 0

The oxidation number of carbon in CH 2 Cl 2 is zero.

So, option (d) is correct.

Q6. Chlorine is in +1 oxidation number in

Answer: (d), Chlorine is in +1 oxidation number in Cl 2 O.

Explanation: To understand it better, foremost we will calculate the oxidation number of HCl, HClO 4 , ICl and Cl 2 O.

The oxidation number of chlorine in HCl will be:

The oxidation number of chlorine in ICl will be:

The oxidation number of chlorine in Cl 2 O will be:

2x + ( -2) = 0

Thus, chlorine is in +1 oxidation number in Cl 2 O.

Q7. When K 2 Cr 2 O 7 is converted to K 2 CrO 4 , the change in the oxidation number of chromium is

Answer: (a), When K 2 Cr 2 O 7 is converted to K 2 CrO 4 , the change in the oxidation number of chromium is zero.

Explanation: To understand it better, foremost we will calculate the oxidation of K 2 Cr 2 O 7 and K 2 CrO 4 .

The oxidation number of chromium in K 2 Cr 2 O 7 is:

2 X 1 + 2x + 7 X ( -2) = 0

2x – 12 = 0

The oxidation number of chromium in K 2 CrO 4 is:

2 X 1 + x + 4 X ( -2) = 0

x – 6 = 0

Thus, the oxidation number of chromium does not change when K 2 Cr 2 O 7 is converted to K 2 CrO 4 .

So, option (a) is correct.

Q8. What is the oxidation number of chlorine in HOCl?

Answer: (a), The oxidation number of chlorine in HOCl is 1.

The oxidation number of chlorine in HOCl will be:

1 + ( -2) + x = 0

The oxidation number of chlorine in HOCl is one.

Q9. Oxidation number of oxygen in O 2 molecule is

Answer: (a), Oxidation number of oxygen in O 2 molecule is zero.

Q10. The process in which oxidation number increases is known as

Answer: (a), The process in which oxidation number increases is known as oxidation.

Q11. Which element in the given compounds has the highest oxidation number?

  • Sulphur in SO 3
  • Carbon in CO 2
  • Aluminum in AlCl 3
  • Sulphur in CaS

Answer: (a) Sulphur in SO 3 will have the highest oxidation number.

Explanation: To understand it better, foremost we will calculate the oxidation number of SO 3 , CO 2 , AlCl 3 and CaS.

The oxidation number of sulphur in SO 3 is

x + 3 X (- 2) = 0

The oxidation number of carbon in CO 2 is

x + 2 X (- 2) = 0

x – 4 = 0

The oxidation number of aluminum in AlCl 3 is

x + 3 X (- 1) = 0

x – 3 = 0

The oxidation number of sulphur in CaS is

x = – 2

Thus, the oxidation number of sulphur in SO 3 is highest.

Q12. Name an element that always shows a negative oxidation number.

Answer: Fluorine always shows a negative ( -1) oxidation number.

Q13. What is the oxidation state?

Answer: The oxidation state is the number of electrons that a specific atom can gain, lose or share with another atom. It explains the degree of oxidation of an atom in a molecule.

Q14. What is the oxidation number?

Answer: The oxidation number is the charge that a central metal atom will have even after all the ligands have been removed from that atom.

Q15. What are the rules for finding oxidation numbers?

Rules for finding the oxidation number:

Rule 1: An atom has a zero oxidation number in its elemental form.

Example: The oxidation number of chlorine in the Cl 2 molecule is zero.

Rule 2: The oxidation number of an ion is equivalent to its charge.

Example: The charge of chlorine ion is -1, so the oxidation number of chlorine ion will be -1.

Rule 3: The oxidation number of alkali metals is +1 , and alkaline earth metal is +2 .

Example: The oxidation number of sodium is +1, while the oxidation number of calcium is +2.

Rule 4: Hydrogen has two probable oxidation numbers, i.e. +1 and -1 .

Example: The oxidation number of hydrogen in NaH is -1, while the oxidation number in HCl is +1.

Rule 5: Oxygen has three probable oxidation numbers: +2, -2 and -1.

Example: The oxidation number of oxygen in H 2 O is -2, while the oxidation number in OF 2 is +2. In contrast, the oxidation number of oxygen is -1 in H 2 O 2 .

Rule 6: The oxidation number of fluorine in any compound is -1.

Example: The oxidation number of fluorine in HF is -1.

Rule 7: The oxidation number of halogen is typically equal to -1 except when bonded with oxygen or fluorine atom.

Example: The oxidation number of chlorine in HCl is -1, while the oxidation number of chlorine in HClO 4 is +7.

Rule 8: The sum of the oxidation numbers of neutral compounds equals zero .

Example: The oxidation number of chlorine is -1 in HCl, while the oxidation number of hydrogen is +1, and their sum is equal to zero.

Practise Questions on Oxidation Number

Q1. Fluorine always shows a -1 oxidation number. Why?

Q2. Differentiate between oxidation state and oxidation number.

Q3. What is the oxidation number of chlorine in the perchlorate ion?

Q4. What is the oxidation number of carbon in carbon suboxide (C 3 O 2 )?

Q5. When Chlorine gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

  • Zero to -1 and Zero to +3
  • Zero to +1 and Zero to -3
  • Zero to +1 and Zero to -5
  • Zero to -1 and Zero to +5

Click the PDF to check the answers for Practice Questions. Download PDF

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COMMENTS

  1. 22.6: Assigning Oxidation Numbers

    In the chlorate ion (ClO−3) ( ClO 3 −), the oxidation number of Cl Cl is +5 + 5, and the oxidation number of O O is −2 − 2. In a neutral atom or molecule, the sum of the oxidation numbers must be 0. In a polyatomic ion, the sum of the oxidation numbers of all the atoms in the ion must be equal to the charge on the ion. Example 22.6.1 22 ...

  2. Using oxidation numbers to identify oxidation and reduction (worked

    By assigning oxidation numbers to the atoms of each element in a redox equation, we can determine which element is oxidized and which element is reduced during the reaction. In this video, we'll use this method to identify the oxidized and reduced elements in the reaction that occurs between I⁻ and MnO₄⁻ in basic solution. Created by Sal Khan.

  3. Oxidation-reduction (redox) reactions (article)

    Oxidation numbers can be assigned to the atoms in a reaction using the following guidelines: An atom of a free element has an oxidation number of 0 . For example, each Cl atom in Cl A 2 has an oxidation number of 0 . The same is true for each H atom in H A 2

  4. Rules for Assigning Oxidation Numbers

    By Anne Marie Helmenstine, Ph.D. Updated on August 16, 2019 Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction.

  5. Oxidation States (Oxidation Numbers)

    Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}). The positive oxidation state is the total number of electrons removed from the elemental state.

  6. 4.3: Oxidation Numbers and Redox Reactions

    Because sodium phosphite is neutral species, the sum of the oxidation numbers must be zero. Letting x be the oxidation number of phosphorus, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3. Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1.

  7. Assigning Oxidation Numbers

    Assigning Oxidation Numbers - Chemistry Tutorial TheChemistrySolution 61.2K subscribers Subscribe Subscribed 385K views 12 years ago This chemistry tutorial discusses how to assign oxidation...

  8. Rules for Assigning Oxidation Numbers to Elements

    Chemistry Articles Rules for Assigning Oxidation Numbers to Elements Updated: 07-16-2021 Explore Book Buy On Amazon Oxidation numbers are bookkeeping numbers. They allow chemists to do things such as balance redox ( red uction/ ox idation) equations.

  9. PDF Assigning Oxidation Numbers

    the next lesson, you will learn that equations of complex redox reactions can be balanced by the use of oxidation-number changes. The set of rules on the ... to assign oxidation numbers to all of the elements in each compound. Students can use the problem-solving strategy outlined in Sample Problem 20.2.

  10. CHEM101: Oxidation Numbers and Redox Reactions

    In order to be able to recognize redox reactions, we need a method for keeping a careful account of all the electrons. This is done by assigning oxidation numbers to each atom before and after the reaction. For example, in NO 3- the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of -2.

  11. Oxidation-Reduction Reactions

    Because oxidation numbers are changing, this is a redox reaction. Note that the total number of electrons lost by the sodium (two, one lost from each atom) is gained by the chlorine atoms (two, one gained for each atom). Test Yourself. Identify what is being oxidized and reduced in this redox equation. C + O2 → CO2.

  12. Oxidation-Reduction Equations

    STEP 1: Write a skeleton equation for the reaction . The skeleton equation for the reaction on which this titration is based can be written as follows. I 3- + S 2 O 32- I - + S 4 O 62-. STEP 2: Assign oxidation numbers to atoms on both sides of the equation.

  13. Oxidation Numbers

    The hydrogen oxidation state in a compound is +1 (except for binary metal hydride when the oxidation number is -1), whereas the oxidation number of hydrogen is +1 (except for elements...

  14. 11.16: Oxidation Numbers and Redox Reactions

    Solution: a) The appropriate oxidation numbers are. The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO 2 has been oxidized by MnO 4-, and so MnO 4- is the oxidizing agent. MnO 4- has been reduced by SO 2, and so SO 2 is the reducing agent.

  15. Balancing Redox Equations Using the Oxidation Number Method

    Step 2. Equalize the changes in oxidation number. Each Zn atom has lost two electrons, and each H atom has gained one electron. You need 2 atoms of H for every 1 atom of Zn. This gives us total changes of +2 and -2. Step 3. Insert coefficients to get these numbers. 1Zn + 2HCl → 1ZnCl2 +1H2. The balanced equation is.

  16. Oxidation Numbers and Rules for Assigning Oxidation Numbers

    The Rules for Assigning Oxidation Number. Oxidation numbers can be positive, negative, or zero. They are assigned based on the following rules: All free elements have an oxidation number zero. The elements could be monoatomic, diatomic, or polyatomic. In a compound, group 1A elements (all alkali metals) have an oxidation number of +1, while ...

  17. How To Calculate Oxidation Numbers

    This chemistry video tutorial provides a basic introduction on how to calculate oxidation numbers. It discusses how to find the oxidation states of elements...

  18. PDF Worksheet 10.1 Oxidation numbers and redox equations

    Rules for assigning oxidation numbers 1 The oxidation number of an element is zero. 2 For a monatomic ion, the oxidation number is the charge on the ion. 3 The oxidation number of combined hydrogen is usually +1. 4 The oxidation number of combined oxygen is usually -2. 5 The sum of all oxidation numbers of atoms in a compound is zero.

  19. 11.1: Oxidation Numbers

    Assigning oxidation numbers for molecular compounds is trickier. The key is to remember rule 6: that the sum of all the oxidation numbers for any neutral species must be zero. Make sure to account for any subscripts which appear in the formula. As an example, consider the compound nitric acid, \(\ce{HNO_3}\). According to rule 4, the oxidation ...

  20. Oxidation numbers calculator

    The oxidation state of any chemically bonded carbon may be assigned by adding -1 for each bond to more electropositive atom (H, Na, Ca, B) and +1 for each bond to more electronegative atom (O, Cl, N, P), and 0 for each carbon atom bonded directly to the carbon of interest. For example: propene: CH3-CH=CH2 lauric acid: CH3 (CH2)10COOH

  21. Example Exercise 17.1 Calculating Oxidation Numbers for Carbon

    We assign oxygen an oxidation number of - 2 and write the equation (c) In SO. 4 2-, the polyatomic anion has a charge of 2 -. We assign oxygen an oxidation number of - 2 and write the equation (d) In S. 2. O. 3 2-, the polyatomic anion has a charge of 2 -. We assign oxygen an oxidation number of - 2 and write the equation. Solution

  22. 7.3: Oxidation-Reduction Reactions

    Rules for Assigning Oxidation States. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero. The oxidation state of a monatomic ion is the same as its charge—for example, Na + = +1, Cl − = −1.; The oxidation state of fluorine in chemical compounds is always −1.

  23. Oxidation Number Questions

    Answer: (a) Sulphur in SO 3 will have the highest oxidation number. Explanation: To understand it better, foremost we will calculate the oxidation number of SO 3, CO 2, AlCl 3 and CaS. The oxidation number of sulphur in SO 3 is. x + 3 X (- 2) = 0. x - 6 = 0. x = 6. The oxidation number of carbon in CO 2 is.