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How to Assign Oxidation Numbers

Assigning Oxidation Numbers

The oxidation number is the positive or negative number of an atom that indicates the electrical charge the atom has if its compound consists of ions. In other words, the oxidation number gives the degree of oxidation (loss of electrons) or reduction (gain of electrons) of the atom in a compound. Because they track the number of electrons lost or gained, oxidation numbers are a sort of shorthand for balancing charge in chemical formulas.

This is a list of rules for assigning oxidation numbers, with examples showing the numbers for elements, compounds, and ions.

Rules for Assigning Oxidation Numbers

Various texts contain different numbers of rules and may change their order. Here is a list of oxidation number rules:

  • Write the cation first in a chemical formula, followed by the anion. The cation is the more electropositive atom or ion, while the anion is the more electronegative atom or ion. Some atoms may be either the cation or anion, depending on the other elements in the compound. For example, in HCl, the H is H + , but in NaH, the H is H – .
  • Write the oxidation number with the sign of the charge followed by its value. For example, write +1 and -3 rather than 1+ and 3-. The latter form is used to indicate oxidation state .
  • The oxidation number of a free element or neutral molecule is 0. For example, the oxidation number of C, Ne, O 3 , N 2 , and Cl 2 is 0.
  • The sum of all the oxidation numbers of the atoms in a neutral compound is 0. For example, in NaCl, the oxidation number of Na is +1, while the oxidation of Cl is -1. Added together, +1 + (-1) = 0.
  • The oxidation number of a monatomic ion is the charge of the ion. For example, the oxidation number of Na + is +1, the oxidation number of Cl – is -1, and the oxidation number of N 3- is -3.
  • The sum of the oxidation numbers of a polyatomic ion is the charge of the ion. For example, the sum of the oxidation numbers for SO 4 2-  is -2.
  • The oxidation number of a group 1 (alkali metal) element in a compound is +1.
  • The oxidation number of a group 2 (alkaline earth) element in a compound is +2.
  • The oxidation number of a group 7 (halogen) element in a compound is -1. The exception is when the halogen combines with an element with higher electronegativity (e.g., oxidation number of Cl is +1 in HOCl).
  • The oxidation number of hydrogen in a compound is usually +1. The exception is when hydrogen bonds with metals forming the hydride anion (e.g., LiH, CaH 2 ), giving hydrogen an oxidation number of -1.
  • The oxidation number of oxygen in a compound is usually -2. Exceptions include OF 2 and BaO 2 .

Examples of Assigning Oxidation Numbers

Example 1: Find the oxidation number of iron in Fe 2 O 3 .

The compound has no electrical charge, so the oxidation numbers of iron and oxygen balance each other out. From the rules, you know the oxidation number of oxygen is usually -2. So, find the iron charge that balances the oxygen charge. Remember, the total charge of each atom is its subscript multiplied by its oxidation number. O is -2 There are 3 O atoms in the compound so the total charge is 3 x -2 = -6 The net charge is zero (neutral), so: 2 Fe + 3(-2) = 0 2 Fe = 6 Fe = 3

Example 2: Find the oxidation number for Cl in NaClO3.

Usually, a halogen like Cl has an oxidation number of -1. But, if you assume Na (an alkali metal) has an oxidation number of +1 and O has an oxidation number of -2, the charges don’t balance out to give a neutral compound. It turns out all of the halogens, except for fluorine, have more than one oxidation number. Na = +1 O = -2 1 + Cl + 3(-2) = 0 1 + Cl -6 = 0 Cl -5 = 0 Cl = -5

  • IUPAC (1997) “Oxidation Number”. Compendium of Chemical Terminology (the “Gold Book”) (2nd ed.). Blackwell Scientific Publications. doi: 10.1351/goldbook
  • Karen, P.; McArdle, P.; Takats, J. (2016). “Comprehensive definition of oxidation state (IUPAC Recommendations 2016)”.  Pure Appl. Chem .  88  (8): 831–839. doi: 10.1515/pac-2015-1204
  • Whitten, K. W.; Galley, K. D.; Davis, R. E. (1992).  General Chemistry  (4th ed.). Saunders.

Related Posts

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

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Chemistry LibreTexts

3.4: Oxidation States

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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, coherent, 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.

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). The reaction is as follows:

\[ Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{3.4.1}\]

Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, 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

\[ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{3.4.2} \]

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 superscript 0 emphasizes the oxidation state of the elements):

\[ 4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{3.4.3}\]

Equation 3.4.1 and Equation 3.4.2 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 3.4.3 , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:

\[ electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{3.4.4a}\]

\[ electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{3.4.4a}\]

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 oxygen in air, 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. Previosuly, 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. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.

bc779aea262f6991d515f6f8928bd425.jpg

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 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 3.4.5 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:

\[ \overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{3.4.5} \]

Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So 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:

\[ electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{3.4.6a}\]

\[ electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{3.4.6b}\]

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 \(\PageIndex{1}\)

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

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

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

Exercise \(\PageIndex{1}\)

  • 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

Redox Reactions of Solid Metals in Aqueous Solution

A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure \(\PageIndex{1}\)). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl − ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

\(Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{3.4.5}\)

In subsequent steps, FeCl 2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH) 3 .

a1821e16fd102a37cb6a5029132c9516.jpg

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

\[metal + acid \rightarrow salt + hydrogen \label{3.4.6}\]

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:

\[ Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{3.4.7}\]

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 3.4.8) and the reduction of silver salts by copper (Equation 3.4.9 and Figure \(\PageIndex{2}\)):

\[ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{3.4.8}\]

\[ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{3.4.9}\]

The reaction in Equation 3.4.8 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.

Precipitation_of_Silver_on_Copper.jpg

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 Zn 2 + . 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:

\[ Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{3.4.10}\]

\[ Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{3.4.11}\]

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).

bc6148e2d176471fe6f7f1b70fe6ebb6.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 \(\PageIndex{2}\)

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 .3.4.4 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.

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

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.

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

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

19ed64d0a413643429e781853dacedcc.jpg

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. 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 \(\PageIndex{3}\)), 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.

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  1. How to Assign Oxidation Numbers

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  2. How do you calculate the oxidation number of an element in a compound

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  3. Solved Assign oxidation numbers to each of the elements in

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  4. Assign oxidation numbers to each element in this compound no

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  5. Assigning Oxidation Numbers

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  6. Rules to find oxidation number of an element in a compound. (English

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    Enter the formula of a chemical compound to find the oxidation number of each element. A net ionic charge can be specified at the end of the compound between { and }. For example: ZnCl4 {2-} or NH2NH3 {+}. Enter just an element symbol to show the common and uncommon oxidation states of the element.

  2. 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 ...

  3. How to Assign Oxidation Numbers

    The oxidation number of oxygen in a compound is usually -2. Exceptions include OF 2 and BaO 2. Examples of Assigning Oxidation Numbers Example 1: Find the oxidation number of iron in Fe 2 O 3. The compound has no electrical charge, so the oxidation numbers of iron and oxygen balance each other out.

  4. Rules for Assigning Oxidation Numbers to Elements

    Rule 1: The oxidation number of an element in its free (uncombined) state is zero — for example, Al (s) or Zn (s). This is also true for elements found in nature as diatomic (two-atom) elements and for sulfur, found as: Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion, for example:

  5. Rules for Assigning Oxidation Numbers

    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.

  6. 4.3: Oxidation Numbers and Redox Reactions

    Rules for Assigning Oxidation Numbers. 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.(Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl 2.)

  7. 11.16: Oxidation Numbers and Redox Reactions

    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.

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

    Transcript 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.

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

    To assign the oxidation numbers to the atoms in each compound, let's follow the guidelines outlined above. (a) We know that the oxidation number of F ‍ is − 1 ‍ (guideline 4). Because the sum of the oxidation numbers of the six F ‍ atoms is − 6 ‍ and SF A 6 ‍ is a neutral compound, the oxidation number of S ‍ must be + 6 ‍ :

  10. Assigning Oxidation Numbers

    This chemistry tutorial discusses how to assign oxidation numbers and includes examples of how to determine the oxidation numbers in a compound following som...

  11. Oxidation numbers calculator

    To calculate oxidation numbers of elements in the chemical compound, enter it's formula and click 'Calculate' (for example: Ca2+, HF2^-, Fe4 [Fe (CN)6]3, NH4NO3, so42-, ch3cooh, cuso4*5h2o ). Formula: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

  12. Assign oxidation numbers to each element in this compound.

    Expert-Verified Answer 2 people found it helpful Eduard22sly The oxidation number of each element in the compound are: 1. The oxidation number of K is +1 2. The oxidation number of O is -2 3. The oxidation number of Mn is +7 The oxidation number of certain elements are constant while others varies. The oxidation number of potassium, K is always +1

  13. Assign oxidation numbers to each element in this compound. NO N = O

    Chemistry Middle School verified answered • expert verified Assign oxidation numbers to each element in this compound. NO N = O = Advertisement Expert-Verified Answer question 8 people found it helpful jamuuj Answer; Oxidation number of O =-2 and N= +2 Explanation;

  14. Oxidation States (Oxidation Numbers)

    Rules to determine oxidation states. 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.

  15. PDF Assigning Oxidation Numbers

    Determine the oxidation number of each element in the following: a. S 2O 3 b. Na 2O 2 c. P 2O 5 d. NO 3 y 11. Determine the oxidation number of chlorine in each of the following substances: KClO b. Cl c. Ca(ClO 4) 2 O Assigning Oxidation Numbers to Atoms What is the oxidation number of each kind of atom in the following ions and compounds? a ...

  16. Assigning Oxidation Numbers (w/ Explanation) Flashcards

    C = +4 O = -2 The oxidation number of oxygen in a compound is usually -2. If, however, the oxygen is in a class of compounds called peroxides (for example, hydrogen peroxide), then the oxygen has an oxidation number of -1. There are two oxygens in this molecule for an overall -4. The sum of all oxidation numbers in a neutral compound is zero.

  17. Assign oxidation numbers to each element in this compound. NO

    As per the rule of assigning oxidation number, the oxidation number of O atom is -2. The sum of the oxidation number of a neutral compound is zero. Let us assume x to be the oxidation number of N. x + (-2) = 0. x - 2 = 0. x = +2. Therefore the oxidation of N in NO is +2 and and the oxidation of O in NO is -2 respectively.

  18. 5.3: Oxidation number

    To determine the oxidation number of Cr in Cr 2 O 7 2-: Oxygen will be -2 (Rule 4), for a total of:-2 × 7 = -14. Since the sum of the oxidation numbers will be -2 (the charge on the entire ion), the total for all Cr must be +12 because: +12 + (-14) = -2. Since there is are two Cr, each Cr will have an oxidation number of +6 = +7

  19. QUIZ 4: CHAPTER REVIEW Flashcards

    HClO. Copper is also oxidized even though no oxygen is present. The reaction 2Cu + O2 2CuO has long been called an oxidation-reaction. Copper also reacts with chlorine Cu + Cl2 CuCl2. In such a reaction: [Pb+2] [Cl-]2. The Ksp expression for the system PbCl2 (s) Pb+2 (aq) + 2 Cl- is: Determine the number of grams of NaCl necessary to make up ...

  20. Solved Assign oxidation numbers to each of the elements in

    Science Chemistry Chemistry questions and answers Assign oxidation numbers to each of the elements in the compounds below. ... This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer

  21. 11.1: Oxidation Numbers

    Notice how the subscript of 2 for the \(\ce{S}\) atom had to be accounted for by dividing the result of the subtraction by 2. When assigning oxidation numbers, you do so for each individual atom. In the above example, the oxidation number of sulfur could also have been determined by looking at just the thiosulfate ion, \(\ce{S_2O_3^{2-}}\).

  22. Solved Assign an oxidation number to each element in the

    Assign an oxidation number to each element in the compound: Na2CO3 Select one: O a. Na = +1, C = +4, O = -2 b. Na = +2, C = +4, 0 = -2 Oc. Na = +2, C = +6, O = -6 O d. Na = +1, C = +6, O = -6 Oe. e. Na = +2, C = -4,0 = -2 Consider the reaction: N2 (g) + 3H, 2 NH+heat We can say that this reaction is and is a reaction. Select one: O a. endothermic;

  23. 3.4: Oxidation States

    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.