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Redox Equations

>> Parts of this equation/concept include:

A - Oxidation Numbers
B - Balancing in Acid
C - Balancing in Base
D - Identifying Components of a Redox Reaction

 

A. Oxidation Numbers

The rules for oxidation numbers follow. (Consider each ion of an ionic compound separately.)

  1. The sum of all oxidation numbers is equal to the charge. Therefore

    1. The oxidation number of a pure element is zero.

    2. The oxidation number of a monoatomic ion is equal to its charge.

  2. The oxidation number of hydrogen is

    1. +1 if combined with a nonmetal.
    2. –1 if combined with a metal.

    3. 0 if combined only with itself.

  3. The oxidation number of oxygen is –2 (unless that contradicts rule 1 or 2).

These are also listed in Chapter 5. More details can be found there.

>> Example 1

What is the oxidation number for each element in the following compounds.

  1. LiH
  2. H2
  3. HNO3
  4. H2O2
  5. KClO3

Solution:

  1. Since hydrogen is combined with a metal, its oxidation number is –1. Since the sum of the oxidation numbers must equal the overall charge, the oxidation number of Li must be +1.
  2. In this example, hydrogen is a pure element. Therefore its oxidation number is zero.
  3. Here hydrogen is combined with nonmetals, so its oxidation number is +1. Oxygen normally has an oxidation number of –2. Since there is no contradictory information, this must be correct. So that the oxidation numbers add up to the overall charge (zero), the oxidation number of nitrogen is calculated from the equation 1 + N + 3(–2) = 0. Solving for N, the oxidation number of nitrogen is +5.
  4. Since hydrogen is combined with a nonmetal, it has an oxidation number of +1. Oxygen normally has an oxidation number of –2, but if that was true, the sum of the oxidation numbers would not be equal to the charge. Since that rule (rule 1) has priority, the oxidation number of oxygen is –1. [From 2(+1) + 2(x) = 0, where x is the oxidation number of oxygen.]
  5. Since this is an ionic compound, it will be easier to split it into ions before determining the oxidation numbers. The ions that make up KClO3 are K+ and ClO3–. According to rule 1b, the oxidation number of potassium ion will be +1. Using rule 3, the oxidation number of oxygen is –2. Using rule 1, the sum of the oxidation numbers equals the charge, and the oxidation number on chlorine is determined from Cl + 3(–2) = –1, so the oxidation number of chlorine is +5.

 

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B. Balancing in Acid

For acidic, aqueous solutions

  1. Separate into the skeletal half–reactions
  2. Balance all elements except hydrogen and oxygen
  3. Add H2O to balance water
  4. Add H+ to balance hydrogen
  5. Add electrons (e–) to balance charge
  6. Multiply each half–reaction so that the number of electrons is the same for each half
  7. Add the half–reactions together
  8. Cancel appropriately

See Chapter 5 for more detailed instructions and more examples.

>> Example 2

Balance the following reaction in acid:

O2 + H2CO H2O + HCO2H

Solution:

  1. The elements changing oxidation state are oxygen and carbon. Therefore the skeletal half–reactions are

    O2 H2O

    H2CO HCO2H

  2. All elements except hydrogen and oxygen are balanced

  3. Balancing oxygen by adding water,

    O2 2 H2O

    H2CO + H2O HCO2H

  4. Balance hydrogen by adding H+:

    O2 + 4 H+ 2 H2O

    H2CO + H2O HCO2H + 2 H+

  5. Balance the charge by adding electrons:

    O2 + 4 H+ + 4 e– 2 H2O

    H2CO + H2O HCO2H + 2 H+ + 2 e–

  6. Multiply the bottom half–reaction by 2 so that both half–reaction have four electrons:

    O2 + 4 H+ + 4 e– 2 H2O

    2 H2CO + 2 H2O 2 HCO2H + 4 H+ + 4 e–

  7. Add the two reactions together:

    O2 + 4 H+ + 4 e– + 2 H2CO + 2 H2O 2 H2O + 2 HCO2H + 4 H+ + 4 e–

    Cancel

    O2 + 2 H2CO 2 HCO2H

>> Example 3

Balance the following reaction in acid:

Sn + H2O2 H2O + Sn2+

Solution:

  1. The elements changing oxidation state are oxygen and tin. The skeletal half–reactions are

    Sn0 Sn2+

    (The superscripted zero is an optional way to remind yourself that the charge of a metal is zero and you didn't just forget to include the charge.)

    H2O2 H2O

  2. All elements except hydrogen and oxygen are balanced

  3. Balance oxygen by adding water. Oxygens are balanced in the tin half, so for the other half

    H2O2 2 H2O

  4. Balance hydrogen by adding H+. It is balanced in the tin reaction, so for the other

    2 H+ + H2O2 2 H2O

  5. Balance the charge by adding electrons (both halves):

    Sn0 Sn2+ + 2 e–

    2 e– + 2 H+ + H2O2 2 H2O

  6. Multiply so that the number of electrons is the same. (They already are, so moving on....)

  7. Add the reactions together,

    Sn0 + 2 e– + 2 H+ + H2O2 2 H2O + Sn2+ + 2 e–

    Cancel

    Sn0 + 2 H+ + H2O2 2 H2O + Sn2+

 

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C. Balancing in Base

For basic, aqueous solutions,

  1. Separate into the skeletal half–reactions
  2. Balance all elements except hydrogen and oxygen
  3. For each oxygen needed, add 2 OH– to that side and one H2O to the other side
  4. For each hydrogen needed, add one H2O to that side and one OH– to the other side
  5. Add electrons (e–) to balance the charge
  6. Multiply reactions so that the number of electrons is the same in each
  7. Add the reactions and cancel appropriately

See Chapter 5 for detailed explanation (and more examples).

>> Example 4

Balance the following reaction in base:

Ni + SO42– Ni(OH)2 + S

Solution:

  1. The elements changing oxidation state are sulfur and nickel. Therefore the skeletal half–reactions are

    Ni Ni(OH)2

    SO42– S

  2. All elements except hydrogen and oxygen are balanced

  3. Balance oxygen by adding hydroxide

    The rule says to add twice as many hydroxides as oxygens needed, but a little common sense should tell you that the easiest way to balance the nickel reaction with hydroxide is

    Ni + 2 OH– Ni(OH)2

    Always let common sense beat out the rules. (Although the rules work, they are more difficult than necessary.)

    SO42– + 4 H2O S + 8 OH–

    The second reaction follows the rules.

  4. Hydrogens are balanced

  5. Balance the charge by adding electrons:

    Ni + 2 OH– Ni(OH)2 + 2 e–

    SO42– + 4 H2O + 6 e– S + 8 OH–

  6. Multiply so that the number of electrons is the same in each half–reaction:

    3 Ni + 6 OH– 3 Ni(OH)2 + 6 e–

    SO42– + 4 H2O + 6 e– S + 8 OH–

  7. Add the reactions together:

    SO42– + 4 H2O + 6 e– + 3 Ni + 6 OH– 3 Ni(OH)2 + 6 e– + S + 8 OH–

    Cancel:

    SO42– + 4 H2O + 3 Ni 3 Ni(OH)2 + S + 2 OH–

 

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D. Identifying Components of a Redox Reaction

Oxidation and reduction can be determined from the change in oxidation numbers. If the oxidation number decreases (less positive, more negative), the substance must have gained electrons and have been reduced. If the oxidation number increases (more positive, less negative), the substance must have lost electrons and have been oxidized.

An easier way is to identify oxidation and reduction from the balanced half–reactions. When electrons are a reactant, the substance is gaining the electrons and therefore is being reduced. When electrons are a product, the substance is losing electrons and therefore is being oxidized.

>> Example 5

Identify the oxidation and reduction half–reactions in the previous examples.

Solution:

The balanced half–reactions from the previous examples are

  1. O2 + 4 H+ + 4 e– 2 H2O

    H2CO + H2O HCO2H + 2 H+ + 2 e–

  2. Sn0 Sn2+ + 2 e–

    2 e– + 2 H+ + H2O2 2 H2O

  3. Ni + 2 OH– Ni(OH)2 + 2 e–

    SO42– + 4 H2O + 6 e– S + 8 OH–

Using these examples, the position of electrons within the half–reaction, oxidation, and reduction can be identified.

  1. O2 + 4 H+ + 4 e– 2 H2O       (reduction; electrons are reactants)

    H2CO + H2O HCO2H + 2 H+ + 2 e–        (oxidation; electrons are products)

  2. Sn0 Sn2+ + 2 e–        (oxidation; electrons are products)

    2 e– + 2 H+ + H2O2 2 H2O        (reduction; electrons are reactants)

  3. Ni + 2 OH– Ni(OH)2 + 2 e–        (oxidation; electrons are products)

    SO42– + 4 H2O + 6 e– S + 8 OH–        (reduction; electron are reactants)

The oxidizing and reducing agents are reactants. The oxidizing agent causes oxidation by gaining electrons, being reduced. The reducing agent causes reduction by losing electrons, being oxidized. Therefore the reactant in the oxidation reaction is the reducing agent and the reactant in the reduction reaction is the oxidizing agent.

>> Example 6

What is the oxidizing and reducing agent in Examples 2–4?

Solution:

  1. In Example 2 the half–reactions are

    O2 + 4 H+ + 4 e– 2 H2O       (reduction)

    H2CO + H2O HCO2H + 2 H+ + 2 e–        (oxidation)

    Therefore O2 is the oxidizing agent. It is the reactant containing the element changing oxidation state. H2CO is the reducing agent, the reactant containing the element undergoing oxidation.

  2. In Example 3 the half–reactions are

    Sn0 Sn2+ + 2 e–        (oxidation)

    2 e– + 2 H+ + H2O2 2 H2O        (reduction)

    Therefore Sn is the reducing agent and H2O2 is the oxidizing agent.

  3. In Example 4 the half–reactions are

    Ni + 2 OH– Ni(OH)2 + 2 e–         (oxidation)

    SO42– + 4 H2O + 6 e– S + 8 OH–        (reduction)

    Ni is losing electrons; it is the reducing agent. Sulfur gains electrons; it is the oxidizing agent.

 

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