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# Balance Reduction and Oxidation (Redox) Reactions

### Skill to develop

• Balance reduction-oxidation (Redox) equations till you have developed a logical method without having to memorize all the steps needed to balance redox equations.

# Balance Oxidation and Reduction Reaction equations

You should have studied the two modules or pages Oxidation states and Half reactions before attempting this module. The steps for balancing Redox equations are:

1. Identify the elements that are oxidized and reduced from their oxidation states. See Oxidation States.
2. Write the oxidation and reduction half-reactions and balance them. See Half Reactions.
3. Add the two half-reactions algebraically such that the electrons in the two half-reaction equations cancel completely. Cancel other species such as H+, OH-, and H2O common to the two sides, if necessary.
4. Check your equation and make certain that numbers of atoms and charge are equal on both sides.

Balance the equation from the following two half-reactions:

Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
H2C2O4 --> 2 CO2 + 2 H+ + 2 e-

Multiply the second equation by 3 and then add them algebraically so that the electrons in the two half-reaction equations cancel completely.

Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
3 H2C2O4 --> 6 CO2 + 6 H+ + 6 e-
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Cr2O72- + 8 H+ + 3 H2C2O4 --> 2 Cr3+ + 7 H2O + 6 CO2

Make sure you check your balanced equation.

Balance the equation from the two half-reactions:

Cd --> Cd2+ + 2 e-
4 H+ + NO3- + 3 e- --> NO + 2 H2O

The first half-reaction has 2 electrons, whereas the second one has 3. The lowest common multiple of 2 and 3 is 6. Thus, you multiply the first equation by 3 and the second one by 2. The half-reaction equations become

3 Cd --> 3 Cd2+ + 6 e-
8 H+ + 2 NO3- + 6 e- --> 2 NO + 4 H2O

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3 Cd + 8 H+ + 2 NO3- --> 3 Cd2+ + 2 NO + 4 H2O

Since the last step in balancing oxidation and reduction reactions is simple, it will be more productive simply to practice the determination of oxidation states and balancing of half-reactions again in the dialogue.

In a basic solution, Fe(OH)2 and Fe(OH)3 are solids. The former may be oxidized by H2O2.

Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O.

Balance this equation.

The balanced half-reactions are:

Fe(OH)2 --> Fe(OH)3 + e-
H2O2 + 2 e- --> 2 OH-

Thus, the balanced equation is

2 Fe(OH)2 + H2O2 --> 2 Fe(OH)3

Note that the balanced equation does not have an H2O in it.

Let us balance the following reaction, which is carried out in an acidic solution:

I- + IO3- --> I2

The half-reactions are:

2 I- --> I2 + 2 e- (oxidized)
2 IO3- + 10 e- --> I2 (reduced)

The balanced equation is

10 I- + 2 IO3- --> 6 I2

You may study Electrochemistry to gain better insight to the oxidation reduction process.

### Confidence Building Questions

• How many electrons should there be in the half-reaction?
Al(s) --> Al3+ + ? e-

If you get... lost

Try... Al3+ has 3 positive charges, and each electron has 1 negative charge.

If you get... 3

Excellent... Al is a group III element.

• How many electrons should there be in the half-reaction.
2 H2O2 --> 2 H2O + O2 + ? e-
This is one of three modes of reaction for H2O2.

If you get... 2

Try... Both O atoms in H2O2 change oxidation states in this reaction.

If you get... lost

Try... Neither side of the equation requires the presence of ionic species.

If you get... 0

Excellent... This is an example of disproportionation. In which the change of oxidation state of one atom is compensated by the change of oxidation state of another atom.
2 H2O2 --> 2 H2O + O2

• How many electrons should there be in the half reaction
H2O2 --> 2 H+ + O2 + ? e-
Note that H2O2 is being oxidized in this reaction.

If you get... 0

Try... There is a change of oxidation state for O in this reaction.

If you get... lost

Try... There are 2 positive charges on the right-hand of the equation.

If you get... 2

Excellent... In this reaction, H2O2 is a reducing agent.

• How many electrons should there be in the half reaction.
H2O2 + ? e- --> 2 OH-
Note thatH2O2 is being reduced in this reaction.

If you get... 0

Try... The oxidation state of O atoms changes from -1 to -2.

If you get... 1

Try... There are two O atoms whose oxidation states have changed.

If you get... lost

Try... There are 2 negative charges on the right-hand-side of the equation.

If you get... 2

Excellent... The oxidation states of 2 O atoms change from -1 to -2.

• How many electrons should there be in the half reaction
S8 + ? e- --> 8 S2-?

If you get... 2

Try... There are 8 S atoms in the equation.

If you get... lost

Try... Eight S atoms changed from 0 to -2.

If you get... 16

Excellent... 8 * (-2) = -16.

• What is the oxidation state of In in In(OH)3?

If you get... 3

Try... Oxidation state is a signed number. Our convention is '+3'.

If you get... lost

Try... Remember the OH- ion?

If you get... +3

Excellent... Read OH as OH-.

• What is the oxidation state of Fe in FeAsO4?

If you get... 3

Try... Use a signed number '+3'.

If you get... lost

Try... Since As belongs to group V A, it has an oxidation state of +5. The common oxidation states of Fe are +2 and +3.

If you get... +3

Excellent... The common oxidation states for Fe are +2 and +3.

• In the reaction
2 Cr3+ + H2O + 6 ClO3- --> Cr2O72- + 6 ClO2 + 2 H+,
identify the reducing agent.
• If you get... Cr

Try... Please be specific, Cr3+ ion is the reducing agent.

If you get... lost

Try... A reducing agent is the one that has been oxidized.

If you get... Cr3+

Excellent... Since Cr3+ is oxidized, it is a reducing agent.

• Identify the reducing agent in the reaction
Cr2O72- + H2C=O --> HCOOH + Cr3+.
The formula H2C=O with a double bond between C and O is an aldehyde, and it may also be written as H2CO. HCOOH is an acid. Write it formally as H2CO2 to determine the oxidation state of C.

If you get... H2CO

Try... Yes, this is equal to H2C=O. But now enter H2C=O to obtain a little more information.

If you get... lost

Try... Since Cr2O72- is reduced, it is an oxidant. It can never be a reductant.

If you get... H2C=O

Excellent... Oxidation of an alcohol gives an aldehyde. Its oxidation gives an acid, which can further be oxidized to give CO2 .

• Identify the reducing agent in the reaction:
7 CN- + 2 OH- + 2 Cu(NH3)42+ --> 2 Cu(CN)32- + 8 NH3 + CNO- + H2O. (give the formula)

If you get... CN

Try... It should really be CN-.

If you get... lost

Try... Consider CN- + 2 OH- --> CNO- + H2O + 2 e-

If you get... CN-

Excellent... You have just answered a fairly difficult question.

• Identify the oxidant in the reaction:
7 CN- + 2 OH- + 2 Cu(NH3)42+ --> 2 Cu(CN)32- + 8 NH3 + CNO- + H2O.

If you get... Cu

Try... You should give Cu(NH3)42+. The charge on the species is important.

If you get... lost

Try... The oxidation state for Cu is +2 in Cu(NH3)42+, +1 in Cu(CN)32-. Justify this statement yourself.

If you get... Cu(NH3)42+

Excellent... This is a difficult question.