How to construct and balance redox equations?

How do we construct and balance redox equations?

The way to construct a redox equation is to write two balanced ionic half-reaction equations, one for the oxidising agent and the other one for the reducing agent, and combine them. The method to balance each half-reaction equation is slightly different from that of a normal stoichiometric equation and involves the following steps:-

    1. Balance elements other than O and H
    2. Balance O by adding H2O
    3. Balance H by adding H+
    4. Balance excess positive charges by adding electrons, e
    5. (For reaction in a basic solution) Balance H+ by adding OHon both sides, combine H+ and OH to form H2O and cancel common terms

For example, the redox equation for the oxidation of acidified iron (II) sulphate by potassium dichromate (VI) is constructed as follows:

1st half-reaction equation

Fe2+ Fe3+

Balancing the equation

Fe2+ Fe3+ + e            eq1

(note that steps 1 through 3 and step 5 do not apply)

2nd half-reaction equation

Cr2O72- Cr3+

Balancing the equation

Cr2O72- 2Cr3+

Cr2O72- → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ + 6e→ 2Cr3+ + 7H2O            eq2

(note that step 5 does not apply)

Next, combine the balanced half-reaction equations, cancel common terms and ensure that all electrons are eliminated in the final equation. For this example, we multiply eq1 by 6 throughout before adding to eq2, giving:

6Fe2+(aq) + Cr2O72- (aq) + 14H+ (aq) → 6Fe3+(aq) + 2Cr3+ (aq) + 7H2O (l)

 

Question

Write the redox equation for the oxidation of sodium formate by potassium manganite (VII) to form sodium carbonate and manganese dioxide in slightly basic conditions.

Answer

1st half-reaction equation

HCO2CO32-

Balancing the equation

HCO2–   + H2O CO32-

HCO2–   + H2O CO32- + 3H+

HCO2–   + H2O CO32- + 3H+ + 2e

HCO2–   + H2+ 3OHCO32- + 3H+ + 2e+ 3OH

HCO2–   + H2+ 3OHCO32- + 3H2O + 2e

HCO2+ 3OHCO32- + 2H2O + 2e

2nd half-reaction equation

MnO4MnO2

Balancing the equation

MnO4MnO2 + 2H2O

MnO4– + 4H+ MnO2 + 2H2O

MnO4– + 4H+ + 3eMnO2 + 2H2O

MnO4– + 4H+ + 3e– + 4OH→ MnO2 + 2H2O + 4OH

MnO4– + 4H2O + 3e→ MnO2 + 2H2O + 4OH

MnO4– + 2H2O + 3e→ MnO2 + 4OH

Combining the balanced half-reaction equations, we have,

3HCO2– (aq) + OH– (aq) + 2MnO4– (aq) → 3CO32- (aq) + 2MnO2 (s) + 2H2O (l)

 

 

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