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)



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.


1st half-reaction equation


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


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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