- An oxidant is reduced and simultaneously the reductant is oxidized. It follows from our above discussion that we should be able to balance oxidation-reduction reactions on the basis of

- Ion-electron method
- Oxidation number method

### Ion-Electron Method

- Ascertain the reactants and products, and their chemical formulas.
- Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.
- If the reaction occurs in acid medium use a requisite number of H⁺ for balancing the number of atoms involved in the partial equation. for alkaline medium use OH⁻ ions.
- Balancing the charge in the partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation or the reduction half-reactions.
- Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.
- Add the partial equations and cancel out species that appear on both sides of the equations.

- An aqueous acid medium, potassium permanganate oxidizes ferrous state to the ferric state

- In this reaction, permanganate ions are the oxidant and ferrous ion the reductant. The left-hand side of the ultimate equation will carry, MnO₄⁻ (KMnO₄), H⁺ (H₂SO₄) and Fe²⁺ (FeSO₄).

- The Right-hand side will have as products, Mn²⁺ (MnSO₄), H₂O and Fe³⁺ [Fe₂(SO₄)₃]. The partial equation representing the reduction of the oxidant MnO₄⁻ will involve,

MnO₄⁻ → Mn²⁺

- Since the reaction occurs in an acid medium we utilize eight H⁺ ions to balance the four oxygen atoms in MnO₄⁻.

Thus, MnO₄⁻+8H⁺ → Mn²⁺+4H₂O

- Since the above partial equation is still unbalanced from the viewpoint of charge, the equation is balanced by bringing in five electrons:

MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O

- If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following:

Fe⁺² ⇆ Fe⁺³ + e

- This equation on multiplication by 5 and then adding to the partial equation of the oxidant gives the final balanced equation:

MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O

5×( Fe⁺² ⇆ Fe⁺³ + e)

MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O

5×( Fe⁺² ⇆ Fe⁺³ + e)

MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O

- Oxidation of potassium iodide by potassium dichromate in dilute acid medium.

- In this reaction, dichromate is reduced to (+6) state to (+3) state and iodide ion is oxidized to elementary iodine. Taking the dichromate side, balancing the atoms and charges provides the partial equation

Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O

(Reduction of Oxidant)

(Reduction of Oxidant)

Considering the case of iodide ion we get

2I⁻ ⇆ I₂ + 2e

(Oxidation of Reductant)

2I⁻ ⇆ I₂ + 2e

(Oxidation of Reductant)

- Multiplying the above equation by 3 and adding these two partial equations we have the following final equation

Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O

3 (2I⁻ ⇆ I₂ + 2e)

Cr₂O₇⁻²+14H⁺+6I⁻→2Cr⁺³+3I₂+7H₂O

3 (2I⁻ ⇆ I₂ + 2e)

Cr₂O₇⁻²+14H⁺+6I⁻→2Cr⁺³+3I₂+7H₂O

- The reaction of permanganate ion with sodium stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and sodium stannate(Na₂SnO₃) are formed.

- The Partial equation representing the reduction of oxidant is, Since the medium is alkaline we put the requisite number of OH⁻ ions to effect atom balance as

MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻

- To balance the charge on both sides three electrons are added on the left-hand side

MnO₄⁻ + 2H₂O + 3e ⇆ MnO₂ + 4OH⁻

For the reductant, the balance partial equation is

SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e

For the reductant, the balance partial equation is

SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e

- To balance the number of electrons involved in the two partial equations the first equation is multiplied by 2 and the second by 3 and on the adding, we have the final form of the ionic equation

2MnO₄⁻ + 4H₂O + 6e ⇆ 2MnO₂ + 8OH⁻3

SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

2MnO₄⁻+3SnO₂⁻²+H₂O→2MnO₂+3SnO₃⁻²+2OH⁻

SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

2MnO₄⁻+3SnO₂⁻²+H₂O→2MnO₂+3SnO₃⁻²+2OH⁻

- Oxidation of Mn⁺² to MnO₄⁻ by PbO₂ or NaBiO₃ in acid medium

Balancing oxidation reaction |

### Oxidation number method

- Assuming that the oxidation number of oxygen remains unchanged manganese in MnO₄⁻ has an oxidation number +7. Calculation of the decrease and increase in oxidation number gives:

Balancing oxidation-reduction reactions |

- Putting the right factors the decreases and increases in oxidation numbers are balanced, giving,

MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³

- Since the reaction medium is acidic and the changes are not equal on the two sides of the above expression, H⁺ ions are added and the requisite number of H₂O written:

MnO₄⁻+8H⁺+5Fe⁺² → Mn⁺²+ 5Fe⁺³+4H₂O

- Oxidation of iodide ion by dichromate ion in acid medium

The reaction represented as Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂

- The oxidation number of Cr decreases by 2 and the oxidation number of Sn increases by 1. Equalizing the increase and decrease in oxidation number,

Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂

- The medium is acidic and because the ionic charges are not equal on the two sides, fourteen H⁺ ions are added to the left-hand side and the requisite number of H₂O added to the left-hand side.

Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ +7H₂O

- Oxidation of sodium stannite to stannate in alkaline medium

This reaction represented as,

MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

- The oxidation number of Mn decreases by 3 and the oxidation number of Sn increases by 2. Equalizing the increase and decrease in oxidation number,

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²

- Because the medium is alkaline and because the ionic charges are not equal on the two sides, two OH⁻ ions are added to the right-hand side.

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

- Since there are two hydrogen atoms on the right and none on the left, H₂O is added to make atom balance. The final equation is

2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

Problem- Express by ion-electron method the reduction of permanganate to manganous stat by hydrogen peroxide in acid medium.

- This Reaction occurs in acid medium and the partial equation is

MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O

(Reduction of Oxidant)

(Reduction of Oxidant)

- In an acid medium, H₂O₂ will give O₂ and the partial equation being

H₂O₂ ⇆ 2H⁺ + O₂ +2e

(Oxidation of Reductant)

(Oxidation of Reductant)

- The first equation is multiplying by 2 and the second is 5 to have electron balanced. We have,

2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O

5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e

2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂

Problem5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e

2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂

- Express by ion-electron method the reduction of nitrate ion to ammonia by aluminum in aqueous NaOH.

- In the alkaline medium, we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is

NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

- Metallic aluminum will go over to aluminate ion, the partial equation being

Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O +3e

- Multiplying by right factors for electron balance we have the balanced equation

3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻

8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O +24e

3NO₃⁻+8Al+2H₂O+5OH⁻→3NH₃+8AlO₂⁻

Problem8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O +24e

3NO₃⁻+8Al+2H₂O+5OH⁻→3NH₃+8AlO₂⁻

- Use the Oxidation Number method to balance the reaction of iodide ion and iodate ion in an acid medium to liberate iodine.

The reaction represented as

I⁻ + IO₃⁻ → I₂

I⁻ + IO₃⁻ → I₂

- Representation of the above equation with the oxidation number of iodine,

I⁻(-1) + IO₃⁻(+5) → I₂(0)

- Since I⁻ (-1) oxidation number of iodide increases by 1 and IO₃⁻(+5) oxidation number decreases by 5. Putting the right factors the decrease and increase in oxidation number balanced

5I⁻ + IO₃⁻ → 3I₂

- Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation

I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O

Problem- Use the oxidation number method to balance the reaction between Sulfurous acid and dichromate in acidic medium.

The reaction represented as,

SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

- Oxidation of two Cr decreases by 2 × (+3) = 6 and the oxidation number of S increases by 2. Equalizing the above equation increase and decrees in oxidation number

3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

- Because the medium is acidic and the ionic charges are not equal on the two sides, eight H⁺ ions are added to the left-hand side.

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³

- Since there are eight hydrogen ion on the left and none on the right, four H₂O added to make atom balance. Thus the final equation is

3SO₃⁻²+Cr₂O₇⁻²+8H⁺→3SO₄⁻²+2Cr⁺³+4H₂O