Question

Balance the redox reaction

FeSO₄ + H₂SO₄ + KMnO₄ ---> Fe₂(SO₄)₃ + K₂SO + MnSO₄ + H₂O

What is the coefficient of FeSO₄ in the balanced equation? Please show work.

Answer 1

Above reaction is an example for a "redox" reaction.

Here, Mn is an oxidizing agent and it undergo reduction (i.e. lowering of charge from 7+ to 2+ state) and Fe is a reducing agent (i.e. increasing of charge from 2+ to 3+ state).

In order to understand the reaction more appropriately, look at its ionic equation.

K⁺ + MnO₄^- + Fe²⁺ + SO₄^2- + 2H⁺ + SO₄^2- 2Fe³⁺ + 3SO₄^2- + Mn²⁺ + SO₄^2- + 2K⁺ + 2SO₄^2- + H₂O

Next, rewrite the reaction without spectator ions
MnO₄^- + Fe²⁺ Mn²⁺ + Fe³⁺

In order to balance a redox problem, split the reaction equation in to an oxidation and reduction half-equations.

Further, balance everything in the equations except of H and O, balance O by adding H₂O to the opposite side, balance H by adding H⁺ to the opposite side. Also, add e^- to balance charge.

Finally, multiply the half-equations by a common term to get e^- lost equal to e^- gained.

Oxidation:
Fe²⁺ Fe³⁺ + e^-

Reduction:
MnO₄^- Mn²⁺ (note: Mn on left side is 7+)
MnO₄^- Mn₂⁺ + 4H₂O
8H⁺ + MnO₄^- Mn²⁺ + 4H₂O
5e^- + 8H⁺ + MnO₄^- Mn²⁺ + 4H₂O

The two half-reactions:
Fe²⁺ Fe³⁺ + e^-
5e^- + 8H⁺ + MnO₄^- Mn²⁺ + 4H₂O

To balancing e^- multiply by factor 5
(Fe²⁺ Fe³⁺ + e^-) x 5

5Fe²⁺ 5Fe³⁺ + 5e^-
5e^- + 8H⁺ + MnO₄^- Mn²⁺ + 4H₂O

Adding back together:
5Fe²⁺ + 5e^- + 8H⁺ + MnO₄^- 5Fe³⁺ + 5e^- + Mn²⁺ + 4H₂O

Upon Simplification we will get balanced net ionic equation
5Fe²⁺ + 8H⁺ + MnO₄^- 5Fe³⁺ + Mn²⁺ + 4H₂O

Then, add back K⁺ and SO₄^2-

4H₂SO₄ + KMnO₄ + 5 FeSO₄MnSO₄ + 5/2 Fe₂(SO₄)₃ + K⁺ + 1/2 SO₄^2- + 4H₂O

Multiply the above equation by factor 2 in order to eliminate the fractional coefficients.

8H₂SO₄ + 2KMnO₄ + 10 FeSO₄ 2MnSO₄ + 5 Fe₂(SO₄)₃ + K₂SO₄ + 8H₂O (Balanced)

Therefore, 10 is the coefficient of FeSO₄ in the balanced equation.