Balance the following redox reaction in acidic solution. Zn (s) + MnO-4 (aq) → Zn+2 (aq) + Mn+2 (aq)

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

Split the reaction into an oxidation and reduction half. By inspecting, balance the half equations with respect to the charges and atoms. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation. Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides. Multiply both equations with appropriate factors to balance the electrons in the two half equations. Add up the balanced half equations and cancel out any specie that occur on both sides. Check to see if the charge and atoms are balanced.

Solution

Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

Zn → Zn²⁺ Oxidation half

MnO₄⁻ → Mn²⁺ Reduction half

Balancing of atoms (in acidic medium)

Zn → Zn²⁺

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

Zn → Zn²⁺ + 2e⁻

MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

Multiply the oxidation half by 5 and reduction half by 2:

5Zn → 5Zn²⁺ + 10e⁻

2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O