Question

Complete and balance the following chemical equations:
(a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \]
(b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]

Hint:

In redox reactions, the oxidizing agent is reduced, and the reducing agent is oxidized. Pay attention to the stoichiometric coefficients when balancing the equation.

Answer 1

1. Complete and Balance the Following Chemical Equations:

(a) Equation:
2MnO₄⁻(aq) + 10I⁻(aq) + 16H⁺(aq) →

Solution:
In acidic medium, MnO₄⁻ gets reduced to Mn²⁺ and I⁻ gets oxidized to I₂.

Half-reactions:
Reduction half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half-reaction: 2I⁻ → I₂ + 2e⁻

Balancing the half-reactions:
Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to equalize the number of electrons transferred.

Reduction: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O

Oxidation: 10I⁻ → 5I₂ + 10e⁻

Balanced Equation:
2MnO₄⁻(aq) + 10I⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 5I₂(aq) + 8H₂O(l)

(b) Equation:
Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) →

Solution:
In acidic medium, Cr₂O₇²⁻ gets reduced to Cr³⁺ and Fe²⁺ gets oxidized to Fe³⁺.

Half-reactions:
Reduction half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation half-reaction: 6Fe²⁺ → 6Fe³⁺ + 6e⁻

Balancing the half-reactions:
The number of electrons is already balanced in the half-reactions.

Balanced Equation:
Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)