The magnitude of the change in oxidising power of the MnO₄ - /Mn²⁺ couple is x × 10⁻⁴ V, if the H⁺ concentration is decreased from 1M to 10⁻⁴ M at 25°C. (Assume concentration of MnO₄ - and Mn²⁺ to be same on change in H⁺ concentration). The value of x is ___________. [ Given :2.303RT/F = 0.059 ]

Solution

Chemistry ma’am · 4 steps

Write the half-reaction first: 8 protons take part, so [H⁺] enters Nernst to the eighth power.

1. Half-reaction. Permanganate is reduced in acid as MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. So n = 5, and the reaction quotient has [H⁺]⁸ in its denominator.

   E = E° - 0.059/5 log[Mn²⁺]/[MnO₄⁻][H⁺]⁸

2. Only H+ changes. [MnO₄⁻] and [Mn²⁺] stay the same, so they cancel when you subtract the two potentials. The change in E depends only on the [H⁺]⁸ term.
3. Put in numbers. At [H⁺] = 1 M the H⁺ part of the log term is log 1 = 0. At 10⁻⁴ M it is log 1/(10⁻⁴)⁸ = log 10³² = 32. So Δ E = 0.059/5 × 32 = 0.0118 × 32 = 0.3776 V.
4. So. The reduction potential falls by 0.3776 V = 3776 × 10⁻⁴ V, so permanganate becomes a weaker oxidant in less acidic solution. That gives x = 3776.

The common slip: 472. Leaving out the power of 8 on [H⁺] gives 0.059/5 × 4 = 0.0472 V, that is 472, not 3776.

The idea to keep. Oxidising power is the reduction potential. When H⁺ takes part in the half-reaction, its coefficient becomes its power in the Nernst equation, so a pH change can shift E a lot.

Answer from the NTA final answer key. Solution by PrepWiser.