Question

The magnitude of the change in oxidising power of the MnO₄⁻ / Mn²⁺ couple is \( x \times 10^{-4} \) V, if the H⁺ concentration is decreased from 1 M 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 __________ .

Hint:

The oxidizing power of Permanganate is highly pH-dependent. Decreasing $[H^+]$ (increasing pH) significantly {reduces} its reduction potential.

Answer 1

Correct Answer: 3776

Step 1: Half reaction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.
Step 2: Nernst equation: $E = E^\circ - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$.
Step 3: Since $[Mn^{2+}] = [MnO_4^-]$, $E = E^\circ - \frac{0.059}{5} \log \frac{1}{[H^+]^8} = E^\circ + \frac{0.059 \times 8}{5} \log [H^+]$.
Step 4: Change $\Delta E = E_2 - E_1 = \frac{0.472}{5} (\log 10^{-4} - \log 1) = 0.0944 \times (-4) = -0.3776 \text{ V}$.
Step 5: Magnitude $= 0.3776 \text{ V} = 3776 \times 10^{-4} \text{ V}$. Thus, $x = 3776$.