Question

MnO4– + 8H+ + 5e–Mn2+ + 4H2O, If H+ concentration is decreased from 1 M to 10–4 M at 25ºC, where as concentration of Mn2+ and MnO4–remain 1 M.

• A. The potential decreases by 0.38 V with decrease in oxidising power
• B. The potential increases by 0.38 V with increase in oxidising power
• C. The potential decreases by 0.25 V with decrease in oxidising power
• D. The potential decreases by 0.38 V without affecting oxidising power

Answer 1

Answer: (A)

The potential decreases by 0.38 V with decrease in

oxidising power

Answer 2

• 8H+ + 5e–$\longrightarrow$ Mn2+ + 4H2O

$E_{2} = E^{0} - \frac{0.0591}{5}\log\frac{\left\lbrack Mn^{2 +} \right\rbrack}{\left\lbrack MnO_{4}^{-} \right\rbrack \times \left( 10^{- 4} \right)^{8}} = - \times \frac{0.0591}{5} = - 0.37824$

E1–E2 = 0.38 Volt.