Q27 / 90 · JEE Main 2023 (01 Feb Shift 1) · Chemistry

At what $pH$ , given half cell ${MnO}_{4}^{-} (0.1 M) ∣ {Mn}^{2 +}$ $(0.001 M)$ will have electrode potential of $1.282 V$ ? (Nearest Integer)
Given $E_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{o} = 1.54 V, \frac{2.303 RT}{F} = 0.059 V$

Correct answer: 3

Solution

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} ⇌ {Mn}^{2 +} + 4 H_{2} O$

$E = E ° - \frac{0.059}{5} \log \frac{[{Mn}^{2 +}]}{[{MnO}_{4}^{-}] {[H^{+}]}^{8}}$

$1.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{- 3}}{10^{- 1} \times {[H^{+}]}^{8}}$

$\frac{0.258 \times 5}{0.059} = \log \frac{10^{- 2}}{{[H^{+}]}^{8}}$

$\Rightarrow 21.86 = - 2 + 8 pH$

$pH = 2.98$

$≃ 3$

Hence, the correct answer is 3.