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 + ? = 1 . 54 V , 2.303 RT F = 0.059 V [2023]

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Solution

Q.
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 $E_{{MnO}_{4}^{-} | {Mn}^{2 +}}^{\circ} = 1.54 V, \frac{2.303 RT}{F} = 0.059 V$ [2023]

Ans.
(3)

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

$10^{- 1} M$ $10^{- 3} M$

$E_{R P} = E_{R P}^{\circ} - \frac{0.059}{5} \log \frac{[{Mn}^{2 +}]}{[{MnO}_{4}^{-}] {[H^{+}]}^{8}}$

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

$- 0.258 = - \frac{0.06}{5} [\log 10 - 2 - 8 \log [H^{+}]]$

$- 0.258 = - \frac{0.06}{5} [- 2 + 8 pH]$

$- 21.5 = 2 - 8 pH$ $\Rightarrow \frac{23.5}{8} = pH$

$pH = 2.9375 \approx 3$

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