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Standard XII

Chemistry

Nernst Equation

The reduction...

Question

The reduction potential of a hydrogen electrode at $p H$ $10$ at $298 K$ is : $(P = 1 a t m)$

0.51

volt

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0

volt

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0.591

volt

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0.059

volt

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Solution

The correct option is C $0.591$ volt
${2 H}^{+} + {2 e}^{-} ⟶ H_{2} p H = 10, [H^{+}] = 10^{- 10}$

$E^{o} = 0$

$E_{r e d} = \frac{0.0591}{2} log \frac{[H^{+}]^{2}}{P_{H_{2}} (g)} = \frac{0.0591}{2} \times 2 log [10^{- 10}]$

$= \frac{0.0591}{2} \times 2 \times [- 10] = - 0.591 V$

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The reduction potential of a hydrogen electrode at pH 10 at 298K is : (P=1atm)

The correct option is C 0.591 volt2H++2e−⟶H2pH=10,[H+]=10−10Eo=0Ered=0.05912log[H+]2PH2(g)=0.05912×2log[10−10]=0.05912×2×[−10]=−0.591V

The reduction potential of a hydrogen electrode at $p H$ $10$ at $298 K$ is : $(P = 1 a t m)$

The correct option is C $0.591$ volt
${2 H}^{+} + {2 e}^{-} ⟶ H_{2} p H = 10, [H^{+}] = 10^{- 10}$

$E^{o} = 0$

$E_{r e d} = \frac{0.0591}{2} log \frac{[H^{+}]^{2}}{P_{H_{2}} (g)} = \frac{0.0591}{2} \times 2 log [10^{- 10}]$

$= \frac{0.0591}{2} \times 2 \times [- 10] = - 0.591 V$