The reduction potential of a hydrogen half-ce | vedclass

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Question

(C) The reduction reaction for a hydrogen half-cell is: $2H^{+} + 2e^{-} \rightarrow H_{2}$Here,$n = 2$ and the reaction quotient $Q = \frac{p(H_{2})}

Vedclass · Accepted Answer

$p(H_{2}) = 2 \ atm$ and $[H^{+}] = 1.0 \ M$

Vedclass · Answer

(C) The reduction reaction for a hydrogen half-cell is: $2H^{+} + 2e^{-} \rightarrow H_{2}$Here,$n = 2$ and the reaction quotient $Q = \frac{p(H_{2})}{[H^{+}]^{2}}$.Using the Nernst equation: $E_{H^{+}/H_{2}} = E^{0}_{H^{+}/H_{2}} - \frac{0.059}{n} \log Q$.Since $E^{0}_{H^{+}/H_{2}} = 0 \ V$,we have $E_{H^{+}/H_{2}} = -\frac{0.059}{2} \log Q$.For $E_{H^{+}/H_{2}}$ to be negative,$\log Q$ must be positive,which means $Q > 1$.Evaluating the options:$(A)$ $Q = \frac{1}{1^{2}} = 1$ (Not negative)$(B)$ $Q = \frac{1}{2^{2}} = 0.25 $(C)$ $Q = \frac{2}{1^{2}} = 2 > 1$ (Negative potential)$(D)$ $Q = \frac{2}{2^{2}} = 0.5 Therefore,the correct option is $(C)$.

The reduction potential of a hydrogen half-cell will be negative if:

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