Consider the single electrode process 2H^{+} | vedclass

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(C) The half-cell reaction is $2H^{+} + 2e^{-} \rightleftharpoons H_2$. Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{H_2}

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$0.1$

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(C) The half-cell reaction is $2H^{+} + 2e^{-} ightleftharpoons H_2$. Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$. Given $E = -0.059 \ V$,$E^{\circ} = 0 \ V$ (for $SHE$),$n = 2$,and $P_{H_2} = 1 \ bar$. Substituting the values: $-0.059 = 0 - \frac{0.059}{2} \log \frac{1}{[H^{+}]^2}$. $-0.059 = -\frac{0.059}{2} 	imes (-2 \log [H^{+}])$. $-0.059 = 0.059 \log [H^{+}]$. $\log [H^{+}] = -1$. $[H^{+}] = 10^{-1} = 0.1 \ M$.

Consider the single electrode process $2H^{+} + 2e^{-} \rightleftharpoons H_2$ catalyzed by a platinum black electrode in $HCl$ electrolyte. The potential of the electrode is $-0.059 \ V$. What is the concentration of the acid in the hydrogen half-cell if the $H_2$ pressure is $1 \ bar$ (in $M$)?

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