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(D) The oxidation reaction for the hydrogen electrode is: $H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-}$.Given $pH = 10$,the concentration of hydrogen ion

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(D) The oxidation reaction for the hydrogen electrode is: $H_{2}(g) ightarrow 2H^{+}(aq) + 2e^{-}$.Given $pH = 10$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-10} \ M$.The pressure of hydrogen gas is $P_{H_{2}} = 1 \ atm$.The oxidation potential is given by the Nernst equation: $E_{ox} = E_{ox}^{0} - \frac{0.0591}{n} \log Q$.For the standard hydrogen electrode,$E_{ox}^{0} = 0 \ V$ and $n = 2$.The reaction quotient $Q = \frac{[H^{+}]^{2}}{P_{H_{2}}} = \frac{(10^{-10})^{2}}{1} = 10^{-20}$.Substituting the values: $E_{ox} = 0 - \frac{0.0591}{2} \log(10^{-20})$.$E_{ox} = -\frac{0.0591}{2} 	imes (-20) = 0.0591 	imes 10 = 0.591 \ V$.

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$A$ hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at $1 \ atm$ pressure. The oxidation potential of the electrode would be $.......... \ V$.

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Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
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$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

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