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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 one atm pressure. The oxidation potential of electrode would be?
A
1.81 V
B
0.059 V
C
0.59 V
D
0.118 V
Detailed Solution
$\frac{1}{2}H_2(g) \rightarrow H^+ + e^-$
$E_{O.P.} = E^\circ_{O.P.} - \frac{0.059}{n}\log\frac{[H^+]}{(P_{H_2})^{1/2}}$
pH = 10, so $[H^+] = 10^{-10}$ M: $E_{O.P.} = 0 - \frac{0.059}{1}\log\frac{10^{-10}}{(1)^{1/2}}$
$E_{O.P.} = 0.59$ V
$E_{O.P.} = E^\circ_{O.P.} - \frac{0.059}{n}\log\frac{[H^+]}{(P_{H_2})^{1/2}}$
pH = 10, so $[H^+] = 10^{-10}$ M: $E_{O.P.} = 0 - \frac{0.059}{1}\log\frac{10^{-10}}{(1)^{1/2}}$
$E_{O.P.} = 0.59$ V