Electrochemistry - Result Question 30
####32. The pressure of $H_2$ required to make the potential of $H_2$-electrode zero in pure water at $298 K$ is :
(a) $10^{-14} atm$
(b) $10^{-12} atm$
(c) $10^{-10} atm$
(d) $10^{-4} atm$
[2016]
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Solution:
- (a) $2 H^{+}$(aq) $+2 e^{-} \to H_2$ (g)
$\therefore \quad E=E^{0}-\frac{0.0591}{2} \log \frac{P _{H_2}}{[H^{+}]^{2}}$
$0=0-0.0295 \log \frac{P _{H_2}}{(10^{-7})^{2}}$
$\frac{P _{H_2}}{(10^{-7})^{2}}=1$
$P _{H_2}=10^{-14} atm$
