Question

The standard reduction potential of the reaction, $ H_2 O + e^- \rightarrow \frac{1}{2} H_2 + OH $ at $298\, K$ is :

• A. $E^{\circ} = \frac{RT}{F} ln \, K_w$
• B. $E^{\circ} = \frac{RT}{F} ln \, [P_{H_2}]^{1/2} [OH^-]$
• C. $E^{\circ} = \frac{RT}{F} ln \, \frac{[P_{H_2}]^{1/2}}{ [H^+]}$
• D. $E^{\circ}=-\frac{R T}{F} \ln K_{w}$

Answer 1

Answer: (B)

$E_{\text {cell }}=0$ at eq $^{m}$
$0=E_{\text {cell }}^{\circ}+\frac{R T}{F} \ln \left[ P _{H_{2}}\right]^{1 / 2}\left[ OH ^{-}\right]$
$E_{\text {cell }}^{\circ}=-\frac{R T}{F} \ln \left[ P _{H_{2}}\right]^{1 / 2}\left[ OH ^{-}\right]$