Q. The standard reduction potential of the reaction, $H_{2}O\ + \ e^{-}\ \ \longrightarrow \ \ 1/2\ \ H_{2} + \ OH$ at 298 K is
A) $E^{{^\circ}} = \ \frac{RT}{F}\ Iln\ K_{w}$
B) $E^{{^\circ}} = \ - \ \frac{RT}{F}\ In\lbrack{P_{H}}_{2}\ \rbrack^{1/2}\lbrack{OH}^{-}\rbrack$
C) $E^{{^\circ}} = - \ \frac{RT}{F}\ In\frac{\lbrack{p_{H}}_{2}\rbrack^{1/2}}{\lbrack H^{+}\rbrack}$
D) $E^{{^\circ}} = - \ \frac{RT}{F}\ In{\ K}_{w}$
Explanation
(a)
At anode; $\frac{1}{2}\ H_{2} + \ {OH}^{-}\ \ \rightarrow \ H_{2}O\ + \ e^{-}\$
At cathode; $H^{+} + \ e^{-}\ \rightleftharpoons \ \frac{1}{2}H_{2}$
Redox change; $H^{+} + \ {OH}^{-}\ \rightleftharpoons \ H_{2}O$
$K\ = \ \frac{\lbrack H_{2}O\rbrack}{\left\lbrack H^{+} \right\rbrack\lbrack{OH}^{-}\rbrack}$
$E^{{^\circ}}\ = \ \frac{RT}{F}In\ K_{w}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.