The standard reduction potential of the reaction, ${{\rm{H}}_2}{\rm{O}} + {e^ - } \to 1/2{{\rm{H}}_2} + {\rm{OH}}$ at 298 K is

${E^ \circ } = \frac{{RT}}{F}{\rm{ln}}{K_w}$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}\left[ {{{\rm{P}}_{\rm{H}}}_2{]^{1/2}}} \right[{\rm{O}}{{\rm{H}}^ - }]$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}\frac{{{{[{p_H}_2]}^{1/2}}}}{{\left[ {{{\rm{H}}^ + }} \right]}}$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}{K_w}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Hard

1 of 1

Analyzing +4 | -1

The standard reduction potential of the reaction,

{\mathrm{H}_2}\mathrm{O} + {e^ - } \to 1/2{\mathrm{H}_2} + \mathrm{OH}

at 298 K is

View answer and explanation

Correct option: A

{E^ \circ } = \frac{{RT}}{F}\mathrm{ln}{K_w}

Option D (

{E^ \circ } = - \frac{{RT}}{F}\mathrm{ln}{K_w}

) is incorrect. The correct answer is A because it relates the standard reduction potential to the equilibrium constant

K_w

for water dissociation.