Question

The standard potential of the reaction $H_{2}\text{O + e- } \rightarrow \frac{1}{2}\text{ H2 + O}\text{H}^{-}$ at 298 K by using Kw (H2O) = 10–14, is :

• A. – 0.828 V
• B. 0.828 V
• C. 0 V
• D. – 0.5 V

Answer 1

Answer: (A)

– 0.828 V

Answer 2

$H^{+}\text{ + }\text{e}^{-} \rightarrow \ \frac{1}{2}\text{ }\text{H}_{2}\text{ , }\text{E}^{0}\text{ = 0 , }\Delta G^{0}\text{ = 0}$

H2O $H^{+}\text{ + O}\text{H}^{-}\text{, }\Delta G^{0}\text{ = - 8.314 }\text{×}\text{ 298 ln 1}\text{0}^{\text{-14}}$H2O +

e–$\longrightarrow$ $\frac { 1 } { 2 }$

$\frac{1}{2}H_{2}\text{ + O}\text{H}^{-}\text{, -1 }\text{×}\text{ }\text{E}^{0}\text{ }\text{×}\text{ 96500 = -8.314 }\text{×}\text{ 298 ln1}\text{0}^{\text{-14}}$E⁰

= – 0.828 Volt