Electrochemistry Result Question 48
48. The standard reduction potential at $25^{\circ} \mathrm{C}$ of the reaction, $2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \rightleftharpoons \mathrm{H}_2+2 \mathrm{OH}^{-}$, is -0.8277 V . Calculate the equilibrium constant for the reaction,
$ 2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-} \text {at } 25^{\circ} \mathrm{C} \text {. } $
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Answer:
Correct Answer: 48. $(1.04 \times 10^{-14})$
Solution:
- $\quad \mathrm{H}_2 \mathrm{O}+e^{-} \rightleftharpoons \frac{1}{2} \mathrm{H}_2+\mathrm{HO}^{-} ; E^{\circ}=-0.8277 \mathrm{~V}$
$ \begin{aligned} & & E^{\circ} & =0.0592 \log \mathrm{K} \\ \Rightarrow & & \log K & =\frac{-0.8277}{0.0592}=-13.98 \\ \Rightarrow & & K & =1.04 \times 10^{-14} . \end{aligned} $
BETA
