Electrochemistry 2 Question 49
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 $-0.8277 \mathrm{~V}$. Calculate the equilibrium constant for the reaction,
$$ 2 \mathrm{H}{2} \mathrm{O} \rightleftharpoons \mathrm{H}{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \text {at } 25^{\circ} \mathrm{C} \text {. } $$
$(1989,3 M)$
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Solution:
- $\mathrm{H}{2} \mathrm{O}+e^{-} \rightleftharpoons \frac{1}{2} \mathrm{H}{2}+\mathrm{HO}^{-} ; E^{\circ}=-0.8277 \mathrm{~V}$
$$ \begin{array}{ccc} & \frac{1}{2} \mathrm{H}{2}+\mathrm{H}{2} \mathrm{O} \rightleftharpoons \mathrm{H}{3} \mathrm{O}^{+}+e^{-} ; \quad E^{\circ}=0 \mathrm{~V} \ \hline 2 \mathrm{H}{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}^{-} \quad E^{\circ}=-0.8277 \mathrm{~V} \ & E^{\circ}=0.0592 \log \mathrm{K} \ \Rightarrow & & \log K=\frac{-0.8277}{0.0592}=-13.98 \ & K & =1.04 \times 10^{-14} . \end{array} $$
