Question

The standard reduction potential at ${25}^{o}C$ of the reaction $H_{2}O+2e^{-}\to H_2+2O{H}^{-}$ is $-\text{0}\text{.8277}\text{V}$. Calculate the equilibrium constant for the reaction $2H_{2}O\to H_3{O}^{+}+O{H}^{-}$ at ${25}^{o}C$.

Answer 1

Consider the given reaction as the net cell reaction two half reaction :

Oxidation :

$H_{2}O+\frac{1}{2}{H}_2⟶H_3{O}^{+}+e^{(-)}$ $E_{cell}^0=0$

Reduction :

$H_{2}O+e^{-}⟶\frac{1}{2}{H}_2\left(g\right)+{OH}^{(-)}$ $E_{cell}^0=-0.8277V$

It is evident from the cell reaction that it involves the transfer of one electron. So that $n=1$

$log{K}_c=\frac{n{E}_{cell}^0}{0.059}$

$\Rightarrow K_c=anti$ $log\frac{1}{0.059}(-0.8277)$

$=-(-14.028)$

$\Rightarrow K_c=anti$ $log\left(14.028\right)$

$=-1.146$