Question

$E_{cell}^{\circ }$ for the reaction, $2H_{2}O\to H_2{O}^{+}+O{H}^{-}$ at $25{}^{\circ }C$ is $-0.8277V$. The equilibrium constant for the reaction is

• A. $10^{-14}$
• B. $10^{-23}$
• C. $10^{-7}$
• D. $10^{-21}$

Answer 1

Answer: (A)

$2H_{2}O\to H_3{O}^{+}+O{H}^{-}$
$E_{cell}^{\circ }=\frac{0.0591}{n}logK$
$logK=\frac{E_{cell}^{\circ }\times n}{0.0591}$
$=\frac{-0.8277\times 1}{0.0591}=-14$
$\Rightarrow K=10^{-14}$