Question

The standard reduction potential at ${25}^{\circ }C$ of the reaction $2H_{2}O+2e^{-}\rightleftharpoons H_2+2\stackrel{\ominus }{O}H$ is -0.8277 V. The equilibrium constant for the reaction $2H_{2}O\rightleftharpoons H_3{O}^{\oplus }+\stackrel{\ominus }{O}$ at ${25}^{\circ }C$ is:

• A. $K_w=9.35\times {10}^{-15}$
• B. $K_w=9.83\times {10}^{-15}$
• C. $K_w=9.14\times {10}^{-15}$
• D. $Noneofthese$

Answer 1

The correct option is A $K_w=9.35\times {10}^{-15}$

Consider an electrode of hydrogen $\left(H_2\right)$ as :

$2H^{\oplus }+2e^{-}\to H_2$ ; $E^{\oplus }=0$

Given electrode is :

$2H_{2}O+2e^{-}\to H_2+2\stackrel{\ominus }{O}H$; $E_{red}^{\ominus }=-0.8227V$

$\because E_{oxid}^{\ominus }$
for $H_{2}O>E_{oxid}^{\ominus }$ for $H_2$

Therefore, cell reactions are :

Anode reaction :

$H_2+2\stackrel{\ominus }{O}H\to 2H_{2}O+2e^{-}$ ; $E_{oxid}^{\ominus }=-0.8277V$.

Cathode reaction :

$2H^{\oplus }+2e^{-}\to H_2$ ; $E{\ominus }_{red}=0$

Net reaction : $2H^{\oplus }+2\stackrel{\ominus }{O}H\rightleftharpoons 2H_{2}O$

Thus, $K=\frac{[H_{2}O{]}^2}{\left[H_2^{\oplus }\right]}[\stackrel{\ominus }{O}H{]}^2$

For the reaction,

$2H_{2}O\rightleftharpoons \left[H_3{O}^{\oplus }\right]\left[\stackrel{\ominus }{O}H\right]$

$K_w=\left[H_3{O}^{\oplus }\right]\left[\stackrel{\ominus }{O}\right]$

$K={\left[\frac{1}{K_w}\right]}^2$

Also, $E_{cell}=E_{oxid\left(H_{2}O\right)}+E_{red\left(H\right)}$

$=E_{oxid\left(H_{2}O\right)}^{\ominus }-\frac{0.059}{2}$ log $\frac{[H_{2}O{]}^2}{\left[H_2\right][\stackrel{\ominus }{O}H{]}^2}+E_{red\left(H^{\oplus }/H_2\right)}^{\ominus }+\frac{0.059}{2}$ log $\frac{[H^{\oplus }{]}^2}{\left[H_2\right]}$

$=0.8277+\frac{0.059}{2}$ log $\frac{[H^{\oplus }{]}^2\cdot [H_2]\cdot [\stackrel{\ominus }{O}H{]}^2}{\left[H2\right]\cdot [H_{2}O{]}^2}$

$=0.8277+\frac{0.059}{2}$ log $\frac{\left[H^{\oplus }{]}^2\right[\stackrel{\ominus }{O}H{]}^2}{[H_{2}O{]}^2}$

$=0.8277+\frac{0.059}{2}log\frac{1}{K}$

From equations (i) and (ii),

$E_{cell}=0.8277+\frac{0.059}{2}$ log $[K_w{]}^2$

At equikibrium, $E_{cell}=0$

$\therefore -0.8277=0.059log{K}_w$

or log $K_w=\frac{0.8277}{0.059}$

or $K_w=9.35\times {10}^{-15}$