Question

The potential of hydrogen electrode having a solution of pH = 4 at 298K is:

• A. $-0.177V$
• B. $-0.236V$
• C. $0.177V$
• D. $0.236V$

Answer 1

Answer: (B)

$-0.236V$

Solution:- (B) $-0.236V$

As we know that,

$pH=-\log \left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-4}(\because pH=4)$

Reaction at electrod-

$H^{+}+e^{-}⟶\frac{1}{2}{H}_2$

Now from nernst equation,

$E_{cell}={E^{\circ }}_{cell}-\frac{0.059}{n}\log \frac{\left[H_2\right]}{\left[H^{+}\right]}$

Here,

${E^{\circ }}_{cell}=0,n=1,P_{H_2}=1atm$

Therefore,

$E_{cell}=-\frac{0.059}{1}\log \frac{1}{{10}^{-4}}$

$\Rightarrow E_{cell}=-0.059(-\log {10}^{-4})$

$\Rightarrow E_{cell}=0.059\times (-4)=-0.236V$