Question

Hydrogen Electrode Is Dipped In A Solution Of $\mathrm{pH}=3\mathrm{at}25°\mathrm{C}$. The potential of the cell would be_______? (the value of $2.303\frac{\mathrm{RT}}{\mathrm{F}}$ is $0.059\mathrm{V}$)

• A. $0.059\mathrm{V}$
• B. $0.087\mathrm{V}.$
• C. $0.177\mathrm{V}.$
• D. $-0.177\mathrm{V}.$

Answer 1

The correct option is D

$-0.177\mathrm{V}.$

Explanation:

Step 1: Calculating the concentration of hydrogen ions

Given, $\mathrm{pH}=3$

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

$\left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{p}\mathrm{H}}={10}^{-3}\mathrm{M}$

Step 2: Calculating the potential of cell

The formula of the Nernst equation for calculating the potential of a cell is given by ;

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{°}-\frac{0.059}{\mathrm{n}}\frac{{\mathrm{P}}^{{\mathrm{H}}_2}}{\log {\left[{\mathrm{H}}^{+}\right]}^2}$

Putting the given data in the above formula, we get;

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0-\frac{0.059}{2}\log \frac{1}{{\left({10}^{-3}\right)}^2} \\ {\mathrm{E}}_{\mathrm{cell}}=-0.177V \end{gathered}$$

Hence, The potential of the cell would be : $-0.177\mathrm{V}$.