Question

The cell $Pt,H_{2\left(g\right)}\left(1\text{atm}\right)/H^{+},pH=x||$ Normal calomel electrode has emf of $0.67\text{V}$ at ${25}^{\circ }\text{C}$. The oxidation potential of calomel electrode on H scale is $-0.28\text{V}$, then $pH$ of solution will be
(Given : $\frac{2.303RT}{F}=0.0591\text{V}$)

• A. $6.6$
• B. $3.3$
• C. $13.2$
• D. $1.1$

Answer 1

The correct option is A $6.6$

Here oxidation occurs in H-electrode and reduction occurs at calomel electrode.
$\therefore H_2\to 2H^{+}+2e^{-}H{g}_{2}C{l}_2+2e^{-}\to 2Hg\left(l\right)+2C{l}^{-}$
Net reaction is $H{g}_{2}C{l}_2\left(s\right)+H_2\left(g\right)\to 2Hg\left(l\right)+2C{l}^{-}\left(aq\right)+2H^{+}\left(aq\right)$
So, $E_{\text{cell}}^{\circ }=E_{\text{Red}}^{\circ }+E_{\text{oxi}}^{\circ }\{\therefore \text{Reduction potential of H-electrode is zero}\}$

$\Rightarrow E_{\text{cell}}^{\circ }=0+0.28V$
$\Rightarrow E_{\text{cell}}^{\circ }=0.28V$

Now the EMF of the cell is
$E_{\text{Cell}}=E_{\text{cell}}^{\circ }-\frac{2.303RT}{nF}\text{log}\frac{\left[H^{+}{]}^2\right[C{l}^{-}{]}^2}{1}$
$\Rightarrow 0.67=0.28-\frac{0.0591}{2}\text{log}\frac{[H{]}^2}{1}.....\left(\right[C{l}^{-}]=1)$
We know that $pH=-\text{log}\left[H^{+}\right]$
$\therefore -\text{log}\left[H^{+}\right]=\frac{(0.67-0.28)\times 2}{2\times 0.0591}=\frac{0.78}{0.1182}=6.6$
$\text{Therefore},pH=6.6$