Question

The cell Pt, $H_2\left(1atm\right)\left|H^{+}\right(pH=x\left)\right||$ Normal calomel Electrode has an EMF of 0.67 V at ${25}^{\circ }C$. The pH of the solution is:
The oxidation potential of the calomel electrode on hydrogen scale is $-0.28V$. (write the value to the nearest integer)

Answer 1

$E_{cell}=E_{calomel}-E_{hydrogen}$
$0.67V=0.28-E_{hydrogen}$
$E_{hydrogen}=-0.39V$
$E_{hydrogen}=E_{hydrogen}^0-\frac{0.0592}{n}log\frac{P_{H2}}{[H^{+}{]}^2}$
$-0.39=0-\frac{0.0592}{2}log\frac{1}{[H^{+}{]}^2}$
$13.175=log\frac{1}{[H^{+}{]}^2}$
$\frac{1}{[H^{+}{]}^2}=1.4985\times {10}^{13}$
$\left[H^{+}\right]=2.58\times {10}^{-7}$
$pH=-log\left[H^{+}\right]=-log2.58\times {10}^{-7}=6.61$
Hence, the pH of the solution is $\mathrm{6.61.}$