Question

The EMF of a galvanic cell consisting of two hydrogen electrodes is $0.17V.$ If the solution of one of the electrodes has $\left[H^{+}\right]=10^{-3}M$, the $pH$ at the other electrode is

• A. 5.87
• B. 4.88
• C. 2.08
• D. 3.08

Answer 1

Answer: (A)

The cell is represented as:

$Pt\left|H_2(1atm)\right|H^{+}\Vert H^{+}\mid H_2(l\text{ atm })\mid Pt$

$E_{\text{cell }}^{\circ }=0.0591\log \frac{{\left[H^{+}\right]}_{\text{cathode }}}{{\left[H^{+}\right]}_{\text{anode }}}$

On substituting the value of $E_{\text{cell }}^{\circ }$ and solving

$-0.17=0.0591\log \frac{\left[H^{+}\right]}{10^{-3}}$

$\frac{-0.17}{0.0591}=\log \frac{\left[H^{+}\right]}{10^{-3}}$

$\log \left[H^{+}\right]=-2.87-3=-5.87$

$pH=-\log \left[H^{+}\right]=-(-5.87)=5.87$