Question

The EMF of a galvanic cell consisting of two hydrogen electrodes is $0.17\, V .$ 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:

$Pt \left| H _{2}(1 atm )\right| H ^{+} \| H ^{+} \mid H _{2}( l \text { atm }) \mid Pt$

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

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 $

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