Question

A cell contains two hydrogen electrodes . The negative electrode is in contact with a solution of 10 power -5 M H+ ions . The emf of the cell is 0.118 V at 298 K . Calculate the concentration of the H+ ions at the positive electrode .

Answer 1

Dear student,

The cell reaction can be written as : H_2(g) + 2H⁺ $\to$ 2H⁺ + H_2(g)
(I) (II) (I) (II) [ Note : (I) is negative and (II) in positive electrode ]
According to Nernst equation : E_cell = E⁰_cell - $\frac{0.0592}{2}\log \frac{[H^{+}{]}^{2}of(I)}{[H^{+}{]}^{2}of(II)}$
E_cell = 0.0 - $\frac{0.0592}{2}2\log \frac{{10}^{-5}}{x}$
0.118 = - 0.0592 log[ 10^-5 / x ]
Then x = 10^-3 M
Regards