Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $10^{-6}$ M hydrogen ion. The EMF of the cell is 0.118 V at 298 K. The concentration of $H^{+}$ ion at the positive electrode is $10^{-x}$ ,The value of 'x' is

Answer 1

Answer: 4

Cell reaction would be like, assume the concentration of hydrogen ion in positive electrode is 'x'

$Pt\left(H_2\right)1atm\left|H^{+}\left(10^{-6}M\right)\right|\left|H^{+}\left(\text{x }M\right)\right|Pt\left(H_2\right)1atm,if{E}_{cell}=0.118V.$

At cathode: $2H^{+}\left(xM\right)+2e^{-}\to H_2(g)$

At anode: $H_2\left(g\right)\to 2H^{+}\left(10^{-6}M\right)+2e^{-}$

$E_{Cell}=E_{\text{Cell }}^o-\frac{0.591}{2}\log \left(\frac{{\left({\left[H^{+}\right]}_{\text{annde }}\right)}^2}{{\left({\left[H^{+}\right]}_{\text{caathode }}\right)}^2}\right)$

$E_{\text{Cell }}=E_{\text{Cell }}^o-\frac{0.591}{2}\log {\left(\frac{\left({\left[H^{+}\right]}_{\text{andde }}\right)}{\left({\left[H^{+}\right]}_{\text{cathode }}\right)}\right)}^2$

$E_{\text{Cell }}=E_{\text{Cell }}^0-\frac{0.0591}{2}\log {\left(\frac{10^{-6}}{x}\right)}^2$

$0.118=0-\frac{0.0591}{2}\log {\left(\frac{10^{-6}}{x}\right)}^2$

$0.118\times 2=-0.0591\log {\left(\frac{10^{-6}}{x}\right)}^2$

$\frac{0.118\times 2}{0.0591}=-\log {\left(\frac{10^{-6}}{x}\right)}^2$

$-\log {\left(\frac{10^{-6}}{x}\right)}^2=\frac{0.118\times 2}{0.0591}$

$-2\log \left(\frac{10^{-6}}{x}\right)=4$

$\log \left(\frac{10^{-6}}{x}\right)=-2$

$\left(\frac{10^{-6}}{x}\right)=10^{-2}$

$x=10^{-4}$