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

Cell reaction would be like, assume the concentration of hydrogen ion in positive electrode is ' $x$ '
Pt $\left( H _{2}\right) 1$ atm $\left| H ^{+}\left(10^{-6} M \right)\right| H ^{+}( xM ) \mid Pt \left( H _{2}\right) 1$ atm , if $E _{\text {cell }}=0.118$
At cathode: $2 H ^{+}(x M )+2 e ^{-} \rightarrow H _{2}( g )$
At anode: $H _{2}( g ) \rightarrow 2 H ^{+}\left(10^{-6} M \right)+2 e ^{-}$
$E _{\text {Cell }}= E _{\text {Cell }}^{0}-\frac{0.591}{2} \log \left(\frac{\left(\left[ H ^{+}\right]_{\text {anode }}\right)^{2}}{\left(\left[ H ^{+}\right]_{\text {cathode }}\right)^{2}}\right)_{2}$
$E _{\text {Cell }}= E _{\text {Cell }}^{0}-\frac{0.591}{2} \log \left(\frac{\left(\left[ H ^{+}\right]_{\text {anode }}\right)}{\left(\left[ H ^{+}\right]_{\text {cathode }}\right)}\right)^{2}$
$E _{\text {Cell }}= E _{\text {Cell }}^{0}-\frac{0.0591}{2} \log \left(\frac{10^{-6}}{ x }\right)^{2}$
$0.118=0-\frac{0.0591}{2} \log \left(\frac{10^{-6}}{ X }\right)^{2}$
$0.118 \times 2=-0.0591 \log \left(\frac{10^{-6}}{x}\right)^{2}$
$\frac{0.118 \times 2}{0.0591}=-\log \left(\frac{10^{-6}}{ X }\right)^{2}$
$-\log \left(\frac{10^{-6}}{ X }\right)^{2}=\frac{0.118 \times 2}{0.0591}$
$-2 \log \left(\frac{10^{-6}}{ X }\right)=4$
$\log \left(\frac{10^{-6}}{ X }\right)=-2$
$\left(\frac{10^{-6}}{ X }\right)=10^{-2}$
$x=10^{-4}$