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'

$Pt \, \left(H_{2}\right) \, 1 \, atm \, \left| \, H^{+} \left(1 0^{- 6} M\right) \, \right|\left| \, H^{+} \left(\text{x } M\right) \, \right| \, Pt \, \left(H_{2}\right) \, 1 \, atm \, , \, if \, E_{c e l l}= \, 0.118 \, V.$

At cathode: $2H^{+}\left(x M\right)+2e^{-} \rightarrow H_{2}\left(\right.g\left.\right)$

At anode: $H_{2}\left(g\right) \rightarrow 2H^{+}\left(1 0^{- 6} M\right)+2e^{-}$

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

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

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

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

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

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

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

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

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

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

$x =10^{-4}$