A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 1 0 - 6 M hydrogen ion. The EMF of the cell is 0.118 V at 298 K. The concentration of H + ion at the po

Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 1 0 - 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 1 0 - x ,The value of 'x' is

TestHub · Accepted Answer

Cell reaction would be like, assume the concentration of hydrogen ion in positive electrode is 'x'
$P t (H_{2}) 1 a t m |H^{+} (1 0^{- 6} M)| |H^{+} (x M)| P t (H_{2}) 1 a t m, i f E_{c e l l} = 0.118 V .$
At cathode: $2 H^{+} (x M) + 2 e^{-} \to H_{2} (g)$
At anode: $H_{2} (g) \to 2 H^{+} (1 0^{- 6} M) + 2 e^{-}$
$E_{C e l l} = E_{C e l l}^{ο} - \frac{0.591}{2} \log (\frac{{({[H^{+}]}_{a n o d e})}^{2}}{{({[H^{+}]}_{c a t h o d e})}^{2}})$
$E_{C e l l} = E_{C e l l}^{ο} - \frac{0.591}{2} {\log (\frac{{({[H^{+}]}_{a n o d e})}^{}}{{({[H^{+}]}_{c a t h o d e})}^{}})}^{2}$
$E_{C e l l} = E_{C e l l}^{ο} - \frac{0.0591}{2} {\log (\frac{1 0^{- 6}}{x})}^{2}$
$0.118 = 0 - \frac{0.0591}{2} {\log (\frac{1 0^{- 6}}{x})}^{2}$
$0.118 \times 2 = - 0.0591 {\log (\frac{1 0^{- 6}}{x})}^{2}$
$\frac{0.118 \times 2}{0.0591} = - {\log (\frac{1 0^{- 6}}{x})}^{2}$
$- {\log (\frac{1 0^{- 6}}{x})}^{2} = \frac{0.118 \times 2}{0.0591}$
$- 2 {\log (\frac{1 0^{- 6}}{x})}^{} = 4$
${\log (\frac{1 0^{- 6}}{x})}^{} = - 2$
${(\frac{1 0^{- 6}}{x})}^{} = 1 0^{- 2}$
$x = 1 0^{- 4}$

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