A galvanic cell was constructed by dipping a zinc electrode (anode) in 0.01 M $Z n {(N O_{3})}_{2}$ and hydrogen electrode (cathode) in which 0.2 M $H I O_{3 (a q)}$ solution is present.
If EMF of cell at $25^{0} C$ is x volt then find the value of x
Given: $K_{a (H I O_{3}) = 0.1, {partial pressure of H}_{2 (g)} = 0.01 bar}$
$E_{Z n^{+ 2} / Z n}^{0} = 0.76 V, \frac{2.303 R T}{F} = 0.06$

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-0.79 V

+ 0.82 V

0.76 V

1.1 V

answer is C.

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Detailed Solution

At anode: $Z n_{(s)} \to Z n_{(a q)}^{2 +} + 2 e^{-}$
At cathode: $2 H^{+} + 2 e^{-} \to H_{2}$
$Z n_{(s)} + 2 H_{a q}^{+} \to Z n_{a q}^{+ 2} + H_{2_{(g)}}$ net reaction
$\begin{array}{l} H I O_{3} ⇄ H^{+} + I O_{3}^{-} \\ C - C α C α C α \end{array}$

$K_{a} = \frac{C α^{2}}{1 - α} \Rightarrow 1 \times 10^{- 1} = 2 \times 10^{- 1} . \frac{α^{2}}{1 - α} \Rightarrow 2 α^{2} + α - 1 = 0$

$∴ α = 0.5 [H^{+}] = 0.1 M E_{c e l l} = E_{c e l l}^{0} - \frac{0.06}{2} \log \frac{[Z n^{2 +}] \times P_{H_{2}}}{{[H^{+}]}^{2}} = 0.76 - \frac{0.06}{2} \log \frac{10^{- 2} \times 10^{- 2}}{{(10^{-})}^{2}} = 0.79 V = x$