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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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For the following half cell $P^+/CH_3COOH/H_2\;Ka_{CH_3COOH}=10^{-5}$ Find $E_{H^+/H_2}$

$\begin{array}{1 1}(a)\;-0.36\\(b)\;-0.18\\(c)\;-0.54\\(d)\;-0.09\end{array}$

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1 Answer

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$CH_3COOH(aq)\rightleftharpoons CH_3COO^-(aq)+H^+(aq)$
$Ka=\large\frac{C\alpha^2}{1-\alpha}$
$1-\alpha\approx 1$
So $\alpha=\sqrt{\large\frac{Ka}{C}}$
So $\alpha=\sqrt{\large\frac{10^{-3}}{10^{-1}}}$
$\Rightarrow 10^{-2}$
So $[H^+]=C\alpha=10^{-3}$
$P^H=-\log [H^+]=3$
$E_{H^+/H_2}=-0.06\times P^H=-0.18V$
Hence (b) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

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