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The EMF of cell is : $Pt/H_2/CH_3COOH(0.1M)\parallel NH_4OH(0.01M)/H_2/Pt$.$K_a$ for $CH_3COOH=1.8\times 10^{-5}$ and $K_b$ for $NH_4OH=1.8\times 10^{-5}$

$\begin{array}{1 1}(a)\;0.4V\\(b)\;0.35V\\(c)\;0.45V\\(d)\;0.3V\end{array}$

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For $CH_3COOH$
$CH_3COOH\leftrightharpoons CH_3COO^-+H^+$
Or $K_a=\large\frac{[H^+]^2}{[CH_3COOH]}$
$H^+=\sqrt{K_a\times [CH_3COOH]}$
$\therefore [H^+]=\sqrt{1.8\times 10^{-5}\times 0.1}$
$\Rightarrow 1.34\times 10^3N$
For $NH_4OH$
$NH_4OH\leftrightharpoons NH_4^++OH^-$
$[OH^-]=\sqrt{K_b\times (NH_4OH)}=\sqrt{1.8\times 10^{-5}\times 0.01}$
$\Rightarrow 4.24\times 10^4M$
$H^+=\large\frac{10^{-14}}{4.24\times 10^{-4}}$
$\Rightarrow 2.357\times 10^{-11}$
Cell reaction
At anode $H_2 \leftrightharpoons 2H^+_{CH_3COOH}+2e^-$
At cathode $2H^+_{NH_4OH}+2e^- \leftrightharpoons H_2$
$2H^+_{NH_4OH} \leftrightharpoons 2H^+_{CH_3COOH}$
On applying Nernst equation
$E_{cell}=E^0_{cell}-\large\frac{0.059}{n}$$\log \large\frac{[H^+]^2_{CH_3COOH}}{[H^+]^2_{NH_4OH}}$
$E_{cell}=0-\large\frac{0.059}{n}$$\log \large\frac{[1.34\times 10^{-3}]^2}{[2.357\times 10^{-11}]^2}$
$\Rightarrow 0.0591\log \large\frac{2.357\times 10^{-11}}{1.34\times 10^{-3}}$
$\Rightarrow 0.45V$
Hence (c) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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