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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Calculate the E.M.F of the electrode-concentration cell : $H_2(P_1),HCl,H_2(P_2)$ at $25^{\large\circ}C$ if $P_1$ = 1atm and $P_2 = 0.25\;atm$.The E.M.F. 'E' is

$\begin{array}{1 1}(A)\;0.016V&(B)\;0.018V\\(C)\;0.017V&(D)0.018V\end{array}$

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1 Answer

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Here,
Right hydrogen electrode,$rx_n$
$2H^++2e^-\rightarrow H_2(P_2)$
Reduction
Left hydrogen electrode,$rx_n$
$H_2(P_1)\rightarrow 2H^++2e^-$
Overall $rx_n$;$H_2(P_1)\rightarrow H_2(P_2)$
$\therefore E=\large\frac{-0.0591}{2}$$\log\large\frac{P_2}{P_1}$
$\Rightarrow \large\frac{-0.0591}{2}$$\log\large\frac{0.25}{2}$
$\Rightarrow 0.0178\approx 0.018V$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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