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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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