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In the calomel half cell,the reduction $rx_n$ is

$\begin{array}{1 1}(A)\;Hg_2Cl_2(s)+2e^-\rightarrow 2Hg(l)+2Cl^-(aq)\\(B)\;Pb(s)+PbO_2+2H_2SO_4\rightarrow 2PbSO_4(s)+2H_2O(l)\\(C)\;2Hg+2Cl^-\rightarrow Hg_2Cl_2+2e^-\\(D)\;Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{+3}(aq)+7H_2O\end{array}$

1 Answer

$Hg_2Cl_2(s)+2e^-\rightarrow 2Hg(l)+2Cl^-(aq)$
Hence (A) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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