ElectrochemistryHard

Question

The calomel electrode is reversible with respect to

Options

A.Hg22+
B.H+
C.Hg2+
D.Cl

Solution

$Hg_{2}Cl_{2}(s) \rightleftharpoons Hg_{2}^{2 +}(aq) + 2Cl^{-}(aq)$

$Hg_{2}^{2 +}(aq) + 2e^{-} \rightleftharpoons 2Hg(I)$

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