Chemistry - Electrochemistry Question with Solution | TestHub

For the reduction of${\mathrm{H}}_2\mathrm{O}$at cathode,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right)+2{\mathrm{OH}}^{-}; {\mathrm{E}}_{\mathrm{red}}^{\circ }=-1.23\mathrm{V}$

For${\mathrm{Ca}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ca}; {\mathrm{E}}_{\mathrm{red}}^{\circ }=-2.87\mathrm{V}$

i.e., water will be reduced at the cathode, in preference to the liberation of$\mathrm{Ca}$.

However, in the case of$\mathrm{Ag}, \mathrm{Cu} \mathrm{and} \mathrm{Cr}$, their standard reduction potentials are higher than$-1.23 \mathrm{V}$and will be liberated/ deposited at the cathode and${\mathrm{H}}_2\mathrm{O}$will not be reduced.