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.