Chemistry - Electrochemistry Question with Solution | TestHub

The reaction at the cathode is:

${\mathrm{Ca}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ca}$

The number of moles of electrons required to deposit one mole of Ca is $n=2$.

The given parameters are $I=0.8\mathrm{A}$, $t=60\min =3600\mathrm{s}$, and $M=40{\mathrm{g}\mathrm{mol}}^{-1}$.

Using Faraday's first law of electrolysis, the mass of Ca deposited is:

$W=\frac{I\times t\times M}{n\times F}$

Substituting the values:

$W=\frac{0.8\times 3600\times 40}{2\times 96500}$

$W=\frac{115200}{193000}$

$W=0.5968\mathrm{g}\approx 0.6\mathrm{g}$

Answer: 0.6 g