Chemistry - Electrochemistry Question with Solution | TestHub

The reaction at the cathode is ${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu}$.

The quantity of charge $Q$ passed through the solution is given by $Q=I\times t$.

Substituting the given values:
$Q=1.5\text{ A}\times (10\times 60)\text{ s}=900\text{ C}$

From Faraday's first law of electrolysis, the mass of copper deposited is:
$m=\frac{M\times Q}{n\times F}$

Here, $M=63{\text{ g mol}}^{-1}$, $n=2$, and $F=96487{\text{ C mol}}^{-1}$.

$m=\frac{63\times 900}{2\times 96487}$

$m=\frac{56700}{192974}\approx 0.2938\text{ g}$

Answer: $0.2938$ g