Question

How many grams of $Cu$ and what volume of $O_2$ gas will be obtained at STP by passing $20$ ampere current for $96.5$ minutes from an aqueous solution of $CuS{O}_4$? The efficiency of the electrolytic cell is $90\%$. $(Cu=63.5)$

Answer 1

$\text{Mass of Cu obtained}=\frac{I\left(A\right)\times t\left(s\right)}{96500\left(C/mol{e}^{-}\right)}\times \text{mole ratio}\times \text{Atomic mass of Cu}$

$\text{Mass of Cu obtained}=\frac{20\left(A\right)\times 96.5\times 60\left(s\right)}{96500\left(C/mol{e}^{-}\right)}\times \frac{\text{1 mol Cu}}{\text{2 mol e}}\times \text{63.5 g/mol}=\text{38.1 g}$

$\text{Moles of O2 obtained}=\frac{I\left(A\right)\times t\left(s\right)}{96500\left(C/mol{e}^{-}\right)}\times \text{mole ratio}$

$\text{Moles of O2 obtained}=\frac{20\left(A\right)\times 96.5\times 60\left(s\right)}{96500\left(C/mol{e}^{-}\right)}\times \frac{\text{1 mol O2}}{\text{4 mol e}}=\text{0.3 mol}$

$\text{Volume of O2 obtained}=\text{0.3 mol}\times \text{22.4 L/mol}=\text{6.72 L}$

However, the efficiency of the electrolytic cell is 90%.

$\text{Mass of Cu obtained}=\text{38.1 g}\times \frac{90}{100}=\text{34.3 g}$

$\text{Volume of O2 obtained}=\text{6.72 L}\times \frac{90}{100}=\text{6.05 L}$