Question

State Faraday's law's of electrolysis. A solution of $Cu{SO}_4$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

Answer 1

Faraday's first law of electrolysis:
The amount of the substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.
Thus mass of the substance produced
$=\frac{\text{I(A)}\times \text{t(s)}}{{\text{96500 C/mol e}}^{-}}\times \text{mole ratio}\times \text{molar mass}$
Here I(A) is current in ampere and t(s) is time in second.
Faraday's second law of electrolysis:
When the same amount of electricity is passed through different cells containing different electrolytes and arranged in series, the amounts of substances oxidized or reduced at the respective electrodes are directly proportional to their chemical equivalent masses.
$\frac{\text{moles of A produced}}{\text{moles of B produced}}=\frac{\text{mole ratio of A half reaction}}{\text{mole ratio of B half reaction}}$
Number of moles of electrons passed
$=\frac{I\left(A\right)\times t\left(s\right)}{96500}=\frac{1.5\times 10\times 60}{96500}=0.009326$ moles of electrons
$C{u}^{2+}+2e^{-}\to Cu$
The number of moles of copper deposited $=\frac{0.009326}{2}=0.004663$ moles.
The molar mass of Cu is 63.5 g/mol.
The mass of Cu deposited $=0.004663mol\times 63.5g/mol=0.296g$