Question

Explain in brief the Faraday's laws of Electrolysis. CuSO$_4$ solution was electrolysed for 20 minutes with 2.0 ampere current. What will be the mass of precipitated copper at cathode?

Hint:

Always match \(n\) to the electrons in the half-reaction. For Cu$^{2+}$, \(n=2\). Use \(t\) in seconds and \(F\) in C\,mol$^{-1}$ for consistent units.

Answer 1

Faraday's First Law: The mass of a substance liberated (or deposited) at an electrode is directly proportional to the total electric charge passed through the electrolyte.
\[ m \propto Q $\Rightarrow$ m = Z\,Q \] where \(m\) is mass deposited, \(Q=It\) is charge, and \(Z\) is the electrochemical equivalent.
Faraday's Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their chemical (equivalent) weights.
Numerical Part: In CuSO$_4$ electrolysis, copper is deposited by \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \] For a metal ion requiring \(n\) electrons, \[ m = \frac{It\,M}{nF} \] where \(I=2.0\,\text{A}\), \(t=20\,\text{min}=1200\,\text{s}\), \(M(\text{Cu})=63.5\,\text{g mol}^{-1}\), \(n=2\), \(F=9.65\times10^4\,\text{C mol}^{-1}\).
\[ m = \frac{(2.0)(1200)\times 63.5}{(2)(9.65\times10^4)} = \frac{2400\times 63.5}{1.93\times10^5} \approx 0.79\,\text{g} \] \[ \boxed{m_{\text{Cu deposited}} \approx 0.79~\text{g}} \]