Question

Explain Faraday's laws of electrolysis.

Hint:

Use \(m=ZIt\) for direct calculations with given \(Z\); otherwise use \(m=\dfrac{It}{F}\dfrac{M}{z}\).
Always match \(z\) to the electrode reaction (e.g., Cu\(^{2+}+2e^-\to\) Cu, so \(z=2\)).

Answer 1

First Law (Mass–Charge law):
The mass \(m\) of a substance liberated (or deposited) at an electrode is directly proportional to the quantity of electricity \(Q\) passed through the electrolyte.
\[ m \propto Q = It $\Rightarrow$ m = ZIt \] where \(I\) is current in ampere, \(t\) is time in seconds, and \(Z\) is the electrochemical equivalent (g C\(^{-1}\)).
Second Law (Equivalent–Weight law):
When the same quantity of electricity passes through different electrolytes, the masses of substances liberated are proportional to their chemical (equivalent) weights.
\[ \frac{m_1}{m_2}=\frac{E_1}{E_2}=\frac{M_1/z_1}{M_2/z_2} \] where \(M\) is molar mass and \(z\) is the electrons transferred per ion (valency).
Equivalently, for a given substance: \[ m=\frac{Q}{F}\,\frac{M}{z}=\frac{It}{F}\,\frac{M}{z} \] with \(F=96485\ \text{C mol}^{-1}\) (Faraday constant).