Question

State Faraday’s second law of electrolysis.

Hint:

Remember: \[ \boxed{m \propto E} \] For the same charge: \[ \boxed{\text{Deposited mass} \propto \text{Equivalent mass}} \] Equivalent mass: \[ \boxed{ E=\frac{\text{Molar mass}}{\text{Number of electrons involved}} } \]

Answer 1

Step 1: Core statement.
Faraday's second law states: when the same quantity of electricity is passed through different electrolytes in series, the masses of substances deposited are directly proportional to their equivalent masses.
Step 2: Mathematical form.
If $m_1$, $m_2$ are masses deposited and $E_1$, $E_2$ are their equivalent masses: \[ m \propto E \implies \frac{m_1}{m_2} = \frac{E_1}{E_2} \]
Step 3: Equivalent mass and implication.
Equivalent mass $= \dfrac{\text{Molar mass}}{\text{Valency (n-factor)}}$. A substance with a higher equivalent mass gets deposited in greater quantity for the same charge passed.
\[ \boxed{\frac{m_1}{m_2} = \frac{E_1}{E_2}} \]