Faraday's Laws of Electrolysis
Quantitative Aspects of Electrolysis | Notes & Practice Quiz
1. Faraday's First Law
Statement: The amount of chemical substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte.
Mathematically:
$$ w \propto Q $$Since $Q = I \times t$ (Current $\times$ Time), we get:
Where:
- $w$ = Mass of substance deposited (grams)
- $I$ = Current in Amperes (A)
- $t$ = Time in Seconds (s)
- $Z$ = Electrochemical Equivalent ($g/C$)
2. Electrochemical Equivalent ($Z$)
$Z$ is the mass deposited when 1 Coulomb of charge passes through the solution.
$$ Z = \frac{\text{Equivalent Weight (E)}}{96500} $$Also, Equivalent Weight ($E$) is related to Molar Mass ($M$) and valency factor ($n$):
$$ E = \frac{M}{n} \quad \Rightarrow \quad Z = \frac{M}{n \times 96500} $$3. Faraday's Second Law
Statement: When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the masses of substances produced at the electrodes are directly proportional to their chemical equivalent weights.
This implies that $\frac{w}{E} = \text{Constant}$, meaning the number of equivalents deposited is the same for all electrolytes in series.
4. The Faraday Constant ($F$)
One Faraday (1F) is the charge carried by 1 mole of electrons.
$$ 1F = N_A \times e^- $$ $$ 1F = (6.022 \times 10^{23}) \times (1.602 \times 10^{-19} C) \approx 96500 \, C/mol $$Calculation shortcut: To deposit 1 mole of a metal $M$ from $M^{n+}$, we need $n$ Faradays of charge.
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