Michael Faraday formulated two quantitative laws that govern the relationship between the quantity of electricity passed through an electrolyte and the mass of chemical substance deposited or liberated at the electrodes.

Statement: The mass ($m$) of a substance deposited, liberated, or dissolved at an electrode during electrolysis is directly proportional to the quantity of electricity ($Q$) passing through the electrolyte.

Mathematical Derivation:

$$ m \propto Q $$ Since quantity of electricity
(Q) = Current (I) × Time (t) $$ Q = It $$ Therefore: $$ m \propto It $$ $$ m = ZIt $$

Where:
• m = Mass of substance deposited or dissolved (in grams, g)
• Q = Quantity of electric charge (in Coulombs, C)
• I = Steady current passed (in Amperes, A)
• t = Duration of electrolysis (in Seconds, s)
• Z = Electrochemical Equivalent (ECE) of the substance (in g/C)

Electrochemical Equivalent (Z): The mass of a substance liberated or deposited by a current of 1 Ampere flowing for 1 Second (i.e., 1 Coulomb of charge).

Apparatus Required: Copper voltameter, direct current (DC) power source, rheostat, ammeter, precision stop-watch, connecting wires, sand paper, and balance.

Procedure:

1. Clean two copper plates thoroughly using sandpaper, rinse with water, dip in alcohol, dry, and weigh the cathode precisely ($m_1$).
2. Set up the circuit with a copper voltameter filled with aqueous $\text{CuSO}_4$ connected in series with an ammeter, rheostat, switch, and DC supply.
3. Pass a steady current ($I_1$) measured by the ammeter for a fixed time ($t$) recorded using the stopwatch.
4. Remove the cathode, wash gently with water, dry thoroughly, and reweigh ($m_2$). The mass deposited is $m = m_2 - m_1$.
5. Repeat the experiment using different values of current ($I_2, I_3$) for the same time interval ($t$), or vary the time duration ($t_1, t_2$) while holding the current ($I$) constant.

Observation & Conclusion:

A plot of mass deposited ($m$) against the product of current and time ($I \times t$ or $Q$) yields a straight line passing through the origin. This confirms that $m \propto Q$, thus verifying Faraday's First Law.

Statement: When the same quantity of electricity ($Q$) is passed through solutions of different electrolytes connected in series, the masses of different substances deposited or liberated at the electrodes are directly proportional to their chemical equivalents (or equivalent masses).

Mathematical Derivation:

$$ m \propto E $$
For two electrolytes connected in series: $$ \frac{m₁}{m₂} = \frac{E₁}{E₂} $$ Where:
$$ E = \frac{M}{\text{v}} $$

Where:
• m₁, m₂ = Masses of substances deposited at the respective cathodes.
• E₁, E₂ = Chemical equivalent masses of the respective ions.
• M = Molar mass (g/mol).
• v = Valency (charge) of the ion.

Faraday's Constant (F): The quantity of electricity carried by one mole of electrons.

$$ \text{1 Faraday (F)} = e × Nₐ $$ $$ ≈ 96,500 C/mol $$ Relationship between Z and E: $$ Z = \frac{E}{F} = \frac{M}{vF} $$ Unified Electrolysis Equation: $$ m = \frac{MIt}{vF} $$

Apparatus Required: Two voltameters connected in series (e.g., a Copper voltameter with $\text{CuSO}_4$ solution and a Silver voltameter with $\text{AgNO}_3$ solution), DC supply, ammeter, rheostat, and stop-watch.

Procedure:

1. Clean and dry the copper cathode and silver cathode, then measure their initial masses ($m_{{Cu_1}}$ and $m_{{Ag_1}}$).
2. Connect the two cells in series so that the exact same current flows through both cells simultaneously.
3. Switch on the current and allow a steady flow for a specific time ($t$).
4. Disconnect the circuit, carefully remove both cathodes, wash, dry, and reweigh them to find their final masses ($m_{{Cu_2}}$ and $m_{{Ag_2}}$).
5. Calculate the mass deposited: $m_{\text{Cu}} = m_{{Cu_2}} - m_{{Cu_1}}$ and $m_{\text{Ag}} = m_{{Ag_2}} - m_{{Ag_1}}$.

Observations and Calculation:

Mass ratio obtained experimentally: $$ \frac{m_{Ag}}{m_{Cu}} = 3 $$ Ratio of chemical equivalents: $$ \frac{E_{Ag}}{E_{Cu}} $$ $$ = \frac{107.87}{1} ÷ \frac{63.55}{2} $$ $$ = \frac{107.87}{31.78} ≈ 3.39 $$ Result:$$ \frac{m_{Ag}}{m_{Cu}} = \frac{E_{Ag}}{E_{Cu}} $$

Conclusion: The ratio of the masses deposited equals the ratio of their chemical equivalent weights. This confirms Faraday's Second Law of Electrolysis.