Q.A current of 1.50 A was passed through an electrolytic cell containing $AgNO_3$ solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow ? (Molar mass of Ag = 108 g mol$^{-1}$, 1F = 96500 C mol$^{-1}$). OR The conductivity of a 0.01 M solution of acetic acid at 298 K is $1.65 \times 10^{-4}$ S cm$^{-1}$. Calculate molar conductivity ($\wedge_m$) of the solution.

Faraday's Laws of Electrolysis – From Intuition to Precision

Imagine you are plating a spoon with silver. You dip it in a silver salt solution, connect it to a battery, and silver metal starts coating the spoon. Two questions naturally arise: How much silver will deposit? And does the amount depend only on the battery's strength, or also on the time?

Faraday answered both with two beautifully simple laws.

The Core Intuition

Electrolysis is about moving electrons. Each silver ion ($\text{Ag}^+$) arriving at the spoon grabs one electron and becomes a neutral silver atom. So the mass of silver deposited is directly proportional to the number of electrons that have flowed — that is, to the total charge passed.

But different ions need different numbers of electrons. A copper ion ($\text{Cu}^{2+}$) needs two electrons to become copper metal. So for the same charge, you get half as many copper atoms as silver atoms. That is why the chemical nature of the substance matters — specifically, its equivalent weight (the mass that reacts with one mole of electrons).

The Two Laws – Precise Statements

Putting Them Together – The Combined Equation

The two laws merge into one powerful formula:

where:

• $m$ = mass deposited (g)
• $Q$ = total charge passed (coulombs) = $I \times t$
• $E$ = equivalent weight (g/eq)
• $F$ = Faraday's constant = 96485 C/mol (charge of one mole of electrons)

Worked Example – Silver Plating

Problem: A current of 2.0 A is passed through a silver nitrate solution for 30 minutes. How much silver deposits? (Atomic mass of Ag = 107.9 g/mol, valency = 1)

Step 1 – Find the charge:

Step 2 – Find equivalent weight:

Step 3 – Apply the combined law:

$m = \frac{Q \times E}{F} = \frac{3600 \times 107.9}{96485} \approx 4.03 \text{ g}$ …