JEE Mains Chemistry · Electrochemistry

Electrolysis and Faraday's Laws

How much substance a given charge deposits or releases, and which product actually forms at each electrode when water competes with the ions.

All Electrochemistry notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, fourteen of them numerical, and one from 2026. Fifteen turn a current and a time into a mass, a gas volume or a number of faradays, or run that backwards; four ask which product forms at each electrode. Two ideas cover the page.

Concept 1 of 2: Faraday's laws of electrolysis

One faraday is one mole of electrons. Count the moles of electrons that flowed, divide by the electrons each particle needs, and you have the moles deposited or released.

Definition

Charge $Q=It$ , with $t$ in seconds. Moles of electrons $=\frac{Q}{F}$ , $F=96500$ C mol⁻¹.
Mass deposited $m=\frac{M\,It}{nF}$ , where $n$ is the electrons per particle.
$n$ per mole: $\mathrm{Ag^+}$ 1, $\mathrm{Cu^{2+}}$ 2, $\mathrm{Ni^{2+}}$ 2, $\mathrm{Al^{3+}}$ 3, Au in $\mathrm{AuCl_4^-}$ 3, $\mathrm{MnO_4^-\to Mn^{2+}}$ 5, $\mathrm{Cr_2O_7^{2-}\to 2Cr^{3+}}$ 6.
Gases: $\mathrm{O_2}$ needs 4 electrons, $\mathrm{H_2}$ and $\mathrm{Cl_2}$ need 2. One mole of water oxidised to $\mathrm{O_2}$ gives 2 moles of electrons.
Use the molar volume the question gives: 22.4 L or 22.7 L.
Second law: the same charge through cells in series gives masses in the ratio of $\frac{M}{n}$ .
Electrochemical equivalent $Z=\frac{M}{nF}$ , the mass deposited by 1 C.
Plating a layer: mass $=$ density $\times$ area $\times$ thickness.

Faraday's first law

m=\frac{M\,I\,t}{n\,F}

Mmolar mass of the substance
nelectrons needed per particle
F96500 C per mole of electrons

Worked example

A current of 1.93 A is passed through

\mathrm{CuSO_4}

solution for 50 minutes. What mass of copper is deposited? (

M=63.5

g mol⁻¹)

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q46Moderate

Example 1 · Electrochemistry · Electrolysis and Faraday's Laws

Electricity is passed through an acidic solution of

{Cu}^{2 +}

till all the

{cu}^{2 +}

was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is

\_\_\_\_

mL. (Nearest integer) [Given :

{Cu}^{2 +}(aq) + 2e^{-} \rightarrow Cu(s)E_{\text{red~}}^{0} = + 0.34\text{ }V

O_{2}(\text{ }g) + 4H^{+} + 4e^{-} \rightarrow 2H_{2}OE_{\text{red~}}^{0} = + 1.23\text{ }V

Molar mass of

Cu = 63.54\text{ }g{\text{ }mol}^{- 1}

Molar mass of

O_{2} = 32\text{ }g{\text{ }mol}^{- 1}

Faraday Constant

= 96500C{mol}^{- 1}

Molar volume at

STP = 22.4\text{ }L

]

Four electrons for oxygen

\mathrm{2H_2O\to O_2+4H^++4e^-}

. One mole of

\mathrm{O_2}

needs 4 F, not 2 F.

Minutes left as minutes

Charge is

I\times t

with

t

in seconds. Convert minutes and hours before multiplying.

The charge on a complex ion's metal

Gold in

\mathrm{AuCl_4^-}

is +3, so each Au atom needs 3 electrons. Read the metal's oxidation state, not the ion's charge.

Drill 14 more on faraday's laws of electrolysis

Concept 2 of 2: Products of electrolysis

In water, every ion must compete with water itself. At the cathode the species with the higher reduction potential is reduced. At the anode the easiest oxidation wins, and an active metal anode can dissolve before anything else.

Definition

Cathode: metal ions above hydrogen in the series ( $\mathrm{Ag^+}$ , $\mathrm{Hg_2^{2+}}$ , $\mathrm{Cu^{2+}}$ ) are deposited, the highest $E^\circ$ first. $\mathrm{Na^+}$ , $\mathrm{K^+}$ , $\mathrm{Mg^{2+}}$ and $\mathrm{Al^{3+}}$ are never deposited from water: $\mathrm{H_2}$ forms instead, and $\mathrm{OH^-}$ is left behind.
Anode, inert (Pt): $\mathrm{Cl^-}$ gives $\mathrm{Cl_2}$ , because oxygen needs an extra overpotential. $\mathrm{NO_3^-}$ and dilute $\mathrm{SO_4^{2-}}$ are not oxidised: water gives $\mathrm{O_2}$ . Concentrated $\mathrm{H_2SO_4}$ gives $\mathrm{S_2O_8^{2-}}$ .
Anode, active (Ag, Cu): the anode metal itself dissolves; no gas forms.
Oxygen can form only at an anode, never at a cathode.
Brine: $\mathrm{Cl_2}$ at the anode, $\mathrm{H_2}$ and $\mathrm{OH^-}$ at the cathode, so the pH rises. Moles of $\mathrm{OH^-}$ formed equal moles of electrons passed.
Once a metal ion is used up, water takes over at the cathode ( $\mathrm{H_2}$ ) while $\mathrm{O_2}$ keeps forming at the anode.

Electrolyte	Electrodes	Cathode	Anode
Molten NaCl	Inert	Na	$\mathrm{Cl_2}$
Aqueous NaCl (brine)	Inert	$\mathrm{H_2}$ , with $\mathrm{OH^-}$ left in solution	$\mathrm{Cl_2}$
Aqueous $\mathrm{AgNO_3}$	Pt	Ag	$\mathrm{O_2}$
Aqueous $\mathrm{AgNO_3}$	Ag	Ag	Ag dissolves as $\mathrm{Ag^+}$
Aqueous $\mathrm{CuSO_4}$	Pt	Cu	$\mathrm{O_2}$
Aqueous $\mathrm{CuSO_4}$	Cu	Cu	Cu dissolves as $\mathrm{Cu^{2+}}$
Dilute $\mathrm{H_2SO_4}$	Pt	$\mathrm{H_2}$	$\mathrm{O_2}$
Concentrated $\mathrm{H_2SO_4}$	Pt	$\mathrm{H_2}$	$\mathrm{S_2O_8^{2-}}$

An active anode dissolves; an inert anode oxidises an anion or water.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q111Moderate

Example 2 · Electrochemistry · Electrolysis and Faraday's Laws

O_{2}

gas will be evolved as a product of electrolysis of : (a) an aqueous solution of

AgNO_{3}

using silver electrodes. (b) an aqueous solution of

AgNO_{3}

using platinum electrodes. (c) a dilute solution of

H_{2}SO_{4}

using platinum electrodes. (d) a high concentration solution of

H_{2}SO_{4}

using platinum electrodes. Choose the correct answer from the options given below :

Depositing sodium from water

Water is reduced long before

\mathrm{Na^+}

\mathrm{Mg^{2+}}

. From an aqueous solution the cathode gives

\mathrm{H_2}

, never the metal.

Forgetting an active anode

With silver or copper electrodes, the anode metal dissolves. An option that gives

\mathrm{O_2}

there is wrong.

Drill 3 more on products of electrolysis

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Faraday's laws of electrolysis
Faraday's first law
$m=\frac{M\,I\,t}{n\,F}$

Reference tables (1)

Products of electrolysis8 rows

Electrolyte	Electrodes	Cathode	Anode
Molten NaCl	Inert	Na	$\mathrm{Cl_2}$
Aqueous NaCl (brine)	Inert	$\mathrm{H_2}$ , with $\mathrm{OH^-}$ left in solution	$\mathrm{Cl_2}$
Aqueous $\mathrm{AgNO_3}$	Pt	Ag	$\mathrm{O_2}$
Aqueous $\mathrm{AgNO_3}$	Ag	Ag	Ag dissolves as $\mathrm{Ag^+}$
Aqueous $\mathrm{CuSO_4}$	Pt	Cu	$\mathrm{O_2}$
Aqueous $\mathrm{CuSO_4}$	Cu	Cu	Cu dissolves as $\mathrm{Cu^{2+}}$
Dilute $\mathrm{H_2SO_4}$	Pt	$\mathrm{H_2}$	$\mathrm{O_2}$
Concentrated $\mathrm{H_2SO_4}$	Pt	$\mathrm{H_2}$	$\mathrm{S_2O_8^{2-}}$

An active anode dissolves; an inert anode oxidises an anion or water.

Watch out for (5)

Four electrons for oxygen
→ Faraday's laws of electrolysis
Minutes left as minutes
→ Faraday's laws of electrolysis
The charge on a complex ion's metal
→ Faraday's laws of electrolysis
Depositing sodium from water
→ Products of electrolysis
Forgetting an active anode
→ Products of electrolysis

Test yourself on Electrochemistry

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