Q.Write Faraday's 2nd law of Electrolysis.
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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 (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 (Cu2+) 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
First Law: The mass of a substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
m∝Qorm=ZQ
where Z is the electrochemical equivalent (mass deposited per unit charge).
Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances liberated are proportional to their chemical equivalents (equivalent weights).
E1m1=E2m2
where E is the equivalent weight (molar mass ÷ valency).
Putting Them Together – The Combined Equation
The two laws merge into one powerful formula:
m=FQ×E
where:
- m = mass deposited (g)
- Q = total charge passed (coulombs) = I×t
- E = equivalent weight (g/eq)
- F = Faraday's constant = 96485 C/mol (charge of one mole of electrons)
A quick way to remember: m=FItE. The charge It is just current × time.
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:
Q=I×t=2.0×(30×60)=3600 C
Step 2 – Find equivalent weight:
E=1107.9=107.9 g/eq
Step 3 – Apply the combined law:
m=FQ×E=964853600×107.9≈4.03 g …
Faraday's second law extends the first law to compare different electrolytes, describing what happens when the same quantity of electricity is passed through several electrolytic cells connected in series. …
With the same charge passed, the amounts of different substances deposited are proportional to their equivalent weights: w1/w2 = E1/E2.
Faraday's second law of electrolysis states:
'When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the masses of the different substances liberated or deposited at the electrodes are directly proportional to their respective chemical equivalent weights (equivalent masses).'
Mathematically, for two electrolytes:
w1 / w2 = E1 / E2
where w1, w2 are the masses deposited and E1, E2 are the chemical equivalent weights (equivalent mass = atomic mass / valency).
Illustration: if the same current is passed in series through solutions of CuSO4 and AgNO3, the masses of copper and silver deposited are in the ratio of their equivalent weights: …
Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set A1 markMCQQ.Which of the following equations represents the Faraday's first law of electrolysis ?(a) mz = c.t(b) m = c.z.t(c) mc = z.t(d) c = m.z.t
›Reveal solutionSolution
Faraday's first law: mass deposited m is proportional to the quantity of charge, m = z x Q = z x c x t (c = current, t = time, z = electrochemical equivalent).
…
- CBSE 2026Set ANNUAL1 markMCQQ.Assertion (A) : Reduction of 1 mole of Cu2+ ions requires 2 Faraday of charge. Reason (R) : 1 Faraday is equal to the charge of 1 mole of electrons.(a) Both (A) and (R) are true and (R) is the correct explanation of (A)(b) Both (A) and (R) are true but (R) is not the correct explanation of (A)(c) (A) is true but (R) is false.(d) (A) is false but (R) is true.
›Reveal solutionSolution
Cu²⁺ + 2e⁻ → Cu needs 2 moles of electrons, i.e. 2 Faraday of charge, and 1 Faraday is defined as the charge carried by 1 mole of electrons — so the reason directly explains the assertion.
Assertion: Reduction of 1 mole of Cu²⁺ requires 2 Faraday of charge.
The reduction half-reaction is:
Cu2+(aq)+2e−→Cu(s)
To reduce 1 mole of Cu²⁺ ions, 2 moles of electrons are needed. Since 1 Faraday (F) is exactly the amount of charge carried by 1 mole of electrons (F = N_A × e ≈ 96500 C/mol), 2 moles of electrons corresponds to 2 Faraday of charge. So the assertion is true.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The charge required to reduce 1 mol of MnO₄⁻ to MnO₂ is-(a)(i) 1F(b)(ii) 3F(c)(iii) 5F(d)(iv) 6F
›Reveal solutionSolution
Mn goes from +7 (in MnO4−) to +4 (in MnO2), a gain of 3 electrons per Mn; 1 mole requires 3 F. Correct option: (ii).
Concept. By Faraday's laws, the charge needed to reduce 1 mole of a species equals (number of electrons gained per ion) × 1 F, where 1 F=96500 C is the charge of 1 mole of electrons.
Steps.
- Oxidation state of Mn in MnO4−: x+4(−2)=−1⇒x=+7.
- Oxidation state of Mn in MnO2: x+2(−2)=0⇒x=+4.
- Change =+7→+4, so each Mn gains 7−4=3 electrons. …
- CBSE 2026Set ANNUAL1 markMCQQ.How much charge is required for the 1 mol Al3+ to Al?(a) 1F(b) 2F(c) 4F(d) 3F
›Reveal solutionSolution
Al3+ + 3e- -> Al, so 1 mol Al needs 3 mol electrons = 3F.
The reduction half-reaction is: Al3+ + 3e- -> Al.
…
- CBSE 2025Set ANNUAL1 markMCQQ.If 96500 coulomb of electricity is passed through CuSO4 solution, it will liberate-(a) 63.5 gm copper(b) 100 gm copper(c) 96500 gm copper(d) None of the these
›Reveal solutionSolution
96500 C = 1 Faraday = 1 mole of electrons, but Cu2+ needs 2 electrons per atom, so only half a mole (31.75 g) of copper is deposited.
At the cathode: Cu2++2e−→Cu.
96500 C=1 F=1 mole of electrons. Since 2 moles of electrons are needed to deposit 1 mole (63.5 g) of copper, 1 mole of electrons (96500 C) deposits only:
…
- CBSE 2025Set D1 markMCQQ.The quantity of electricity required to liberate 32 g of oxygen is(a) 1 faraday(b) 2 faraday(c) 3 faraday(d) 4 faraday
›Reveal solutionSolution
32 g O2 = 1 mol; the electrode reaction transfers 4 electrons per O2, so 4 faraday are needed.
32 g of oxygen (O2, molar mass 32 g/mol) is 1 mole of O2 molecules.
At the anode, oxygen is liberated by:
2H2O → O2 + 4H+ + 4e- (or 4OH- → O2 + 2H2O + 4e-)
…
- CBSE 2025Set ANNUAL1 markMCQQ.The number of electrons with one coulomb of charge will be:(a) 6.29 x 10^11(b) 1.6 x 10^19(c) 6.24 x 10^18(d) 5.46 x 10^29
›Reveal solutionSolution
Since the charge on one electron is 1.6 × 10^-19 C, the number of electrons carrying 1 C of charge is 1 / (1.6 × 10^-19) ≈ 6.24 × 10^18.
The charge on a single electron is e = 1.6 × 10^-19 coulomb (this value itself is option b, which is a distractor — it is the charge of ONE electron, not the count of electrons).
To find how many electrons (n) are needed to make up a total charge of 1 C:
n = Total charge / charge per electron = 1 / (1.6 × 10^-19) = 6.25 × 10^18, which rounds to the listed 6.24 × 10^18.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The amount of electricity required to deposit 1 mol of aluminium from a solution of AlCl3 will be(a) 0.33 faraday(b) 1 faraday(c) 3 faraday(d) 1 ampere
›Reveal solutionSolution
Aluminium exists as Al3+ in AlCl3, so depositing one mole of Al at the cathode needs 3 moles of electrons, i.e. 3 faradays.
Reasoning
The cathodic reduction half-reaction is:
Al3++3e−→Al
By Faraday's first law, the quantity of electricity needed to deposit 1 mole of a substance equals (number of electrons transferred per ion) ×F, where F=96500 C mol−1.
Here n=3, so the charge required is:
Q=nF=3F
…
- CBSE 2024Set D1 markMCQQ.A charge of 96500 coulomb liberates .............. from the solution of CuSO4.(a) 63.5 gm copper(b) 31.76 gm copper(c) 96500 gm copper(d) 100 gm copper
›Reveal solutionSolution
96500 C = 1 faraday = 1 mole of electrons. Depositing Cu requires 2 electrons per Cu atom, so 1 F deposits 63.5/2 = 31.76 g Cu.
Electrode reaction: Cu2+ + 2e- -> Cu.
To deposit 1 mole of copper (63.5 g) you need 2 moles of electrons = 2 x 96500 C = 193000 C.
Therefore the charge passed here, 96500 C (1 faraday, i.e. 1 mole of electrons), deposits half a mole of copper: …
- CBSE 2024Set B1 markQ.Fill in the blank: One Faraday electricity equals to ______ coulomb.
›Reveal solutionSolution
1 Faraday = charge carried by one mole of electrons = 96,500 C (more precisely 96,487 C, usually rounded to 96,500 C).
One Faraday (F) is defined as the quantity of electric charge carried by one mole (Avogadro's number, 6.022x10^23) of electrons:
1F=NA×e=6.022×1023×1.602×10−19 C≈96,500 C mol−1
…
- CBSE 2024Set ANNUAL1 markQ.State Faraday's first law of Electrolysis.
›Reveal solutionSolution
More charge passed through an electrolytic cell means proportionally more substance deposited at the electrode.
Faraday's first law of electrolysis states: the mass (w) of a substance produced (deposited or liberated) at an electrode during electrolysis is directly proportional to the quantity of electricity (charge, Q) that passes through the electrolyte.
w ∝ Q
Since charge Q = current (I) × time (t), this can be written:
w = Z × I × t
…
- CBSE 2023Set ANNUAL1 markQ.A solution of MgSO4 is electrolysed to carry out a deposition of 24.3 g of magnesium at cathode. How many electrons pass through the solution during the process ?
›Reveal solutionSolution
Depositing 1 mole of Mg²⁺ needs 2 moles of electrons (per Faraday's law), i.e. 2NA electrons.
Magnesium is deposited at the cathode by the two-electron reduction:
Mg2++2e−→Mg(s)
Moles of Mg deposited =24.3 g/mol24.3 g=1 mol
Since each mole of Mg requires 2 moles of electrons:
moles of electrons=1×2=2 mol
…
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