ChemistryISC Class 12 · Electrochemistry

Electrolysis and Faraday's Laws: ISC Class 12 Numericals

Updated 15 October 2026 · Fact-checked

Faraday's first law says the mass deposited at an electrode is proportional to the charge passed: m = ZQ = ZIt. The second law says that the same charge deposits masses in the ratio of equivalent masses. One faraday, 96 500 C, is the charge on one mole of electrons and deposits one gram-equivalent.

Electrons travel in the wire, ions carry current in the liquid; cations are reduced at the cathode, and m = ZIt, with 1 F = 96 500 C.

Illustration drawn for QuestionBuddy.

Understand Electrolysis and Faraday's Laws

Electrolysis is the decomposition of an electrolyte, molten or in solution, by passing a direct current through it. In an electrolytic cell electrical energy drives a non-spontaneous reaction. The electrode joined to the negative terminal is the cathode, where cations gain electrons (reduction); the electrode joined to the positive terminal is the anode, where anions or the anode itself lose electrons (oxidation). Electrons flow in the external wire; ions carry the current in the solution.

Faraday's first law. The mass $m$ of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $Q$ passed: $m = Z Q = Z I t$ , where $I$ is the current in amperes, $t$ the time in seconds and $Z$ the electrochemical equivalent, the mass deposited by 1 coulomb.

The faraday. A mole of electrons carries $N_{A} \times e = 6.022 \times 1 0^{23} \times 1.602 \times 1 0^{- 19} \approx 96500$ C. This charge is 1 faraday (F). From the electrode equation you can read how many faradays one mole of product needs: $A g^{+} + e^{-} \to A g$ needs 1 F per mole; $C u^{2 +} + 2 e^{-} \to C u$ needs 2 F; $A l^{3 +} + 3 e^{-} \to A l$ needs 3 F. So one faraday deposits one gram-equivalent of any substance (molar mass divided by the charge number), and $Z = \frac{equivalent mass}{96 500}$ . Combining: $m = \frac{M I t}{n F}$ , where $n$ is the number of electrons per formula unit.

Faraday's second law. When the same quantity of electricity passes through different electrolytes (cells connected in series), the masses of the substances deposited are proportional to their equivalent masses: $\frac{m _{1}}{m _{2}} = \frac{E _{1}}{E _{2}}$ . For example, the charge that deposits 1 mol of Ag (108 g) deposits only ½ mol of Cu (31.75 g).

Products of electrolysis. What forms depends on the electrolyte, its state, the electrodes and competing reactions of water. In molten NaCl, Na is formed at the cathode and Cl₂ at the anode. In aqueous NaCl, water is reduced in preference to Na⁺ at the cathode ( $E^{\circ}$ for $2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is −0.83 V against −2.71 V for Na⁺), so H₂ is formed; at the anode Cl₂ is released rather than O₂ because oxygen evolution needs an extra voltage (overpotential). In aqueous CuSO₄ with platinum electrodes, Cu is deposited at the cathode and O₂ is released at the anode, and the solution slowly becomes acidic and loses its blue colour. With copper electrodes, copper dissolves from the anode ( $C u \to C u^{2 +} + 2 e^{-}$ ) as fast as it deposits at the cathode, so the concentration stays constant: this is the basis of electrorefining and electroplating.

Electrolytic versus galvanic. A galvanic cell turns chemical energy into electrical energy (spontaneous, $Δ G < 0$ , anode negative); an electrolytic cell uses electrical energy to drive a non-spontaneous change ( $Δ G > 0$ , anode positive). In both, oxidation occurs at the anode.

Electrolysis and Faraday's Laws: formulas and key facts

Faraday's first law: $m = Z Q = Z I t$; $I$ in A, $t$ in s, $Q$ in C; $Z$ in g C⁻¹.
Electrochemical equivalent: $Z = \frac{E}{F} = \frac{M}{n F}$; $E$ = equivalent mass $= M / n$ ; $F = 96500$ C mol⁻¹.
Mass from current and time: $m = \frac{M I t}{n F}$; $n$ = electrons in the electrode equation per formula unit.
Faraday's second law: $\frac{m _{1}}{m _{2}} = \frac{E _{1}}{E _{2}}$; Same charge, cells in series.
The faraday: $1 F = N_{A} e \approx 96500 C mol^{- 1}$; Charge on one mole of electrons.
Gas volume at the electrode: $V_{S T P} = \frac{Q}{n F} \times 22.4 L$; $n = 2$ for H₂ and Cl₂, $n = 4$ for O₂.

How to solve Electrolysis and Faraday's Laws questions

Every Faraday numerical is a conversion between charge and moles of electrons, then between moles of electrons and moles of product. Write the electrode equation first and the rest follows.

1Write the electrode reaction ( $A g^{+} + e^{-} \to A g$ , $C u^{2 +} + 2 e^{-} \to C u$ , $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ ) and read off $n$ .
2Find the charge: $Q = I t$ , with $t$ converted to seconds.
3Moles of electrons $= \frac{Q}{96 500}$ ; moles of product $= \frac{Q}{n F}$ .
4Mass $=$ moles × molar mass; gas volume at STP $=$ moles × 22.4 L.
5Working backwards (time or current), start from the mass, find moles, then $Q = n \times moles \times F$ , then $t = Q / I$ .
6For cells in series, find moles of electrons from the cell with known data and use the same moles of electrons in every other cell (second law).

Electrolysis and Faraday's Laws in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electrorefining output depends only on charge

Suppose a small copper refinery runs one cell with an impure copper anode, a pure copper cathode and acidified CuSO₄ solution, passing a steady 100 A for an 8-hour shift. Assume all the current deposits copper (Cu = 63.5).

Calculate the mass of pure copper deposited on the cathode in one shift, and explain why the CuSO₄ concentration hardly changes.

Try it first, then show the solutionSolution

$Q = I t = 100 \times 8 \times 3600 = 2.88 \times 1 0^{6}$ C.
Moles of electrons $= \frac{2.88 \times 1 0 ^{6}}{96 500} = 29.84$ mol; $C u^{2 +} + 2 e^{-} \to C u$ , so moles of Cu $= 14.92$ mol.
Mass $= 14.92 \times 63.5 = 948$ g.
At the anode, copper dissolves ( $C u \to C u^{2 +} + 2 e^{-}$ ) at the same rate, so Cu²⁺ ions lost at the cathode are replaced.

Answer: About 948 g (nearly 1 kg) of pure copper per shift; the solution stays nearly constant because the anode supplies Cu²⁺ as fast as the cathode removes it.

Memory hook: Charge in, metal out: count the coulombs and you know the copper.

Tricks to remember Electrolysis and Faraday's Laws

Short memory aids. Say each one aloud once, then use it on a question the same day.

AN OX, RED CAT
Oxidation at the ANode, REDuction at the CAThode, in electrolytic and galvanic cells alike.
Use it when: Writing electrode reactions in product questions.
One faraday, one equivalent
96 500 C always deposits 1 gram-equivalent: 108 g Ag, 31.75 g Cu, 9 g Al, 1 g H.
Use it when: Quick checks and second-law ratios.
m equals M I t over n F
Read it as 'Mighty Indians toil, not fail': $m = \frac{M I t}{n F}$ .
Use it when: The single formula behind most Faraday numericals.

Electrolysis and Faraday's Laws: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Charge to reduce dichromate · 1 mark · MCQBoard level

The charge required to reduce 1 mol of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acid solution is:
(a) 3 F
(b) 6 F
(c) 2 F
(d) 12 F

Try it first, then show the solutionSolution

Cr changes from +6 to +3: 3 electrons per Cr atom.
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ : 6 mol electrons per mol of dichromate.

Answer: (b) 6 F.

Example 2: Copper deposited by a known current · 2 marks · Short answerBoard level

A steady current of 2.5 A flows through aqueous CuSO₄ for 25 minutes. Calculate the mass of copper deposited at the cathode. (Cu = 63.5, $F = 96500$ C mol⁻¹)

Try it first, then show the solutionSolution

$Q = I t = 2.5 \times 25 \times 60 = 3750$ C (½ mark).
$C u^{2 +} + 2 e^{-} \to C u$ : $m = \frac{M I t}{n F} = \frac{63.5 \times 3750}{2 \times 96 500}$ (1 mark).
$m = 1.23$ g (½ mark).

Answer: About 1.23 g of copper.

Example 3: Time to nickel-plate · 3 marks · Short answerHarder board level

How long will a current of 1.2 A take to deposit 1.0 g of nickel from a NiSO₄ bath? Also calculate the electrochemical equivalent of nickel. (Ni = 58.7, $F = 96500$ C mol⁻¹)

Try it first, then show the solutionSolution

$N i^{2 +} + 2 e^{-} \to N i$ ; moles of Ni $= \frac{1.0}{58.7} = 0.01704$ (½ mark).
$Q = 2 \times 0.01704 \times 96500 = 3288$ C (1 mark).
$t = \frac{Q}{I} = \frac{3288}{1.2} = 2740$ s, about 45.7 minutes (½ mark).
$Z = \frac{58.7}{2 \times 96 500} = 3.04 \times 1 0^{- 4}$ g C⁻¹ (1 mark).

Answer: About 2740 s (45.7 min); $Z \approx 3.04 \times 1 0^{- 4}$ g C⁻¹.

Example 4: Second law: silver and gold in series · 3 marks · Short answerHarder board level

Two electrolytic cells, one containing AgNO₃ and the other AuCl₃ solution, are connected in series. When a current is passed, 2.16 g of silver is deposited in the first cell. Calculate (a) the mass of gold deposited and (b) the charge passed. (Ag = 108, Au = 197)

Try it first, then show the solutionSolution

Moles of electrons $=$ moles of Ag $= \frac{2.16}{108} = 0.0200$ (1 mark).
(a) $A u^{3 +} + 3 e^{-} \to A u$ : moles of Au $= \frac{0.0200}{3}$ ; mass $= \frac{0.0200}{3} \times 197 = 1.31$ g (1 mark).
(b) $Q = 0.0200 \times 96500 = 1930$ C (1 mark). Check by the second law: $\frac{m _{A u}}{m _{A g}} = \frac{197/3}{108} = 0.608$ , and $2.16 \times 0.608 = 1.31$ g.

Answer: (a) About 1.31 g of gold. (b) 1930 C.

Example 5: Electrolysis of copper sulphate with different electrodes · 5 marks · Long answerHOTS / competency

Aqueous CuSO₄ is electrolysed between platinum electrodes with a current of 0.75 A for 1.5 hours. (Cu = 63.5, $F = 96500$ C mol⁻¹, molar volume at STP = 22.4 L)
(a) Write the reactions at the cathode and anode. [1]
(b) Calculate the mass of copper deposited and the volume of gas (at STP) released at the anode. [3]
(c) How would the anode reaction and the solution change if copper electrodes were used instead? [1]

Try it first, then show the solutionSolution

(a) Cathode: $C u^{2 +} + 2 e^{-} \to C u$ . Anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ (1 mark).
(b) $Q = 0.75 \times 1.5 \times 3600 = 4050$ C; moles of electrons $= \frac{4050}{96 500} = 0.04197$ (1 mark).
Cu: $\frac{0.04197}{2} = 0.02098$ mol $= 1.33$ g (1 mark).
O₂: $\frac{0.04197}{4} = 0.01049$ mol, volume $= 0.01049 \times 22.4 = 0.235$ L $= 235$ mL (1 mark).
(c) With copper electrodes the anode dissolves ( $C u \to C u^{2 +} + 2 e^{-}$ ), no O₂ is released, and the CuSO₄ concentration and blue colour stay the same (1 mark).

Answer: (a) Cu deposited at cathode; O₂ from water at anode. (b) 1.33 g Cu; 235 mL O₂ at STP. (c) Copper anode dissolves; no gas; concentration unchanged.

Where students lose marks in Electrolysis and Faraday's Laws

The slips examiners see most often, and what to write instead.

Mistake: Wrong substitution in $m = \frac{M I t}{n F}$
ISC examiners noted incorrect substitution in a 2025 question on electroplating aluminium.
Wrong working
$Q = \frac{4.75}{27} \times 96500$ (using $n = 1$ for Al)
Right working
$Q = \frac{4.75}{27} \times 3 \times 96500$
Do this: Read $n$ from the electrode equation ( $A l^{3 +} + 3 e^{-}$ ) and write it in.
Usually costs: 1 to 2 marks.
Mistake: Answering in the wrong unit
Examiners reported answers in faradays when coulombs were asked (2025).
Do this: Check what the question asks for; multiply faradays by 96 500 to get coulombs.
Usually costs: 1 mark.
Mistake: Time left in minutes or hours
Students substitute $t = 30$ for 30 minutes.
Wrong working
$Q = 2.5 \times 25 = 62.5$ C
Right working
$Q = 2.5 \times 1500 = 3750$ C
Do this: Convert time to seconds before multiplying by current.
Usually costs: The answer mark.
Mistake: Using different charges in series cells
Students recalculate the charge for each cell instead of using the same moles of electrons.
Do this: Cells in series carry the same current for the same time; use the same $Q$ (or moles of electrons) throughout.
Usually costs: 1 mark per metal.

Exam tips

Write the electrode equation in every numerical; it shows $n$ and earns the method mark.
Use $F = 96500$ C mol⁻¹ unless the paper gives another value.
Give masses to three significant figures with 'g', time in seconds (and minutes if helpful), and gas volumes with 'at STP'.
For product questions, state the electrode, the half-reaction and the product, in that order.
Remember that in electrolysis the anode is positive, but oxidation still happens there.

Practice questions on Electrolysis and Faraday's Laws

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Very short answerBoard level
State Faraday's first law of electrolysis.
Hint
Mass and quantity of electricity.
Answer
The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte ( $m = Z Q$ ).
Q2 · 2 marks · Short answerBoard level
Define electrochemical equivalent. Calculate it for silver. (Ag = 108, $F = 96500$ C mol⁻¹)
Hint
$Z = E / F$ .
Answer
The mass of a substance deposited or liberated by 1 coulomb of electricity. For silver, $Z = \frac{108}{96 500} = 1.12 \times 1 0^{- 3}$ g C⁻¹.
Q3 · 2 marks · Short answerHarder board level
How many electrons flow through a wire carrying a current of 0.40 A for 10 minutes? ( $e = 1.602 \times 1 0^{- 19}$ C)
Hint
$Q = I t$ , then divide by the charge of one electron.
Answer
$Q = 0.40 \times 600 = 240$ C. Number of electrons $= \frac{240}{1.602 \times 1 0 ^{- 19}} = 1.50 \times 1 0^{21}$ .
Q4 · 3 marks · Short answerHarder board level
A current of 3.0 A is passed through dilute sulphuric acid between platinum electrodes for 20 minutes. Calculate the volumes of hydrogen and oxygen released at STP.
Hint
2 electrons per H₂, 4 electrons per O₂.
Answer
$Q = 3.0 \times 1200 = 3600$ C; moles of electrons $= 0.03731$ . H₂: $0.01865$ mol $\times 22.4 = 0.418$ L. O₂: $0.009326$ mol $\times 22.4 = 0.209$ L (half the volume of H₂).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When the same quantity of electricity is passed through AgNO₃ and CuSO₄ solutions in series, the masses of silver and copper deposited are different.
Reason (R): The masses deposited are proportional to the equivalent masses of the metals.
(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.
Hint
Faraday's second law.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Aqueous NaCl is electrolysed between inert electrodes. Name the products at the cathode and the anode, and explain, using $E^{\circ}$ values, why sodium metal is not formed. ( $E^{\circ}$ : Na⁺/Na = −2.71 V; $2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ , −0.83 V)
Hint
The reduction with the higher E° happens first.
Answer
Cathode: H₂ (and OH⁻ in solution). Anode: Cl₂. Water has a much higher reduction potential (−0.83 V) than Na⁺ (−2.71 V), so water is reduced in preference and Na is not deposited. At the anode Cl₂ is released rather than O₂ because oxygen evolution needs an overpotential. The solution becomes NaOH.
Q7 · 3 marks · Short answerHOTS / competency
Suppose an aluminium plant produces 1.0 tonne (10⁶ g) of aluminium per day by electrolysis of molten alumina. Calculate the charge needed per day and the average current. (Al = 27, $F = 96500$ C mol⁻¹)
Hint
$A l^{3 +} + 3 e^{-} \to A l$ ; one day = 86 400 s.
Answer
Moles of Al $= \frac{1 0 ^{6}}{27} = 3.70 \times 1 0^{4}$ . $Q = 3 \times 3.70 \times 1 0^{4} \times 96500 = 1.07 \times 1 0^{10}$ C. $I = \frac{1.07 \times 1 0 ^{10}}{86 400} = 1.24 \times 1 0^{5}$ A.
Q8 · 3 marks · Short answerHOTS / competency
Three electrolytic cells containing ZnSO₄, CuSO₄ and AgNO₃ solutions are connected in series. A steady current of 2.0 A deposits 1.308 g of zinc in the first cell. Calculate (a) the masses of copper and silver deposited and (b) the time for which the current flowed. (Zn = 65.4, Cu = 63.5, Ag = 108)
Hint
Find the moles of electrons from the zinc cell first.
Answer
Moles of Zn $= 0.0200$ , so moles of electrons $= 0.0400$ . (a) Cu: 0.0200 mol $= 1.27$ g; Ag: 0.0400 mol $= 4.32$ g. (b) $Q = 0.0400 \times 96500 = 3860$ C; $t = \frac{3860}{2.0} = 1930$ s (about 32 min).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Value of 1 faraday?Show answer
About 96 500 C mol⁻¹, the charge on one mole of electrons.
2.State Faraday's second law.Show answer
When the same quantity of electricity passes through different electrolytes, the masses deposited are proportional to their equivalent masses.
3.How many faradays deposit 1 mol of Al?Show answer
3 F.
4.Product at the anode when aqueous CuSO₄ is electrolysed with Pt electrodes?Show answer
O₂ gas.
5.Relation between Z and equivalent mass?Show answer
$Z = E / F$ .
6.Why does the concentration of CuSO₄ stay constant with copper electrodes?Show answer
The copper anode dissolves at the same rate as copper is deposited at the cathode.

Electrolysis and Faraday's Laws: frequently asked questions

What are Faraday's laws of electrolysis?

First law: mass deposited is proportional to the charge passed, m = ZIt. Second law: for the same charge, masses deposited are proportional to equivalent masses.

What is the difference between electrochemical equivalent and equivalent mass?

Electrochemical equivalent is the mass deposited by 1 coulomb; equivalent mass is the mass deposited by 1 faraday (96 500 C). So Z = E/96\,500.

Why is hydrogen, not sodium, formed at the cathode in aqueous NaCl electrolysis?

Water has a much higher reduction potential than Na⁺, so water is reduced to H₂ in preference.

How do you find the time needed to deposit a given mass?

Convert mass to moles, multiply by n and F to get the charge, then divide the charge by the current.

Is the anode positive or negative in electrolysis?

Positive in an electrolytic cell (negative in a galvanic cell). Oxidation occurs at the anode in both.