ICSE Class 10 Electrolysis — the Practical Questions Answered

By Aniket Bhardwaj · 25 September 2026 · CBSE / ICSE

Electrolysis questions in ICSE Class 10 are highly predictable. You are given a cell, told what the electrodes are made of, and asked four things: which ions are present, which one is discharged at each electrode, what you would observe, and what happens to the solution. This guide answers that set of four for every cell the syllabus uses, then works the numericals.

The vocabulary, defined precisely

Term	Exact meaning
Electrolyte	A substance that conducts electricity in the molten state or in aqueous solution, and is chemically decomposed by it
Non-electrolyte	A substance that does not conduct in either state — sugar solution, alcohol, pure water (very nearly)
Strong electrolyte	Almost completely dissociated or ionised in solution — HCl, NaOH, NaCl
Weak electrolyte	Only partly ionised, so an equilibrium exists — acetic acid, ammonium hydroxide, carbonic acid
Cathode	The electrode joined to the negative terminal. Cations go here and gain electrons (reduction)
Anode	The electrode joined to the positive terminal. Anions go here and lose electrons (oxidation)

ICSE also separates two words that look interchangeable. Electrolytic dissociation is what happens to an ionic compound such as NaCl: the ions already exist in the lattice and simply become free when it melts or dissolves. Ionisation is what happens to a polar covalent compound such as HCl: the ions are formed when the molecule meets water. Getting this pair right is a standard two-mark answer.

How current is actually carried

In the wires and electrodes → by electrons
In the electrolyte → by moving ions

Electrons never travel through the liquid. Saying they do is one of the commonest ways students lose a mark on an otherwise correct answer.

Preferential discharge — the rule that decides everything

When more than one kind of ion could be discharged, the one that is discharged depends on three things: position in the electrochemical series, the concentration of the ions, and the nature of the electrode.

Cations, in increasing ease of discharge:
K⁺ < Ca²⁺ < Na⁺ < Mg²⁺ < Al³⁺ < Zn²⁺ < Fe²⁺ < Pb²⁺ < H⁺ < Cu²⁺ < Ag⁺ < Au³⁺

Anions, in increasing ease of discharge:
SO₄²⁻ < NO₃⁻ < Cl⁻ < Br⁻ < I⁻ < OH⁻

Read it as: the harder a metal is to extract, the harder its ion is to discharge. Potassium sits at the far left of both lists for the same reason it needs electrolysis to be extracted at all. And remember that in any aqueous solution, water supplies H⁺ and OH⁻ ions that join the competition — which is why the products of electrolysing a molten salt and its solution are often different.

Cell 1 — molten lead(II) bromide, graphite electrodes

Ions present: Pb²⁺ and Br⁻ only. There is no water, so there is no competition.

Cathode: Pb²⁺ + 2e⁻ → Pb
Anode: 2Br⁻ − 2e⁻ → Br₂
Overall: PbBr₂ → Pb + Br₂

Observations: a greyish-white bead of molten lead collects below the cathode; reddish-brown bromine vapour with a pungent smell appears at the anode.

Why must it be molten? In the solid the ions are locked in the lattice and cannot move, so solid lead bromide does not conduct. Melting frees them.

Cell 2 — acidified water, platinum electrodes

Ions present: H⁺ and OH⁻ from water, plus H⁺ and SO₄²⁻ from the dilute sulphuric acid. Sulphate is at the bottom of the anion list, so hydroxide is discharged instead.

Cathode: 2H⁺ + 2e⁻ → H₂
Anode: 4OH⁻ − 4e⁻ → 2H₂O + O₂
Overall: 2H₂O → 2H₂ + O₂

Observation: colourless gas at both electrodes, but twice as much by volume at the cathode. Hydrogen relights nothing but pops with a lighted splint; oxygen rekindles a glowing splint.

Why acidify? Pure water is such a weak electrolyte that it barely conducts. The acid supplies ions; the acid itself is not used up, so the solution slowly becomes more concentrated as water is removed.

Cell 3 — copper sulphate solution with copper electrodes

This is the electro-refining cell. The anode is an active electrode, so it dissolves in preference to any ion being discharged there.

Cathode: Cu²⁺ + 2e⁻ → Cu (pink-brown copper deposits)
Anode: Cu − 2e⁻ → Cu²⁺ (the anode dissolves and gets thinner)

What happens to the solution? Nothing. As many Cu²⁺ ions enter the solution at the anode as leave it at the cathode, so the concentration and the blue colour stay the same. That is precisely why the method works for refining.

Cell 4 — copper sulphate solution with platinum electrodes

Now the anode is inert, so it cannot dissolve and an anion must be discharged.

Cathode: Cu²⁺ + 2e⁻ → Cu
Anode: 4OH⁻ − 4e⁻ → 2H₂O + O₂

What happens to the solution? Copper ions are removed and are not replaced, so the blue colour fades; the H⁺ and sulphate ions left behind mean the solution slowly becomes acidic. Comparing Cells 3 and 4 is the classic five-mark question in this chapter: same electrolyte, different electrodes, different result.

Cell 5 — concentrated sodium chloride solution (brine)

Ions present: Na⁺, H⁺, Cl⁻, OH⁻. Hydrogen is discharged in preference to sodium because it is far easier to discharge; chloride is discharged in preference to hydroxide because the brine is concentrated — the concentration factor overrides the position in the series here.

Cathode: 2H⁺ + 2e⁻ → H₂
Anode: 2Cl⁻ − 2e⁻ → Cl₂
Left in solution: Na⁺ and OH⁻, that is sodium hydroxide

Electroplating — what to write

Three fixed rules, whatever the metal:

The article to be plated is the cathode, cleaned thoroughly first.
The pure plating metal is the anode, so it dissolves and keeps the electrolyte concentration constant.
The electrolyte contains ions of the plating metal.

For nickel plating the electrolyte is nickel sulphate solution. For silver plating, ICSE specifies sodium argentocyanide rather than silver nitrate, because a solution containing free Ag⁺ ions deposits silver too quickly, giving a coarse, loose, non-adherent coat; the complex salt releases silver ions slowly and the layer is smooth and sticks. Electroplating is done to prevent corrosion and to improve appearance.

Worked example 1 — reading the volumes

Q. During the electrolysis of acidified water, 40 cm³ of gas is collected at the cathode. Name both gases, give the volume at the anode, and give a test for each.

Cathode gas is hydrogen; anode gas is oxygen.

From 2H₂O → 2H₂ + O₂, hydrogen and oxygen are produced in a 2 : 1 ratio by volume, so the anode gives
40 ÷ 2 = 20 cm³ of oxygen.

Tests: hydrogen burns with a pop when a lighted splint is brought near; oxygen relights a glowing splint.

Worked example 2 — linking mass at the cathode to gas at the anode

Q. Copper sulphate solution is electrolysed with platinum electrodes until 0.635 g of copper has been deposited. What volume of oxygen, measured at S.T.P., is liberated at the anode? (Cu = 63.546)

Step 1 — moles of copper.
n(Cu) = 0.635 ÷ 63.546 = 0.00999 mol

Step 2 — electrons involved. From Cu²⁺ + 2e⁻ → Cu, each mole of copper needs 2 moles of electrons:
n(e⁻) = 2 × 0.00999 = 0.01999 mol

Step 3 — oxygen at the anode. From 4OH⁻ − 4e⁻ → 2H₂O + O₂, four moles of electrons release one mole of oxygen:
n(O₂) = 0.01999 ÷ 4 = 0.004997 mol

Step 4 — volume.
V = 0.004997 × 22.4 = 0.1119 L = 112 cm³ at S.T.P.

(That uses the 22.4 L/mol molar volume at 273.15 K and 1 atm, the convention ICSE papers use. On the 1-bar convention the molar volume is 22.7 L/mol and the answer is 113 cm³. Say which one you are using.)

Worked example 3 — a full "state the observations" answer

Q. Copper sulphate solution is electrolysed using copper electrodes. State what is seen at each electrode and what happens to the colour of the solution. Then say how your answer would change if both electrodes were platinum.

With copper electrodes: a fresh pink-brown deposit builds up on the cathode; the anode becomes thinner and its mass falls. The blue colour is unchanged, because copper ions are replaced at the anode as fast as they are removed at the cathode.

With platinum electrodes: the cathode still gains copper, but the inert anode cannot dissolve, so hydroxide ions are discharged and colourless oxygen bubbles off. The blue colour fades as the Cu²⁺ ions are used up, and the solution turns acidic because H⁺ and SO₄²⁻ ions remain.

Mistakes that cost marks

Saying electrons flow through the electrolyte. Ions carry the current in the liquid; electrons only move in the external circuit and the electrodes.
Assuming the cathode is negative in every kind of cell. In an electrolytic cell the cathode is negative and the anode positive. In a galvanic cell the signs are the other way round. Reduction still happens at the cathode in both.
Forgetting the ions from water. Every aqueous electrolysis has H⁺ and OH⁻ in the mixture, and they very often win.
Writing that solid lead bromide conducts. It does not — its ions are not free to move until it melts.
Not distinguishing active from inert electrodes. The same copper sulphate solution gives two completely different answers depending on the electrode material.
Balancing the anode half-equation wrongly. It is 4OH⁻ − 4e⁻ → 2H₂O + O₂. Check the oxygens: four on the left, two in the water plus two in the O₂ on the right.
Using silver nitrate for silver plating. ICSE wants sodium argentocyanide, and wants the reason.

Where this chapter is examined

Board / exam	How it appears
ICSE Class 10	A named chapter. Expect a described cell, ion lists, half-equations, observations and the electroplating reasoning
CBSE Class 10	Appears inside "Metals and Non-metals" and "Chemical Reactions" — electrolytic reduction and electrolysis of water
Class 12	The same cells return with Faraday's laws, electrode potentials and the Nernst equation attached
Competitive papers	Products-of-electrolysis questions, which are just the preferential discharge order applied carefully

Qualitatively, this chapter rewards precision more than it rewards volume of writing. Every mark here is for a specific word — cathode, reduction, active electrode, concentrated — so write short exact sentences rather than long vague ones.

Practise the mass-to-mole step from Worked Example 2. Every electrolysis numerical starts by turning a deposited mass into moles, and ends by turning moles of gas back into a volume. The Mass ↔ Mole calculator does the conversion for any formula so you can check each step of your own working.

Open the Mass ↔ Mole Calculator →

Heading into Class 11–12 chemistry, where these cells come back with Faraday's laws attached? ABC Chemistry runs Class 11–12 coaching at the Gurugram centre and online classes across India — details at abcchemistry.in. For one-to-one home tuition in Delhi, Noida or Gurgaon, see delhihometutor.com.