ChemistryICSE Class 10 · Electrolysis

Selective Discharge of Ions in Electrolysis (ICSE Class 10)

Q: What decides which ion is discharged during electrolysis?

Mainly the ion's position in the activity series, then its concentration and the nature of the electrode. Ions of less reactive metals (Cu²⁺, Ag⁺) are discharged before H⁺; OH⁻ is discharged before SO₄²⁻ and NO₃⁻.

Q: Why does solid lead bromide not conduct electricity?

Its ions are held in fixed positions in the crystal lattice and cannot move. Melting frees the ions.

Q: Why is graphite used for the electrolysis of molten lead bromide?

Graphite conducts electricity and is inert: it does not react with the bromine formed at the anode.

Q: Is the anode positive or negative in electrolysis?

Positive. It is connected to the positive terminal of the battery, and anions (negative ions) move to it.

Updated 15 October 2026 · Fact-checked

When an electrolyte has more than one kind of cation or anion, only one is usually discharged at each electrode. This is selective discharge. It depends on the ion's position in the activity series (lower metals' ions discharge more easily), its concentration, and whether the electrode is inert or active.

Understand Selective Discharge of Ions

During electrolysis cations move to the cathode and anions to the anode. At the cathode a cation gains electrons (reduction); at the anode an anion loses electrons (oxidation). That is why the anode is called the oxidising electrode and the cathode the reducing electrode. An ion that has gained or lost its charge has been discharged and becomes a neutral atom or molecule.

One ion of each kind: molten lead bromide. Solid PbBr₂ does not conduct because its ions are locked in the crystal. When it is heated until molten, the ions become free: $P b B r_{2} \to P b^{2 +} + 2 B r^{-}$ . With graphite electrodes, which are inert and not attacked by bromine, Pb²⁺ is reduced at the cathode ( $P b^{2 +} + 2 e^{-} \to P b$ , silvery grey lead) and Br⁻ is oxidised at the anode ( $B r^{-} - e^{-} \to B r$ , then $B r + B r \to B r_{2}$ , reddish-brown vapour). There is no choice to make.

More than one ion: selective discharge. In aqueous solutions water supplies H⁺ and OH⁻ ions as well as the ions of the dissolved compound. Only one cation and one anion are normally discharged. The choice depends on three factors:

1. Position in the activity series. A metal high in the series (K, Na, Ca, Mg) forms ions easily and its ions hold on to their charge; a metal low in the series (Cu, Ag) forms ions reluctantly and its ions accept electrons easily. Ease of discharge of cations increases in the order K⁺ < Na⁺ < Ca²⁺ < Mg²⁺ < Al³⁺ < Zn²⁺ < Fe²⁺ < Pb²⁺ < H⁺ < Cu²⁺ < Ag⁺. So in water Cu²⁺ is discharged before H⁺, but H⁺ before Na⁺. Among anions, OH⁻ is discharged in preference to SO₄²⁻ and NO₃⁻, which are very hard to discharge.
2. Concentration. An ion present in much greater concentration may be discharged even if it is slightly less favoured.
3. Nature of the electrode. An inert electrode (platinum, graphite) only passes electrons. An active anode, such as copper in copper sulphate solution, is itself oxidised ( $C u - 2 e^{-} \to C u^{2 +}$ ) because copper atoms lose electrons more readily than OH⁻ or SO₄²⁻ ions; no anion is discharged.

Putting it together. For any aqueous electrolysis: list every ion (from the compound and from water), send cations to the cathode and anions to the anode, pick the ion lowest in the activity series at the cathode and OH⁻ (not SO₄²⁻ or NO₃⁻) at an inert anode, then check whether the anode is active.

Selective Discharge of Ions: formulas and key facts

Molten lead bromide (graphite electrodes): $Cathode: P b^{2 +} + 2 e^{-} \to P b; Anode: B r^{-} - e^{-} \to B r, B r + B r \to B r_{2}$
Hydroxide ion at an inert anode: $4 O H^{-} - 4 e^{-} \to 2 H_{2} O + O_{2}$
Hydrogen ion at the cathode: $2 H^{+} + 2 e^{-} \to H_{2}$
Ease of discharge of cations (increasing): Hardest to discharge K⁺, Na⁺, Ca²⁺, Mg²⁺
Harder than hydrogen Al³⁺, Zn²⁺, Fe²⁺, Pb²⁺
Hydrogen ion H⁺
Easiest to discharge Cu²⁺, Ag⁺
Factors for selective discharge: Activity series the lower the metal, the more easily its ion is discharged
Concentration a much more concentrated ion may be discharged
Electrode an active anode dissolves instead of an anion being discharged
Electrode terms: Cathode negative electrode; cations gain electrons (reduction)
Anode positive electrode; anions lose electrons (oxidation)
Inert electrode platinum or graphite, takes no part
Active electrode e.g. copper in CuSO₄, dissolves at the anode

Hardest to discharge	K⁺, Na⁺, Ca²⁺, Mg²⁺
Harder than hydrogen	Al³⁺, Zn²⁺, Fe²⁺, Pb²⁺
Hydrogen ion	H⁺
Easiest to discharge	Cu²⁺, Ag⁺

Activity series	the lower the metal, the more easily its ion is discharged
Concentration	a much more concentrated ion may be discharged
Electrode	an active anode dissolves instead of an anion being discharged

Cathode	negative electrode; cations gain electrons (reduction)
Anode	positive electrode; anions lose electrons (oxidation)
Inert electrode	platinum or graphite, takes no part
Active electrode	e.g. copper in CuSO₄, dissolves at the anode

How to solve Selective Discharge of Ions questions

Predicting products is a four-step routine. Write the ions first; the rest follows from the activity series.

1Write the ions from the electrolyte and, for an aqueous solution, H⁺ and OH⁻ from water.
2Send cations to the cathode and anions to the anode.
3At the cathode, discharge the cation lowest in the activity series (Cu²⁺ or Ag⁺ before H⁺; H⁺ before Na⁺, K⁺, Mg²⁺).
4At an inert anode, discharge OH⁻ in preference to SO₄²⁻ or NO₃⁻ (a halide ion in a molten halide).
5If the anode is the same metal as the cation (copper in CuSO₄), the anode dissolves instead.
6Write both electrode equations with electrons, and state the observation at each electrode.

Selective Discharge of Ions 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.

Recovering copper selectively from a mixture

Suppose a recycling unit has waste solution containing both copper(II) sulphate and zinc sulphate, and electrolyses it with inert electrodes to recover copper.

Which metal is deposited first at the cathode, and why is zinc not deposited while copper ions remain? What is given off at the anode?

Try it first, then show the solutionSolution

Cations present: Cu²⁺, Zn²⁺ and H⁺ (from water).
Copper is lowest in the activity series of the three, so Cu²⁺ accepts electrons most readily and copper is deposited first.
Zinc is above hydrogen, so Zn²⁺ is harder to discharge than Cu²⁺ (and H⁺).
At the inert anode OH⁻ is discharged in preference to SO₄²⁻, giving oxygen.

Answer: Copper deposits first; zinc stays in solution because Zn²⁺ is harder to discharge; oxygen is evolved at the anode.

Memory hook: Lower in the series, lighter to discharge: the least reactive metal comes out first.

Tricks to remember Selective Discharge of Ions

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

AN OX, RED CAT
ANode: OXidation; REDuction: CAThode.
Use it when: Labelling electrode reactions and justifying 'oxidising electrode'.
Kings Never Come, Many Are Zealous For Pleasure; Here Comes Agent
K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Ag: ions get easier to discharge as you go right.
Use it when: The cation discharge order.
Sulphate and nitrate stay put
SO₄²⁻ and NO₃⁻ are almost never discharged in solution; OH⁻ goes instead and gives oxygen.
Use it when: Anode product in aqueous sulphate or nitrate solutions with inert electrodes.

Selective Discharge of Ions: solved examples

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

Example 1: Ion discharged at the anode · 1 mark · MCQBoard level

During the electrolysis of acidified water with platinum electrodes, the ion discharged at the anode is:
(a) H⁺
(b) SO₄²⁻
(c) OH⁻
(d) Pt²⁺

Try it first, then show the solutionSolution

Anions OH⁻ and SO₄²⁻ reach the anode; platinum is inert.
OH⁻ is discharged in preference to SO₄²⁻.

Answer: (c) OH⁻.

Example 2: Molten, with graphite · 2 marks · Short answerBoard level

In the electrolysis of lead(II) bromide, give a reason why (a) the lead bromide must be molten, (b) graphite electrodes are used.

Try it first, then show the solutionSolution

(a) In solid PbBr₂ the ions are held in fixed positions; melting sets them free to move to the electrodes.
(b) Graphite is inert: it conducts but is not attacked by the bromine vapour formed at the anode.

Answer: (a) Ions must be free to move. (b) Graphite is inert and unaffected by bromine.

Example 3: Choosing between ions in copper sulphate solution · 3 marks · Short answerHarder board level

Aqueous copper(II) sulphate is electrolysed with platinum electrodes. (a) List the ions present. (b) Which ion is discharged at each electrode, and why? (c) Write the electrode reactions.

Try it first, then show the solutionSolution

(a) Cu²⁺, SO₄²⁻ (from CuSO₄) and H⁺, OH⁻ (from water).
(b) Cathode: Cu²⁺, because copper is below hydrogen in the activity series. Anode: OH⁻, discharged in preference to SO₄²⁻.
(c) Cathode: $C u^{2 +} + 2 e^{-} \to C u$ . Anode: $4 O H^{-} - 4 e^{-} \to 2 H_{2} O + O_{2}$ .

Answer: Cu²⁺ at the cathode (copper deposited); OH⁻ at the anode (oxygen evolved).

Example 4: Oxidising and reducing electrodes · 3 marks · Short answerHarder board level

Using the electrolysis of molten lead bromide, explain why the anode is called the oxidising electrode and the cathode the reducing electrode. Write the relevant equations.

Try it first, then show the solutionSolution

Oxidation is loss of electrons; reduction is gain of electrons.
Br⁻ ions migrate to the anode and lose electrons: $B r^{-} - e^{-} \to B r$ , so oxidation happens at the anode.
Pb²⁺ ions migrate to the cathode and gain electrons: $P b^{2 +} + 2 e^{-} \to P b$ , so reduction happens at the cathode.

Answer: Anions lose electrons at the anode (oxidation); cations gain electrons at the cathode (reduction).

Example 5: A mixture of cations · 4 marks · Long answerHOTS / competency

A solution contains sodium nitrate and copper(II) nitrate. It is electrolysed for a long time with inert electrodes. (a) What is deposited at the cathode first? (b) What forms at the cathode after all the copper ions are used up? (c) Is sodium ever deposited? (d) What forms at the anode throughout? Give reasons.

Try it first, then show the solutionSolution

Cations: Na⁺, Cu²⁺, H⁺. Anions: NO₃⁻, OH⁻.
(a) Copper, since Cu²⁺ is the easiest cation to discharge (lowest in the series).
(b) Hydrogen: once Cu²⁺ is gone, H⁺ is the next easiest, $2 H^{+} + 2 e^{-} \to H_{2}$ .
(c) No: Na⁺ is far harder to discharge than H⁺, which water keeps supplying.
(d) Oxygen, from OH⁻, which is discharged in preference to NO₃⁻.

Answer: (a) Copper (b) hydrogen (c) no (d) oxygen.

Example 6: Why the copper anode dissolves · 3 marks · Short answerHOTS / competency

In the electrolysis of copper(II) sulphate solution, oxygen is evolved at a platinum anode, but nothing is evolved at a copper anode. Explain, with the anode equation in each case.

Try it first, then show the solutionSolution

Platinum is inert, so an anion must be discharged: $4 O H^{-} - 4 e^{-} \to 2 H_{2} O + O_{2}$ .
Copper atoms lose electrons more readily than OH⁻ or SO₄²⁻ ions, so the copper anode itself is oxidised: $C u - 2 e^{-} \to C u^{2 +}$ .
No anion is discharged at the copper anode, so no gas is given off; the anode becomes thinner.

Answer: Inert Pt: OH⁻ discharged, oxygen evolved. Active Cu: the anode dissolves as Cu²⁺, no gas.

Where students lose marks in Selective Discharge of Ions

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

Mistake: Saying the anode itself loses electrons
In 2025 (Q3(ii)) examiners noted that candidates wrote 'the anode loses electrons' instead of 'anions migrate to the anode and lose electrons'.
Do this: Write: 'anions migrate to the anode and lose electrons there, so oxidation occurs at the anode'.
Usually costs: 1 mark.
Mistake: Oxidation defined by oxygen
The same report found oxidation and reduction explained as gain or loss of oxygen or hydrogen instead of electron transfer.
Do this: In electrolysis use the electron definition: oxidation is loss of electrons; reduction is gain.
Usually costs: 1 mark.
Mistake: No reason for graphite
Examiners noted that candidates named graphite electrodes for molten lead bromide but gave no reason (2025 Q3(ii)).
Do this: Graphite is inert and is not affected by the bromine vapour.
Usually costs: 1 mark.
Mistake: Discharging Na⁺ or SO₄²⁻ in water
Students pick the ion from the dissolved salt and forget H⁺ and OH⁻ from water.
Wrong working
Aqueous Na₂SO₄: cathode gives Na
Right working
Aqueous Na₂SO₄: cathode gives H₂ (H⁺ discharged); anode gives O₂ (OH⁻ discharged)
Do this: Always add H⁺ and OH⁻ to the ion list for aqueous solutions.
Usually costs: 1 to 2 marks.

Exam tips

In 'justify' questions, write the migration step as well as the electron change: the examiner looks for both.
Electrode equations may be written with '− e⁻' on the left (ICSE style) or '+ e⁻' on the right; both are accepted if balanced.
State observations, not just names: 'reddish-brown vapours', 'silvery grey metal', 'colourless gas that relights a glowing splint'.
The 2025 report lists preferential discharge as a commonly confused idea; practise predicting products for several solutions.

Practice questions on Selective Discharge of Ions

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 · MCQBoard level
Which of these cations is discharged most readily at the cathode?
(a) Na⁺
(b) Mg²⁺
(c) H⁺
(d) Ag⁺
Hint
Lowest in the activity series.
Answer
(d) Ag⁺.
Q2 · 1 mark · Very short answerBoard level
What is meant by selective discharge of ions?
Hint
More than one ion at an electrode.
Answer
When two or more ions of similar charge reach an electrode, only one of them is discharged in preference to the others.
Q3 · 1 mark · Very short answerBoard level
Name the three factors that decide which ion is discharged at an electrode.
Hint
Series, amount, electrode.
Answer
Position of the ion in the activity series, concentration of the ion, and the nature of the electrode (inert or active).
Q4 · 2 marks · Short answerHarder board level
For the electrolysis of molten lead bromide, write the cathode and anode reactions and the observation at the anode.
Hint
Bromine is formed at the anode.
Answer
Cathode: $P b^{2 +} + 2 e^{-} \to P b$ . Anode: $B r^{-} - e^{-} \to B r$ ; $B r + B r \to B r_{2}$ . Observation: reddish-brown vapours of bromine at the anode.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Hydrogen is not liberated at the cathode during the electrolysis of aqueous copper(II) sulphate with platinum electrodes.
Reason (R): Copper is below hydrogen in the activity series, so Cu²⁺ ions accept electrons more readily than H⁺ ions.
(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
Compare Cu²⁺ and H⁺.
Answer
(a).
Q6 · 2 marks · Short answerHarder board level
Aqueous sodium sulphate is electrolysed using inert electrodes. Why is hydrogen, not sodium, obtained at the cathode?
Hint
Where is sodium in the activity series?
Answer
Sodium is far above hydrogen in the activity series, so Na⁺ ions hold their charge and are hard to discharge. H⁺ ions (from water) are discharged in preference: $2 H^{+} + 2 e^{-} \to H_{2}$ .
Q7 · 2 marks · Short answerHarder board level
Arrange Cu²⁺, Na⁺, Fe²⁺ and H⁺ in increasing order of ease of discharge at the cathode, and explain the basis of your order.
Hint
Use the activity series.
Answer
Na⁺ < Fe²⁺ < H⁺ < Cu²⁺. The more reactive a metal (higher in the activity series), the more readily it forms ions and the harder its ions are to discharge.
Q8 · 3 marks · Short answerHOTS / competency
Silver nitrate solution is electrolysed with platinum electrodes. Predict the product at each electrode and how the solution changes, giving reasons.
Hint
Apply the same rules as for copper sulphate.
Answer
Cathode: Ag⁺ is discharged before H⁺ (silver is below hydrogen): $A g^{+} + e^{-} \to A g$ , silver deposited. Anode: OH⁻ is discharged rather than NO₃⁻, giving oxygen. Ag⁺ and OH⁻ are removed, leaving H⁺ and NO₃⁻, so the solution becomes acidic (nitric acid).

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.Which ion is discharged at a platinum anode in CuSO₄ solution?Show answer
OH⁻ (giving oxygen).
2.Why is the cathode called the reducing electrode?Show answer
Cations gain electrons there.
3.Product at the cathode in molten PbBr₂?Show answer
Lead (silvery grey).
4.Which is discharged more easily, H⁺ or Cu²⁺?Show answer
Cu²⁺.
5.Name an active anode.Show answer
Copper in copper(II) sulphate solution.

Selective Discharge of Ions: frequently asked questions

What decides which ion is discharged during electrolysis?

Mainly the ion's position in the activity series, then its concentration and the nature of the electrode. Ions of less reactive metals (Cu²⁺, Ag⁺) are discharged before H⁺; OH⁻ is discharged before SO₄²⁻ and NO₃⁻.

Why does solid lead bromide not conduct electricity?

Its ions are held in fixed positions in the crystal lattice and cannot move. Melting frees the ions.

Why is graphite used for the electrolysis of molten lead bromide?

Graphite conducts electricity and is inert: it does not react with the bromine formed at the anode.

Is the anode positive or negative in electrolysis?

Positive. It is connected to the positive terminal of the battery, and anions (negative ions) move to it.