Electrode reactions

Selective discharge tells you which ion reaches an electrode and gets picked; electrode reactions tell you what happens the instant it arrives — the balanced half-equation showing electrons gained or lost as the ion becomes a neutral atom or molecule. This is the step where electrolysis stops being a story and becomes chemistry you can write down: a cation meeting the cathode picks up electrons and is reduced, an anion meeting the anode gives electrons up and is oxidised, and the two half-equations must transfer exactly the same number of electrons for the circuit to balance. At ICSE Class 10 this comes down to three standard electrolytes — molten lead bromide, acidified water, and aqueous copper sulphate — each with its own twist. Molten salts have no water to complicate things; acidified water hides its acid ions behind selective discharge; and copper sulphate splits into two entirely different anode stories depending on whether the electrode is inert or made of copper itself. Master the balancing pattern once and all three fall into place.

The core insight

An electrode reaction is a charge-balanced half-equation — cations gain electrons and are reduced at the cathode, anions lose electrons and are oxidised at the anode — and the electrons released at the anode must exactly equal the electrons consumed at the cathode.

The core idea

Two general patterns. Whatever the electrolyte, every electrode reaction follows one of two templates. At the cathode (reduction), a cation gains electrons to become a neutral atom:

Mn++n e−→MM^{n+} + n\,e^- \rightarrow M

where: Mn+M^{n+} is the cation of charge n+n+, nn is the number of electrons transferred, and MM is the neutral metal (or hydrogen) atom deposited or released. At the anode (oxidation), an anion loses electrons to become a neutral atom or molecule:

Xn−→X+n e−X^{n-} \rightarrow X + n\,e^-

where: Xn−X^{n-} is the anion of charge n−n- and XX is the neutral product; for elements that exist as diatomic molecules — Cl₂, Br₂, O₂ — two anions combine so the coefficients must be doubled (2X⁻ → X₂ + 2e⁻), never written as a single atom with a fractional charge.

Balancing rule. Charge on the ion side must equal the electrons transferred, and atoms must balance on both sides. This is nothing more than conservation of charge applied to a chemical equation — the reasoning behind it is worked out fully below.

Case (a): molten binary salt (e.g. molten PbBr₂), inert electrodes. Only two ions exist, so there is no competition:

Cathode: Pb2++2e−→PbAnode: 2Br−→Br2+2e−\text{Cathode: } \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \qquad \text{Anode: } 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-

Case (b): acidified/dilute water, platinum electrodes. The dilute sulphuric acid only makes the water conducting — its own ions are not discharged (selective discharge favours water's ions here):

Cathode: 2H++2e−→H2Anode: 4OH−→O2+2H2O+4e−\text{Cathode: } 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \qquad \text{Anode: } 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-

This is simply water decomposing into hydrogen and oxygen in a 2:1 volume ratio; the acid is regenerated, not consumed.

Case (c): aqueous copper sulphate — the electrode material decides everything. With inert electrodes (platinum or graphite): Cu²⁺ is discharged at the cathode (it sits lower than H⁺ in the electrochemical series), and since SO₄²⁻ is a stable oxy-anion that resists discharge, OH⁻ is discharged at the anode instead:

Cathode: Cu2++2e−→CuAnode: 4OH−→O2+2H2O+4e−\text{Cathode: } \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \qquad \text{Anode: } 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-

With a copper anode (an active, attackable electrode): no anion is discharged at all — the copper metal itself dissolves into solution, releasing electrons directly:

Anode: Cu→Cu2++2e−\text{Anode: } \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-

while the cathode still deposits Cu²⁺ + 2e⁻ → Cu. Copper is deposited at the same rate it dissolves, so the blue colour and concentration of the solution stay practically unchanged — this is the working principle of electroplating and electro-refining.

Figure
An electrolytic cell showing the cathode on the left where a cation gains electrons and is reduced and the anode on the right where an anion loses electrons and is oxidised with the external circuit carrying electrons from anode to cathode
Electrode reactions: reduction at the cathode, oxidation at the anode electrolyte solution or melt CATHODE (−) ANODE (+) Mⁿ⁺ Mⁿ⁺ + ne⁻ → M Xⁿ⁻ Xⁿ⁻ → X + ne⁻ electrons flow anode to cathode through the external wire Electrons lost at the anode always equal electrons gained at the cathode
Derivation AWhy an electrode half-equation must balance electrons and atoms

The goal is to show, using nothing but conservation of charge and conservation of atoms, why an electrode half-equation such as 2Br⁻ → Br₂ + 2e⁻ needs exactly those coefficients and not, say, Br⁻ → Br₂ + e⁻ — a mistake students make constantly when an element forms a diatomic molecule at the electrode.

Step 1. Fix what must balance. A half-equation is not a free sketch; it must obey two independent conservation laws at once — the number of atoms of each element must be the same on both sides, and the total charge must be the same on both sides (electrons count as charge of −1-1 each). Both checks are required; passing one does not excuse failing the other.

Step 2. Start from the ion and its neutral product. Bromide is discharged at the anode as bromine gas, which exists as the diatomic molecule Br₂, not as free Br atoms. So the atom count already forces at least two Br⁻ ions to appear on the left, since one Br₂ molecule on the right contains two bromine atoms: 2Br−→Br22\text{Br}^- \rightarrow \text{Br}_2.

Step 3. Now balance the charge. The left side carries a total charge of 2×(−1)=−22 \times (-1) = -2 (two singly-charged bromide ions). The right side, before adding electrons, is neutral Br₂ with charge 00. For charge to balance, the missing −2-2 must appear as two electrons released on the right: 2Br−→Br2+2e−2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-. Check: left charge −2-2; right charge 0+2(−1)=−20 + 2(-1) = -2. Equal, as required.

Step 4. Generalise the pattern. For any anion Xn−X^{n-} forming a diatomic element X2X_2, the same two steps give 2Xn−→X2+2n e−2X^{n-} \rightarrow X_2 + 2n\,e^-; for a cation Mn+M^{n+} discharged directly to the neutral metal, only charge balance is needed since one ion already gives one atom: Mn++n e−→MM^{n+} + n\,e^- \rightarrow M. This is exactly why the cathode equations in this article never need a doubled coefficient (Pb²⁺, Cu²⁺, H⁺ paired with H₂ already balances atom-for-atom when H⁺ is doubled to match H₂'s two atoms), while diatomic anode products such as Br₂ and O₂ always do. Every electrode half-equation in this article can be re-checked by this same two-step test: count atoms first, then count charge.

Worked example

Question: Molten lead bromide, PbBr₂, is electrolysed using graphite electrodes. Write the balanced electrode reactions and state what is observed at each electrode.

Step 1. Identify the ions present. Molten PbBr₂ dissociates into Pb²⁺ and Br⁻ ions only — there is no water, so there is no competing ion to consider; the metal cation and the halide anion are the only species that can be discharged.

Step 2. Write the cathode reaction. Pb²⁺ migrates to the negative electrode and is reduced by gaining two electrons, giving a neutral lead atom directly (one ion, one atom, so only charge balance is needed):

Pb2++2e−→Pb\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}

Observation: a grey, shiny bead of molten lead metal collects at the cathode.

Step 3. Write the anode reaction. Br⁻ migrates to the positive electrode; since bromine exists as the diatomic molecule Br₂, two Br⁻ ions must combine, each losing one electron:

2Br−→Br2+2e−2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-

Observation: reddish-brown fumes of bromine vapour are given off at the anode.

Step 4. Check the electron balance. The cathode consumes 2 electrons; the anode releases 2 electrons. The transfer is equal, confirming the circuit is complete and both half-equations are correctly balanced. Answer: cathode Pb²⁺ + 2e⁻ → Pb (grey lead bead); anode 2Br⁻ → Br₂ + 2e⁻ (reddish-brown bromine fumes).

Competitive stretch

Question: Aqueous copper sulphate is electrolysed once with platinum electrodes and separately with copper electrodes. Write both anode reactions and state what happens to the blue colour of the solution in each case.

Step 1. With platinum (inert) electrodes, SO₄²⁻ is not discharged (it is a stable oxy-anion), so OH⁻ from water is discharged instead: 4OH−→O2+2H2O+4e−4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-, while the cathode still deposits Cu²⁺ + 2e⁻ → Cu. Copper leaves the solution steadily and nothing replaces it, so the blue colour fades as electrolysis proceeds.

Step 2. With copper (active) electrodes, the anode itself dissolves instead of any anion being discharged: Cu→Cu2++2e−\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-. The cathode reaction is unchanged. Answer: with platinum electrodes the blue colour fades over time; with copper electrodes the colour stays essentially unchanged, because copper dissolving at the anode exactly replaces the copper deposited at the cathode — the basis of electro-refining.

Common traps

  • Writing an unbalanced diatomic half-equation such as Br⁻ → Br₂ + e⁻, forgetting that two bromide ions and two electrons are needed for one Br₂ molecule.
  • Forgetting to check that electrons lost at the anode equal electrons gained at the cathode — this equality is what completes the circuit and confirms a half-equation pair is correct.
  • Writing SO₄²⁻ as discharged at the anode in aqueous copper sulphate — sulphate is a stable oxy-anion and is not discharged; OH⁻ (or the electrode itself, if active) reacts instead.
  • Forgetting that electrode material changes the anode reaction entirely: an inert electrode forces an ion to be discharged, while an active electrode like copper reacts itself and no anion equation is needed at all.

Ask the tutor: Want to practise balancing electrode reactions for a new electrolyte, or need the electron-counting steps walked through one more time?

Why does bromide need a coefficient of 2 but lead does not, when both are simple ions

Because bromine is discharged as the diatomic molecule Br₂, so one Br₂ needs two bromine atoms and therefore two Br⁻ ions on the left — the atom count forces the doubling before you even look at charge. Lead, by contrast, is discharged directly to the neutral metal atom Pb, and one Pb²⁺ ion already gives exactly one Pb atom, so only the two electrons need adding for charge balance. The rule of thumb: check whether the neutral product is a lone atom (metals, hydrogen written per atom) or a diatomic molecule (Cl₂, Br₂, O₂) — diatomic products almost always need doubled ion coefficients.

Why is sulphate never discharged in aqueous copper sulphate electrolysis, with either type of electrode

Sulphate, SO₄²⁻, is a stable polyatomic oxy-anion that is very difficult to oxidise under these conditions — it simply is not one of the ions that gets picked at the anode. With inert electrodes, OH⁻ (from water) is discharged instead, releasing oxygen; with a copper anode, the electrode metal itself dissolves, so no anion needs to be discharged at all. In both cases SO₄²⁻ remains in solution unchanged throughout electrolysis, which is exactly why it never appears in either anode half-equation you are asked to write at this level.

Do I need to memorise every electrode reaction separately for the exam

No — memorise the two templates, Mn++ne−→MM^{n+} + ne^- \rightarrow M at the cathode and Xn−→X+ne−X^{n-} \rightarrow X + ne^- at the anode (doubled for diatomic products), plus which ion selective discharge actually picks for the given electrolyte and electrode material. Once you know the discharged ion, writing the balanced half-equation is just applying the atom-then-charge check from the derivation above. The three standard cases here (molten salt, acidified water, copper sulphate with inert versus active electrodes) cover every combination ICSE Class 10 expects.

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