How do I decide which ions are discharged in electrolysis?

Look at the anode first: any metal other than gold or platinum (and not graphite) is oxidised itself. At an inert anode the anions discharge in the order S²⁻ > I⁻ > Br⁻ > Cl⁻ > OH⁻ > oxo-anions such as SO₄²⁻ and NO₃⁻ > F⁻. The cathode material does not react; cations are reduced in the order Ag⁺ > Fe³⁺ > Cu²⁺ > H⁺(acid) > Pb²⁺ > Sn²⁺ > Fe²⁺ > Zn²⁺ > H⁺(water) > Al³⁺ > Mg²⁺ > Na⁺ > Ca²⁺ > K⁺. Ions from Al³⁺ onwards are never reduced from water, which is why Na, Mg and Al are made from molten compounds. When H⁺ or OH⁻ comes from water, write H₂O in the half-equation.

How do you write fuel-cell half-equations?

The fuel is oxidised at the negative electrode and O₂ reduced at the positive one. Decide the products (carbon ends up as CO₃²⁻ in alkali, CO₂ otherwise), count electrons (CxHyOz loses 4x + y − 2z, so 8 for CH₄ and 6 for CH₃OH), balance charge with the electrolyte’s ion (H⁺ in acid, OH⁻ in alkali, CO₃²⁻ in molten carbonate, O²⁻ in solid oxide), then balance H and O with water. Alkaline methane: CH₄ + 10OH⁻ → CO₃²⁻ + 7H₂O + 8e⁻ and O₂ + 2H₂O + 4e⁻ → 4OH⁻, overall CH₄ + 2O₂ + 2OH⁻ → CO₃²⁻ + 3H₂O.

What are the lead–acid battery reactions on discharge and charge?

On discharge the negative plate reacts Pb + SO₄²⁻ → PbSO₄ + 2e⁻ and the positive plate PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O; overall Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O, so both plates gain mass and the acid gets weaker. On charging the reactions reverse: the former negative plate is the cathode (PbSO₄ + 2e⁻ → Pb + SO₄²⁻) and the former positive plate is the anode (PbSO₄ + 2H₂O → PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻).

How do you restore a solution after electrolysis?

Add back exactly what left the solution, combined into one compound. Electrolysing brine releases H₂ and Cl₂, so bubble in HCl gas (not hydrochloric acid, which adds water). Electrolysing CuSO₄ removes Cu and O₂, so add CuO or CuCO₃ — not Cu(OH)₂. For AgNO₃ add Ag₂O; for dilute H₂SO₄, NaOH or Na₂SO₄ only water was electrolysed, so add water. This does not apply when an active anode dissolves.

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🔋Electrode Half-Equation Writer - Electrolysis, Batteries, Fuel Cells & Corrosion

This tool writes the half-equations that school and first-year chemistry ask for, and checks every one of them for atom and charge balance. Electrolysis mode lets you pick the electrolyte (saturated brine, CuSO₄, CuCl₂, dilute H₂SO₄ or HCl, NaOH, Na₂SO₄, AgNO₃, KI, or molten NaCl, MgCl₂ and Al₂O₃ in cryolite) and the anode (inert graphite/platinum, or active Cu, Fe, Ag, impure copper). It applies the discharge order — an active anode first, then S²⁻ > I⁻ > Br⁻ > Cl⁻ > OH⁻ > oxo-anions; at the cathode Ag⁺ > Fe³⁺ > Cu²⁺ > H⁺(acid) > … > Zn²⁺ > H⁺(water) > Al³⁺ — and gives the anode and cathode equations, the ionic and full equations, the type of electrolysis, the pH change in the solution and near each electrode, what to add to restore the electrolyte, observations, and product masses and gas volumes for any amount of charge, with a cell diagram showing electron flow and ion migration. For example, electrolysis of brine with inert electrodes: anode 2Cl⁻ → Cl₂ + 2e⁻, cathode 2H₂O + 2e⁻ → H₂ + 2OH⁻, overall 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂; for CuSO₄, 2CuSO₄ + 2H₂O → 2Cu + O₂ + 2H₂SO₄, restored by adding CuO. Battery mode covers the Zn–Cu cell (acid and salt bridge), Mg–Al in NaOH and Cu–Al in concentrated HNO₃, alkaline and Leclanché cells, the lead–acid battery on discharge and charge (Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O), Ni–Cd, Ni–MH, Li-ion, silver oxide and aluminium–air cells. The fuel-cell writer takes any CxHyOz (or N-containing) fuel in acidic, alkaline, molten-carbonate or solid-oxide electrolyte and shows the steps — e.g. alkaline methane: CH₄ + 10OH⁻ → CO₃²⁻ + 7H₂O + 8e⁻; acidic methanol: CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻. A corrosion section compares hydrogen-evolution and oxygen-absorption rusting and judges sacrificial-anode and impressed-current protection.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Electrode reactions (cells & electrolysis)

1Choose Electrolysis, Batteries & cells, Fuel-cell equations or Corrosion, or click an example such as chlor-alkali, copper plating or aluminium extraction
2Pick the electrolyte and electrode materials, a battery (with discharge/charge), or a fuel and electrolyte
3Read the cell diagram, half-equations, overall equation, pH change, restoring reagent, observations and discharge-order analysis; copy any equation with proper sub- and superscripts

FAQ

How do I decide which ions are discharged in electrolysis?: Look at the anode first: any metal other than gold or platinum (and not graphite) is oxidised itself. At an inert anode the anions discharge in the order S²⁻ > I⁻ > Br⁻ > Cl⁻ > OH⁻ > oxo-anions such as SO₄²⁻ and NO₃⁻ > F⁻. The cathode material does not react; cations are reduced in the order Ag⁺ > Fe³⁺ > Cu²⁺ > H⁺(acid) > Pb²⁺ > Sn²⁺ > Fe²⁺ > Zn²⁺ > H⁺(water) > Al³⁺ > Mg²⁺ > Na⁺ > Ca²⁺ > K⁺. Ions from Al³⁺ onwards are never reduced from water, which is why Na, Mg and Al are made from molten compounds. When H⁺ or OH⁻ comes from water, write H₂O in the half-equation.
How do you write fuel-cell half-equations?: The fuel is oxidised at the negative electrode and O₂ reduced at the positive one. Decide the products (carbon ends up as CO₃²⁻ in alkali, CO₂ otherwise), count electrons (CxHyOz loses 4x + y − 2z, so 8 for CH₄ and 6 for CH₃OH), balance charge with the electrolyte’s ion (H⁺ in acid, OH⁻ in alkali, CO₃²⁻ in molten carbonate, O²⁻ in solid oxide), then balance H and O with water. Alkaline methane: CH₄ + 10OH⁻ → CO₃²⁻ + 7H₂O + 8e⁻ and O₂ + 2H₂O + 4e⁻ → 4OH⁻, overall CH₄ + 2O₂ + 2OH⁻ → CO₃²⁻ + 3H₂O.
What are the lead–acid battery reactions on discharge and charge?: On discharge the negative plate reacts Pb + SO₄²⁻ → PbSO₄ + 2e⁻ and the positive plate PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O; overall Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O, so both plates gain mass and the acid gets weaker. On charging the reactions reverse: the former negative plate is the cathode (PbSO₄ + 2e⁻ → Pb + SO₄²⁻) and the former positive plate is the anode (PbSO₄ + 2H₂O → PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻).
How do you restore a solution after electrolysis?: Add back exactly what left the solution, combined into one compound. Electrolysing brine releases H₂ and Cl₂, so bubble in HCl gas (not hydrochloric acid, which adds water). Electrolysing CuSO₄ removes Cu and O₂, so add CuO or CuCO₃ — not Cu(OH)₂. For AgNO₃ add Ag₂O; for dilute H₂SO₄, NaOH or Na₂SO₄ only water was electrolysed, so add water. This does not apply when an active anode dissolves.

Features

12 electrolytes with inert, active or impure-copper anodes, products chosen by discharge order
Ionic and full overall equations with electrolysis and cryolite conditions
Type of electrolysis, pH change overall and near each electrode, how to restore the solution, observations
Moles, mass and STP gas volume of every product for a given n(e⁻)
Cell diagrams with power supply, electron flow, ion migration, ion-exchange membrane and salt bridge
13 batteries and cells, with charging reactions for lead–acid, Ni–Cd, Ni–MH and Li-ion
Fuel-cell writer for any CxHyOzNw fuel in four electrolytes, step by step
Rusting, sacrificial anodes and impressed current; every equation auto-checked for balance

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