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Electrolysis

What happens at each electrode during electrolysis, and how much product does a current make?

IntermediateRedox & ElectrochemistryLast reviewed 5 October 2026
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What is it?

Electrolysis uses electricity to drive a redox reaction that would not happen on its own. It is the reverse of a galvanic cell: there, a spontaneous reaction makes electricity; here, electricity forces a non-spontaneous reaction.

An electrolytic cell has:

  • a DC power supply, which pumps electrons;
  • two electrodes, often inert (graphite or platinum);
  • an electrolyte: an ionic compound that is molten or dissolved, so that its ions can move.

The anode is connected to the positive terminal: negative ions (anions) move to it and are oxidized. The cathode is connected to the negative terminal: positive ions (cations) move to it and are reduced.

Key idea

In every cell, oxidation at the anode, reduction at the cathode. In electrolysis the anode is positive and the cathode negative, the opposite signs to a galvanic cell, because the power supply decides the charges.

Why does it matter?

  • Metals. All aluminium, and the most reactive metals (sodium, magnesium), are extracted by electrolysis because no chemical reducing agent is strong enough.
  • Chemicals. Electrolysis of brine makes chlorine, hydrogen and sodium hydroxide, used in plastics (PVC), bleach, soap and water treatment.
  • Purity and protection. Copper for electrical wiring is purified by electrolysis, and electroplating coats objects with chromium, nickel, silver or gold.

How does it work?

1. Molten electrolytes

A molten salt contains only its own ions, so the products are the elements:

  • molten NaCl: cathode NaX++eX−→Na\ce{Na+ + e- -> Na}; anode 2 ClX−→ClX2+2 eX−\ce{2Cl- -> Cl2 + 2e-}.

2. Aqueous electrolytes: water competes

In solution, water molecules can also be reduced or oxidized, so the more easily discharged species wins:

ElectrodeProductRule
Cathode (−)the metalif the metal is less reactive than hydrogen (Cu, Ag, Au)
Cathode (−)hydrogenif the metal is reactive (Na, K, Mg, Al): 2 HX2O+2 eX−→HX2+2 OHX−\ce{2H2O + 2e- -> H2 + 2OH-}
Anode (+)the halogenfor concentrated chloride, bromide or iodide
Anode (+)oxygenfor sulfate, nitrate (and dilute chloride): 2 HX2O→OX2+4 HX++4 eX−\ce{2H2O -> O2 + 4H+ + 4e-}

3. Three important industrial processes

  • Brine (concentrated NaCl solution): chlorine at the anode, hydrogen at the cathode, and sodium hydroxide left in solution.
  • Aluminium: aluminium oxide is dissolved in molten cryolite at about 950 °C. Cathode: AlX3++3 eX−→Al\ce{Al^3+ + 3e- -> Al}. Oxygen forms at the carbon anodes, which burn away to COX2\ce{CO2} and must be replaced.
  • Copper refining: an impure copper anode dissolves (Cu→CuX2++2 eX−\ce{Cu -> Cu^2+ + 2e-}), and pure copper deposits on the cathode; impurities such as silver and gold fall to the bottom as “anode sludge”.

4. Faraday’s law

The amount of product depends only on the charge passed:

Q=Itn(e−)=QFF=96 485 C/mol\begin{aligned} &Q = It \\[2pt] &n(e^-) = \frac{Q}{F} \\[2pt] &F = 96\,485\ \text{C/mol} \end{aligned}

Then use the half-equation to convert moles of electrons into moles of product: 1 mol Ag\ce{Ag} needs 1 mol e⁻, 1 mol Cu\ce{Cu} needs 2 mol e⁻, 1 mol Al\ce{Al} needs 3 mol e⁻, and 1 mol ClX2\ce{Cl2} releases 2 mol e⁻.

Think of it like this

A galvanic cell is a ball rolling downhill and turning a wheel on the way. Electrolysis is pushing the ball back up the hill with a motor: it only goes as long as you keep supplying energy, and the amount moved depends on how hard and how long you push (current × time).

More precisely

Which ion is discharged in solution depends on the electrode potentials, the concentrations and the electrode material. Concentrated chloride gives chlorine (not oxygen) even though oxygen has the less negative potential, because oxygen formation needs a large extra voltage (an overpotential) at most electrodes. The minimum voltage for electrolysis equals −E°cell-E°_\text{cell} of the reverse, spontaneous reaction, but real cells need more, to overcome resistance and overpotentials.

Visualise it

Electrolysis of concentrated brine. A DC supply is connected to two electrodes in salt solution. Electrons flow from the anode to the positive terminal and from the negative terminal to the cathode. At the anode (positive, oxidation), chloride ions are attracted and chlorine gas bubbles off: 2Cl− → Cl2 + 2e−. At the cathode (negative, reduction), sodium ions are attracted but water is reduced, giving hydrogen gas: 2H2O + 2e− → H2 + 2OH−. Na+ and OH− stay in solution as sodium hydroxide.
Brine gives three products: chlorine, hydrogen and sodium hydroxide.

Worked example

Worked example: Predicting the products

Question: Give the products at each electrode (inert electrodes) for: (a) molten PbBrX2\ce{PbBr2} (b) aqueous CuSOX4\ce{CuSO4} (c) dilute HX2SOX4\ce{H2SO4}.

  1. (a) Molten: cathode lead (PbX2++2 eX−→Pb\ce{Pb^2+ + 2e- -> Pb}), anode bromine (2 BrX−→BrX2+2 eX−\ce{2Br- -> Br2 + 2e-}).
  2. (b) Cathode copper (less reactive than hydrogen); anode oxygen (sulfate is not discharged). The blue colour fades as CuX2+\ce{Cu^2+} is used up, and the solution becomes acidic.
  3. (c) Cathode hydrogen, anode oxygen, in a 2 : 1 volume ratio: overall, water is split, 2 HX2O→2 HX2+OX2\ce{2H2O -> 2H2 + O2}.

Worked example: Aluminium from a cell

Question: What mass of aluminium is produced by a cell running at 1.00 × 10⁵ A for 24.0 h? (AlX3++3 eX−→Al\ce{Al^3+ + 3e- -> Al}; Al = 26.98 g/mol)

  1. t=24.0 h×3600 s/h=8.64×104 st = 24.0\ \text{h} \times 3600\ \text{s/h} = 8.64 \times 10^{4}\ \text{s}
  2. Q=It=(1.00×105 C/s)(8.64×104 s)=8.64×109 CQ = It = (1.00 \times 10^{5}\ \text{C/s})(8.64 \times 10^{4}\ \text{s}) = 8.64 \times 10^{9}\ \text{C}
  3. n(e−)=8.64×109 C96 485 C/mol=8.955×104 moln(e^-) = \dfrac{8.64 \times 10^{9}\ \text{C}}{96\,485\ \text{C/mol}} = 8.955 \times 10^{4}\ \text{mol}
  4. n(Al)=8.955×104 mol e−×1 mol Al3 mol e−=2.985×104 moln(\ce{Al}) = 8.955 \times 10^{4}\ \text{mol}\ e^- \times \dfrac{1\ \text{mol Al}}{3\ \text{mol}\ e^-} = 2.985 \times 10^{4}\ \text{mol}
  5. m=2.985×104 mol×26.98 g/mol=8.05×105 g=m = 2.985 \times 10^{4}\ \text{mol} \times 26.98\ \text{g/mol} = 8.05 \times 10^{5}\ \text{g} = 805 kg

Worked example: Volume of chlorine from brine

Question: A current of 2.00 A passes through brine for 1.00 h. What volume of ClX2\ce{Cl2} forms at 25.0 °C and 1.00 atm?

  1. Q=(2.00 C/s)(3600 s)=7200 CQ = (2.00\ \text{C/s})(3600\ \text{s}) = 7200\ \text{C}; n(e−)=7200 C96 485 C/mol=0.07462 moln(e^-) = \dfrac{7200\ \text{C}}{96\,485\ \text{C/mol}} = 0.07462\ \text{mol}

  2. 2 ClX−→ClX2+2 eX−\ce{2Cl- -> Cl2 + 2e-}: n(ClX2)=0.07462 mol2=0.03731 moln(\ce{Cl2}) = \dfrac{0.07462\ \text{mol}}{2} = 0.03731\ \text{mol}

  3. Ideal gas law:

    V=nRTP=(0.03731 mol)×(0.08206 L atmmol K)×(298.15 K)÷(1.00 atm)=0.913 L\begin{aligned} &V = \frac{nRT}{P} \\[4pt] &= (0.03731\ \text{mol}) \\[4pt] &\quad \times \left(0.08206\ \tfrac{\text{L atm}}{\text{mol K}}\right) \\[4pt] &\quad \times (298.15\ \text{K}) \\[4pt] &\quad \div (1.00\ \text{atm}) \\[4pt] &= \textbf{0.913 L} \end{aligned}

Common mistake

Common mistake: Using galvanic-cell signs

In electrolysis the anode is positive; in a galvanic cell it is negative. What never changes is that oxidation happens at the anode.

Common mistake: Expecting a reactive metal from its solution

Electrolysis of aqueous NaCl does not give sodium: water is reduced instead, giving hydrogen. Sodium is made only from molten sodium chloride.

Common mistake: Forgetting the electron ratio

After finding moles of electrons, divide by the number of electrons in the half-equation: 3 for AlX3+\ce{Al^3+}, 2 for CuX2+\ce{Cu^2+} and ClX2\ce{Cl2}. Skipping this step makes the answer two or three times too large.

Notation note

  • 1 A = 1 C/s; 1 faraday = 96 485 C, the charge on one mole of electrons.
  • “Inert” electrodes (graphite, platinum) conduct electricity but do not react.
  • “Discharged” means an ion gains or loses electrons at an electrode and becomes a neutral atom or molecule.

Remember this

Remember this

  • Electrolysis: electricity forces a non-spontaneous redox reaction. Anode (+): oxidation; cathode (−): reduction.
  • Molten salts give the elements. In water: cathode gives the metal (if below hydrogen) or H₂; anode gives the halogen (concentrated halide) or O₂.
  • Brine → Cl₂, H₂, NaOH. Aluminium from molten Al₂O₃ in cryolite. Copper refined with an impure copper anode.
  • Q=ItQ = It, n(e−)=Q/Fn(e^-) = Q/F, then the half-equation ratio.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is electrolysis?
    Answer

    Using electricity to drive a non-spontaneous redox reaction.

  2. Question
    Signs and reactions at the electrodes in electrolysis?
    Answer

    Anode (+): oxidation. Cathode (−): reduction.

  3. Question
    Products of electrolysing molten NaCl?
    Answer

    Sodium at the cathode, chlorine at the anode.

  4. Question
    When does a metal NOT form at the cathode in aqueous solution?
    Answer

    When the metal is more reactive than hydrogen (e.g. Na, K, Mg, Al): water is reduced to H₂ instead.

  5. Question
    What forms at the anode in aqueous solution?
    Answer

    The halogen from a concentrated halide; otherwise oxygen (e.g. with sulfate or nitrate).

  6. Question
    Products of electrolysing concentrated brine?
    Answer

    Chlorine (anode), hydrogen (cathode) and sodium hydroxide (left in solution).

  7. Question
    How is aluminium extracted?
    Answer

    Electrolysis of Al₂O₃ dissolved in molten cryolite (about 950 °C); Al³⁺ + 3e⁻ → Al; carbon anodes burn away.

  8. Question
    How is copper purified?
    Answer

    Impure copper anode dissolves; pure copper deposits on the cathode; impurities fall as anode sludge.

  9. Question
    Faraday's-law steps?
    Answer

    Q = It; n(e⁻) = Q ÷ F (F = 96 485 C/mol); use the half-equation ratio; then mass or volume.

  10. Question
    Moles of electrons per mole of Ag, Cu, Al and Cl₂?
    Answer

    Ag 1, Cu 2, Al 3, Cl₂ 2.

Quiz

7 questions

  1. Question 1EasyIn an electrolytic cell, the anode is:
    Show answer

    Answer: positive, where oxidation happens

    Oxidation always happens at the anode; in electrolysis it is connected to the positive terminal.

  2. Question 2MediumWhat forms at the cathode when aqueous KNO₃ is electrolysed (inert electrodes)?
    Show answer

    Answer: hydrogen

    Potassium is far more reactive than hydrogen, so water is reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻.

  3. Question 3MediumWhat forms at the anode when aqueous CuSO₄ is electrolysed (inert electrodes)?
    Show answer

    Answer: oxygen

    Sulfate is not discharged; water is oxidized: 2H₂O → O₂ + 4H⁺ + 4e⁻. Copper forms at the cathode.

  4. Question 4MediumWhy is sodium made from molten NaCl and not from salt solution?
    Show answer

    Answer: in solution, water is reduced instead of Na⁺

    Na⁺ is very hard to reduce; in water, hydrogen forms at the cathode instead.

  5. Question 5MediumA current of 1.00 A flows for 965 s. How many moles of electrons pass?
    Show answer

    Answer: 0.0100 mol

    Q = 1.00 C/s × 965 s = 965 C; n = 965 C ÷ 96 485 C/mol = 0.0100 mol.

  6. Question 6HardThe same charge deposits 0.0300 mol of Ag. How many moles of Cu would it deposit?
    Show answer

    Answer: 0.0150 mol

    0.0300 mol Ag needs 0.0300 mol e⁻; Cu²⁺ needs 2 e⁻ per atom, so 0.0300 ÷ 2 = 0.0150 mol Cu.

  7. Question 7EasyIn copper refining, the impure copper is:
    Show answer

    Answer: the anode

    The impure anode is oxidized and dissolves (Cu → Cu²⁺ + 2e⁻); pure copper is deposited on the cathode.

Notes and downloads

  • Worksheet

    Electrolysis Worksheet

    9 questions on electrode products, half-equations, industrial electrolysis, and Faraday's-law mass, time, gas-volume and energy calculations. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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