Lesson 5 · 30 min

Electrolysis: driving reactions with electricity

Apply a large enough voltage and a reaction runs against its spontaneous direction. What reacts at each electrode depends on potentials and overpotentials; how much forms follows Faraday’s laws.

In this lesson

Distinguish an electrolytic from a galvanic cell: electrode polarity, direction and energy conversion
Predict the products of molten-salt and aqueous electrolysis from electrode potentials plus overpotentials
Estimate the minimum decomposition voltage
Use Faraday’s laws to relate charge, mass or volume of product, time and current

In the mid-nineteenth century aluminium cost more than gold: Napoleon III served his most honoured guests with aluminium cutlery while lesser guests made do with gold. Aluminium is the third most abundant element in the crust, yet carbon cannot smelt it from alumina — it clings to oxygen too tightly. In 1886 Hall and Héroult independently used electric current to split molten alumina, and the price collapsed. Apply a large enough external voltage and a reaction runs against its spontaneous direction: that is electrolysis.

Predict first

Copper(II) sulfate solution is electrolysed with graphite electrodes. What forms at the anode?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

\ce{SO2}

gas

\ce{O2}

gas Copper Sulfur

The electrolytic cell: pushing electrons back

Electrodes are defined exactly as in a galvanic cell: oxidation at the anode, reduction at the cathode. What differs is the driving force: the power supply pulls electrons out of the anode and pushes them into the cathode, so in electrolysis the cathode is connected to the negative terminal and the anode to the positive one — the signs are the reverse of a galvanic cell.

When several substances could react, which goes first?

Cathode: the reduction with the highest electrode potential (easiest to reduce) happens. $\ce{Cu^2+}$ (+0.34 V) beats $\ce{H+}$ ; $\ce{Na+}$ (−2.71 V) is far behind water, so aqueous solutions give hydrogen, not sodium.
Anode: the oxidation with the lowest electrode potential (easiest to oxidise) happens. $\ce{Cl-}$ , water and a metal anode itself (such as copper) all compete.

The difference between the two actual electrode potentials is the minimum voltage needed. Equilibrium potentials are not the whole story: gas-evolving reactions usually need an extra overpotential, and that can change who wins.

E_\text{min} = \left(E_\text{anode} + \eta_\text{a}\right) - \left(E_\text{cathode} - \eta_\text{c}\right)

Minimum decomposition voltage: the actual oxidation potential at the anode minus the actual reduction potential at the cathode. For water the equilibrium difference is 1.229 V; with overpotentials it takes about 1.8 V in practice.

$E_\text{anode},\ E_\text{cathode}$: Nernst potentials of the two electrode reactions
$\eta_\text{a},\ \eta_\text{c}$: anodic and cathodic overpotentials (positive magnitudes)

What reacts at each electrode, and how much charge?

Electrolysis cell

Electrolyte Current · 1.00 A Time · 0.0 min Include overpotentials

Electrodes: cathode steel, anode RuO₂/TiO₂-coated anode

Charge Q: 0 C
Electrons n(e⁻): 0 mol
At the cathode: H₂ 0 mL
At the anode: Cl₂ 0 mL
Minimum decomposition voltage: 2.52 V (thermodynamic part 2.14 V)
Energy at the minimum voltage: 0 kJ

Hydrogen leaves the cathode and NaOH stays behind; chlorine leaves the anode. The ion-exchange membrane lets only Na⁺ through and keeps Cl₂ away from the NaOH.

Cathode (to −): reduction
Competing half-reactions	Equilibrium E	Overpotential η	Effective E	Wins
$\ce{2H2O + 2e- -> H2 + 2OH-}$	−0.828 V	0.35 V	−1.178 V
$\ce{Na+ + e- -> Na}$	−2.667 V	0.05 V	−2.717 V

Anode (to +): oxidation
Competing half-reactions	Equilibrium E	Overpotential η	Effective E	Wins
$\ce{2Cl- -> Cl2 + 2e-}$	1.315 V	0.03 V	1.345 V
$\ce{2H2O -> O2 + 4H+ + 4e-}$	1.052 V	0.60 V	1.652 V

At the cathode the half-reaction with the highest potential (easiest to reduce) wins; at the anode the one with the lowest potential (easiest to oxidise) wins.

Compare them in turn: molten NaCl gives sodium; water (Na₂SO₄) gives H₂ and O₂ in a 2 : 1 volume ratio; copper sulfate with a copper anode becomes electrorefining. In saturated brine, switch off “Include overpotentials”: thermodynamics alone predicts O₂ at the anode — the opposite of industrial fact.

Faraday’s laws: charge is stoichiometry

In 1833 Faraday found that the amount of an electrolysis product is proportional to the charge passed, and that the same charge liberates different substances in proportion to their “equivalents”. In today’s language: a mole of electrons carries $F = 96\,485$ C; depositing 1 mol of copper takes 2 mol of electrons, 1 mol of aluminium 3 mol, 1 mol of oxygen gas 4 mol. So current, time and yield are linked by one direct chain.

n(\ce{e-}) = \frac{It}{F},\qquad m = \frac{It}{zF}\,M

Faraday’s laws: the charge Q = It divided by F gives moles of electrons; divide again by the electrons z per formula unit of product.

$I$: current, A
$t$: time, s
$z$: electrons per formula unit of product
$M$: molar mass of the product, g/mol

Current × time → charge → electrons → product

Faraday’s-law calculator

Product

Find

$\ce{Cu^2+ + 2e- -> Cu}$

Current I / A Time t / min

Charge $Q = It = 2\ \text{A} \times 1800\ \text{s}$ 3600 C
Moles of electrons $n(e^-) = Q/F$ 0.03731 mol
Moles of product $n = n(e^-)/2$ 0.01866 mol
Mass $m = n \times 63.55\ \text{g/mol}$ 1.185 g

The chain shows every step. Switch to silver (z = 1) and aluminium (z = 3): the same charge gives moles that differ threefold. Then choose “Time needed” and run the problem backwards.

Worked example

Copper plating: how much in half an hour

Copper is plated at 2.00 A for 30.0 min (assume 100 % current efficiency). What mass of copper deposits on the cathode?

Concept check

In a membrane cell electrolysing saturated brine, what is the main cathode product?

Sodium metal Hydrogen gas Chlorine gas Oxygen gas

Try the calculation

How many minutes at 0.500 A does it take to deposit 1.00 g of silver?

Key takeaways

Electrolysis uses a power supply to drive a non-spontaneous redox reaction; oxidation is still at the anode and reduction at the cathode, but the anode is connected to the positive terminal.
The easiest reduction wins at the cathode and the easiest oxidation at the anode; in water, Na⁺ loses to water.
Overpotentials can change the winner: saturated brine gives Cl₂, not O₂, at the anode.
Faraday’s laws: n(e⁻) = It/F, and moles of product = n(e⁻)/z.

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Data and references

D. Pletcher, F. C. Walsh, Industrial Electrochemistry, 2nd ed., Chapman & Hall (1990), chs. 1–3 (electrode kinetics, chlor-alkali, electrometallurgy).
A. J. Bard, L. R. Faulkner, Electrochemical Methods, 2nd ed., Wiley (2001), ch. 3 (kinetics of electrode reactions and overpotential).
C. E. Housecroft, A. G. Sharpe, Inorganic Chemistry, 5th ed., Pearson (2018), chs. 11 and 13 (the chlor-alkali process; the Hall–Héroult process).
P. Vanýsek, “Electrochemical Series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010).