Redox and electrochemistry in QCE Chemistry, explained

Published 2026-05-26 · Updated 2026-05-26

Redox and electrochemistry close out QCE Chemistry Unit 3 (Equilibrium, Acids and Redox Reactions). It is the part of the unit students find most mechanical, which is good news: once the system clicks, it is reliable marks. The catch is that it has a lot of moving parts, oxidation numbers, half-equations, two kinds of cell, electrode potentials and Faraday's law, and they only make sense once you see how they connect.

This guide walks through that chain from the ground up: what oxidation and reduction actually are, how to balance redox half-equations, how galvanic and electrolytic cells differ, how to read electrode potentials from the data book, and how to do electrolysis calculations.

Oxidation and reduction: the core idea

Redox is electron transfer. In any redox reaction, one species loses electrons and another gains them, at the same time.

• Oxidation is the loss of electrons (the oxidation number increases).
• Reduction is the gain of electrons (the oxidation number decreases).

Two memory hooks: OIL RIG (Oxidation Is Loss, Reduction Is Gain) and LEO the lion says GER (Lose Electrons Oxidation, Gain Electrons Reduction).

The agents are where students slip:

• The oxidising agent is the species that causes oxidation, so it is itself reduced.
• The reducing agent is the species that causes reduction, so it is itself oxidised.

The agent is always the opposite of what it does.

Oxidation numbers

Oxidation numbers are the bookkeeping that lets you spot redox and balance it. The rules you apply in order:

• Free elements (O₂, Na, Cl₂) are 0.
• Simple ions equal their charge (Na⁺ is +1, S²⁻ is −2).
• Oxygen is usually −2 (but −1 in peroxides like H₂O₂).
• Hydrogen is usually +1 (but −1 in metal hydrides like NaH).
• The oxidation numbers in a neutral compound sum to 0; in a polyatomic ion they sum to the ion's charge.

If an element's oxidation number changes across a reaction, it is a redox reaction. The element going up is oxidised; the element going down is reduced.

Balancing half-equations in acidic conditions

QCE asks you to balance redox equations in acidic solution. Split the reaction into two half-equations and balance each with this sequence:

1. Balance the main atoms (everything except O and H).
2. Balance oxygen by adding H₂O.
3. Balance hydrogen by adding H⁺.
4. Balance charge by adding electrons (e⁻).

Then scale the two half-equations so the electrons cancel, and add them. The electrons must cancel exactly; if they do not, you have an error.

Worked outline for MnO₄⁻ being reduced in acid:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxygen balanced with 4 H₂O, hydrogen with 8 H⁺, and the 5 electrons make the charge balance (left: −1 + 8 − 5 = +2; right: +2). That last charge check is the step most worth doing twice.

Galvanic cells: turning redox into electricity

A galvanic (voltaic) cell uses a spontaneous redox reaction to produce electrical energy. The two half-reactions are physically separated so the electrons must travel through an external wire, and that flow is the current.

Two rules hold in every cell, galvanic or electrolytic:

• Oxidation always happens at the anode.
• Reduction always happens at the cathode.

The mnemonic AN OX and RED CAT (ANode OXidation, REDuction CAThode) holds for both cell types. What changes is the polarity.

In a galvanic cell:

• The anode is negative (electrons are pushed out from oxidation).
• The cathode is positive (electrons arrive and reduction happens).
• Electrons flow anode → cathode through the wire.
• The salt bridge completes the circuit and keeps each half-cell electrically neutral, with anions moving toward the anode and cations toward the cathode.

Cell diagram notation records this compactly:

anode | anode solution || cathode solution | cathode

A single bar is a phase boundary; the double bar is the salt bridge. By convention the anode (oxidation) is written on the left.

Standard electrode potentials

Every half-reaction has a standard electrode potential E°, measured in volts against the standard hydrogen electrode (defined as exactly 0.00 V). The QCAA data book lists these as reduction potentials. A more positive E° means a stronger tendency to be reduced (a stronger oxidising agent).

To find the cell voltage:

E°cell = E°cathode − E°anode

using the reduction potentials straight from the table. A positive E°cell means the reaction is spontaneous (a galvanic cell). The half-reaction with the more positive E° is the one that runs as reduction (the cathode); the other runs in reverse as oxidation (the anode). Knowing where each value sits in the data book is a real, transferable exam skill, since the same book is with you in the IA1 Data Test and the external exam.

Electrolytic cells: forcing the reaction backwards

An electrolytic cell does the opposite of a galvanic cell. It uses an external power supply to drive a non-spontaneous redox reaction (a negative E°cell). The same AN OX / RED CAT rule applies, but the polarity flips:

Molten salts electrolyse simply: the cation is reduced at the cathode and the anion is oxidised at the anode. Aqueous solutions are trickier because water itself can be oxidised or reduced, so it competes with the dissolved ions. Predicting the products means comparing the electrode potentials of every species that could react at each electrode and picking the one that occurs more readily, while keeping concentration effects in mind.

Faraday's law and electrolysis calculations

The amount of substance produced in electrolysis is proportional to the charge passed. The chain of formulas:

• Charge: Q = It (charge in coulombs = current in amps × time in seconds).
• Moles of electrons: n(e⁻) = Q / F, where the Faraday constant F = 96 500 C mol⁻¹.
• Then use the half-equation to convert moles of electrons into moles, mass or volume of product.

The half-equation is the bridge. Depositing one mole of Cu from Cu²⁺ + 2e⁻ → Cu needs two moles of electrons; one mole of Ag from Ag⁺ + e⁻ → Ag needs only one. Reading the electron ratio off the half-equation is the step that most often goes wrong.

Common mistakes that cost marks

1. Confusing the agent with the species oxidised. The reducing agent is the one that is oxidised, and vice versa.
2. Getting anode polarity wrong. Oxidation is always at the anode, but the anode is negative in a galvanic cell and positive in an electrolytic cell.
3. Forgetting to balance charge with electrons when writing half-equations.
4. Reversing the E°cell formula. It is cathode minus anode, both as reduction potentials.
5. Ignoring water in aqueous electrolysis. Water competes with the dissolved ions at both electrodes.
6. Skipping the half-equation in Faraday's law problems, which loses the electron ratio.

How to prepare

Electrochemistry rewards a fixed routine. For half-equations, always run the same five steps and finish with the charge check. For cells, always state oxidation/reduction, anode/cathode, polarity and electron direction in that order. For calculations, write the half-equation first, then Q = It, then convert. Repetition turns these into automatic marks.

The bottleneck is usually a small consistency slip you cannot spot in your own work. Avocado is an AI-powered Chemistry tutor built specifically for the QCE syllabus, so you can drill balancing half-equations, build galvanic and electrolytic cell diagrams, and work electrolysis calculations with specific feedback on exactly where a sign, polarity or electron ratio slipped.

Frequently asked questions

What is the difference between oxidation and reduction? Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease. They always happen together.

Where do oxidation and reduction happen in a cell? Oxidation is always at the anode and reduction is always at the cathode, in both galvanic and electrolytic cells (AN OX, RED CAT).

Why does the anode change sign between cell types? Oxidation still occurs at the anode in both, but a galvanic cell produces electrons there (negative), while an electrolytic cell has the power supply pull electrons away (positive).

How do I calculate cell potential? E°cell = E°cathode − E°anode, using the standard reduction potentials from the data book. A positive value means the reaction is spontaneous.

What is the Faraday constant? 96 500 coulombs per mole of electrons. It links the charge passed (Q = It) to the moles of electrons, which you then convert to product using the half-equation.

Content aligned to the QCAA Chemistry General Senior Syllabus, Unit 3 (Equilibrium, Acids and Redox Reactions). Always confirm current syllabus detail with your teacher and the QCAA website.