Redox and electrochemistry: WACE Year 12 Chemistry

Syllabus dot point

“Assign oxidation numbers, write and balance half-equations, and use the standard electrode potential series to predict and calculate cell potentials for galvanic and electrolytic cells”

9 min readUpdated 2 Jun 2026

Quick answer

Oxidation is loss of electrons (oxidation number increases) and reduction is gain of electrons (oxidation number decreases). In a galvanic cell a spontaneous redox reaction produces electricity, with oxidation at the anode and reduction at the cathode; in an electrolytic cell electricity drives a non-spontaneous reaction. The standard electrode potential series lets you predict which species reacts and calculate the cell potential $E^{\circ}_{cell}$ .

What this dot point is asking

Redox reactions involve the transfer of electrons. Oxidation is the loss of electrons; reduction is the gain of electrons (remember OIL RIG: Oxidation Is Loss, Reduction Is Gain). Every redox reaction couples one of each: the species oxidised is the reducing agent, and the species reduced is the oxidising agent.

Oxidation numbers

Oxidation numbers track electron distribution and reveal what is oxidised and reduced. The key rules are: elements in their standard state are 0; oxygen is usually $-2$ (peroxides $-1$ ); hydrogen is $+1$ with non-metals and $-1$ in metal hydrides; a monatomic ion equals its charge; and the sum of oxidation numbers equals the overall charge of the species. An increase in oxidation number is oxidation; a decrease is reduction.

Oxidation and reduction

Oxidation is the loss of electrons, shown by an increase in oxidation number. Reduction is the gain of electrons, shown by a decrease in oxidation number. The two always occur together.

Writing half-equations

A redox reaction splits into an oxidation half-equation and a reduction half-equation. To balance one (in acidic solution): balance the atoms being oxidised or reduced, balance oxygen with $\text{H}_2\text{O}$ , balance hydrogen with $\text{H}^+$ , then balance charge with electrons. For example the reduction of dichromate:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

To combine two half-equations, multiply each so the electrons cancel, then add them.

Standard electrode potentials

The standard electrode potential ( $E^{\circ}$ ) of a half-reaction measures its tendency to be reduced, relative to the standard hydrogen electrode ( $E^{\circ} = 0\ \text{V}$ ), measured at $25^{\circ}\text{C}$ , $100\ \text{kPa}$ and $1\ \text{mol L}^{-1}$ concentrations. The SCSA data booklet lists these as reduction half-equations. A more positive $E^{\circ}$ means a stronger oxidising agent (more readily reduced); a more negative $E^{\circ}$ means a stronger reducing agent (more readily oxidised). The species with the higher (more positive) $E^{\circ}$ is reduced; the other is oxidised.

Galvanic (voltaic) cells

A galvanic cell converts the energy of a spontaneous redox reaction into electrical energy. Two half-cells are connected by an external wire (electron path) and a salt bridge (ion path that keeps each half-cell neutral). Oxidation occurs at the anode (negative electrode in a galvanic cell) and reduction at the cathode (positive electrode); electrons flow through the wire from anode to cathode. The cell potential is

E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}

A positive $E^{\circ}_{cell}$ confirms the reaction is spontaneous.

Standard cell potential

E^{\circ}_{cell} = E^{\circ}_{\text{reduction (cathode)}} - E^{\circ}_{\text{oxidation half as a reduction (anode)}}

A positive value means the overall reaction is spontaneous as written.

Electrolytic cells

An electrolytic cell uses an external power supply to drive a non-spontaneous redox reaction (for example electrolysis of molten salts or aqueous solutions, electroplating, and electrorefining). Here the anode is positive and the cathode is negative, the reverse polarity of a galvanic cell, but oxidation still occurs at the anode and reduction at the cathode.

Anode and cathode

In both cell types, oxidation is at the anode and reduction is at the cathode. What changes is the sign: anode is negative in a galvanic cell but positive in an electrolytic cell.

Calculating a cell potential

A galvanic cell combines $\text{Zn}^{2+}/\text{Zn}$ ( $E^{\circ} = -0.76\ \text{V}$ ) and $\text{Cu}^{2+}/\text{Cu}$ ( $E^{\circ} = +0.34\ \text{V}$ ). Find $E^{\circ}_{cell}$ and write the overall equation.

Step 1: Identify cathode and anode

The species with the more positive standard electrode potential is reduced at the cathode. Copper has the more positive $E^{\circ}$ (+0.34 V), so $\text{Cu}^{2+}$ is reduced at the cathode and zinc is oxidised at the anode.

Step 2: Calculate the cell potential

Apply the formula cathode minus anode, using the reduction potentials as listed:

E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = (+0.34) - (-0.76) = +1.10\ \text{V}

The positive value confirms the reaction is spontaneous as written.

Step 3: Write the overall equation

Write the half-equations and add them so the electrons cancel. Anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ . Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ . Adding:

\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)

Final answer: $E^{\circ}_{cell} = +1.10\ \text{V}$

When answering redox questions in the WACE examination, read half-equations from the data booklet exactly as written (reductions), identify cathode and anode by comparing $E^{\circ}$ values, and state the electron and ion flow directions explicitly.

Subtracting potentials the wrong way

A common error is computing $E^{\circ}_{cell}$ as anode minus cathode, giving a negative value for a spontaneous cell. Always use cathode minus anode. Another frequent slip is multiplying $E^{\circ}$ values when scaling half-equations to balance electrons; $E^{\circ}$ is an intensive property and is never multiplied. A third is assuming the cathode is always negative; it is positive in a galvanic cell and negative in an electrolytic cell.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

A galvanic cell is constructed from a

\text{Sn}^{2+}/\text{Sn}

half-cell (

E^\circ = -0.14\ \text{V}

) and a

\text{Fe}^{3+}/\text{Fe}^{2+}

half-cell (

E^\circ = +0.77\ \text{V}

) with a platinum electrode. (a) Identify the oxidising agent. (b) Write the balanced overall ionic equation. (c) Calculate

E^\circ_{\text{cell}}

and predict whether the cell is spontaneous.

Show worked answer →

A 7 mark question rewards identifying the agents, balancing with electrons, and the cell potential.

(a) The $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple has the more positive $E^\circ$ , so $\text{Fe}^{3+}$ is reduced and is the oxidising agent.

(b) Cathode: $\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}$ (multiply by 2 to balance electrons). Anode: $\text{Sn} \rightarrow \text{Sn}^{2+} + 2\text{e}^-$ . Adding:

\text{Sn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Fe}^{2+}(aq).

(c)

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.77) - (-0.14) = +0.91\ \text{V}.

The positive value confirms the reaction is spontaneous as written. Note

E^\circ

is not multiplied when the half-equation is scaled.

Markers reward $\text{Fe}^{3+}$ as oxidising agent, the balanced overall equation, and $E^\circ_{\text{cell}} = +0.91\ \text{V}$ with a spontaneity statement.

WACE 2020-style5 marks

Explain the similarities and differences between a galvanic cell and an electrolytic cell. Your answer should refer to spontaneity, energy conversion, the location of oxidation, and electrode polarity.

Show worked answer →

A 5 mark answer needs the shared role plus three contrasts.

Similarity: In both cells oxidation occurs at the anode and reduction at the cathode, and electrons flow through the external circuit from anode to cathode.
Spontaneity and energy: A galvanic cell runs a spontaneous reaction ( $E^\circ_{\text{cell}} > 0$ ) and converts chemical energy into electrical energy. An electrolytic cell uses an external supply to drive a non-spontaneous reaction ( $E^\circ_{\text{cell}} < 0$ ), converting electrical energy into chemical energy.
Polarity: In a galvanic cell the anode is negative and the cathode positive. In an electrolytic cell the supply makes the anode positive and the cathode negative.

Markers reward the oxidation-at-anode similarity and the three contrasts (spontaneity, energy direction, electrode polarity).

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA