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How is electron transfer used to generate electricity in galvanic cells and to drive reactions in electrolytic cells?

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

A focused answer to the WACE Year 12 Chemistry dot point on oxidation numbers, half-equations, galvanic and electrolytic cells and standard electrode potentials, with a worked cell-potential calculation and common mistakes.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksA 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 20205 marksExplain 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).