SACE Board SACE 2022-style practice: A galvanic cell is built from a zinc electrode in Zn^2+ solution and a copper electrode in Cu^2+ solution. Standard reduction potentials: Zn^2+ + 2e^- → Zn, E^ = -0.76\ V; Cu^2+ + 2e^- → Cu, E^ = +0.34\ V. (a) Identify the anode and cathode. (b) Calculate the standard cell potential. (c) Write the overall cell reaction.

(a) The more negative potential is oxidised: zinc is the **anode** (oxidation), copper is the **cathode** (reduction). (1 mark) (b) E^_cell = E^_cathode - E^_anode = (+0.34) - (-0.76) = +1.10\ V. (2 marks) (c) Zn + Cu^2+ → Zn^2+ + Cu (zinc is oxidised, copper(II) is reduced). (2 marks)

SACE Board SACE 2020-style practice: Using the standard reduction potentials Ag^+ + e^- → Ag, E^ = +0.80\ V and Ni^2+ + 2e^- → Ni, E^ = -0.25\ V: (a) calculate the cell potential for a nickel-silver cell; (b) state whether the reaction Ni^2+ + 2Ag → Ni + 2Ag^+ is spontaneous and justify your answer.

(a) Silver has the higher (more positive) potential, so it is the cathode and nickel the anode: E^_cell = (+0.80) - (-0.25) = +1.05\ V for the spontaneous nickel-silver cell. (2 marks) (b) The reaction as written (Ni^2+ reduced, Ag oxidised) is the **reverse** of the spontaneous cell, so its potential is -1.05\ V. A negative E^_cell means the reaction is **non-spontaneous** as written. (2 marks)

§-Syllabus dot point

SAChemistrySyllabus dot point

How does a galvanic cell convert a redox reaction into electrical energy?

Describe the operation of galvanic cells and use standard electrode potentials to calculate cell potential and predict spontaneity.

The structure and operation of galvanic cells, the role of the salt bridge, using standard electrode potentials to calculate cell EMF and predict spontaneity, and worked SACE-style cell-potential calculations.

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
Lead worked calculation
How a galvanic cell works
The salt bridge
Standard electrode potentials
Predicting spontaneity
Why it matters for managing resources

What this dot point is asking

SACE expects you to describe cell operation, identify electrodes and electron flow, calculate $E^\circ_\text{cell}$ from standard reduction potentials, and predict spontaneity.

Lead worked calculation

Calculating cell potential and spontaneity

A cell pairs magnesium and lead. Standard reduction potentials: $\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}$ , $E^\circ = -2.37\ \text{V}$ ; $\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$ , $E^\circ = -0.13\ \text{V}$ . Find the anode, cathode, cell potential and overall reaction.

Decide which is oxidised

The half-reaction with the more negative potential proceeds as oxidation. Magnesium ( $-2.37\ \text{V}$ ) is more negative than lead ( $-0.13\ \text{V}$ ), so magnesium is the anode (oxidised) and lead the cathode (reduced).

Calculate the cell potential

E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = (-0.13) - (-2.37) = +2.24\ \text{V}

Write the overall reaction

\text{Mg} + \text{Pb}^{2+} \rightarrow \text{Mg}^{2+} + \text{Pb}

The positive $E^\circ_\text{cell}$ confirms the reaction is spontaneous, so this cell will deliver current.

How a galvanic cell works

A memory aid: at the anode, oxidation (AN OX); at the cathode, reduction (RED CAT). In a galvanic cell the anode is negative and the cathode positive (the reverse of electrolysis).

The salt bridge

As the cell runs, the anode solution gains positive ions (metal dissolving) and the cathode solution loses them (metal depositing), so charge would build up and stop the reaction. The salt bridge (a tube of inert electrolyte such as $\text{KNO}_3$ ) lets ions migrate to neutralise this build-up: anions move toward the anode and cations toward the cathode. Without it, no sustained current flows.

Standard electrode potentials

Predicting spontaneity

The sign of $E^\circ_\text{cell}$ tells you whether a redox reaction proceeds on its own:

$E^\circ_\text{cell} > 0$ : the reaction is spontaneous (the cell delivers energy; a galvanic cell).
$E^\circ_\text{cell} < 0$ : the reaction is non-spontaneous as written (it would need an external energy source, an electrolytic cell).

To check a specific reaction, identify which species is reduced and which oxidised, then compute $E^\circ_\text{cell}$ . If a proposed reaction is the reverse of the spontaneous direction, its potential is simply the negative.

Why it matters for managing resources

Galvanic cells are the basis of every battery, converting chemical energy into portable electrical energy. Electrode potentials also predict which metals corrode and which protect others (sacrificial anodes), directly informing how we manage and conserve metal resources.

Exam-style practice questions

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

SACE 20225 marksA galvanic cell is built from a zinc electrode in

\text{Zn}^{2+}

solution and a copper electrode in

\text{Cu}^{2+}

solution. Standard reduction potentials:

\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}

E^\circ = -0.76\ \text{V}

;

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

E^\circ = +0.34\ \text{V}

. (a) Identify the anode and cathode. (b) Calculate the standard cell potential. (c) Write the overall cell reaction.

Show worked answer →

(a) The more negative potential is oxidised: zinc is the anode (oxidation), copper is the cathode (reduction). (1 mark)

(b) $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = (+0.34) - (-0.76) = +1.10\ \text{V}$ . (2 marks)

(c) $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ (zinc is oxidised, copper(II) is reduced). (2 marks)

SACE 20204 marksUsing the standard reduction potentials

\text{Ag}^+ + e^- \rightarrow \text{Ag}

E^\circ = +0.80\ \text{V}

and

\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}

E^\circ = -0.25\ \text{V}

: (a) calculate the cell potential for a nickel-silver cell; (b) state whether the reaction

\text{Ni}^{2+} + 2\text{Ag} \rightarrow \text{Ni} + 2\text{Ag}^+

is spontaneous and justify your answer.

Show worked answer →

(a) Silver has the higher (more positive) potential, so it is the cathode and nickel the anode: $E^\circ_\text{cell} = (+0.80) - (-0.25) = +1.05\ \text{V}$ for the spontaneous nickel-silver cell. (2 marks)

(b) The reaction as written ( $\text{Ni}^{2+}$ reduced, $\text{Ag}$ oxidised) is the reverse of the spontaneous cell, so its potential is $-1.05\ \text{V}$ . A negative $E^\circ_\text{cell}$ means the reaction is non-spontaneous as written. (2 marks)