Galvanic cells and electrode potentials (SACE Stage 2 Chemistry)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

A galvanic (voltaic) cell converts the energy of a spontaneous redox reaction into electrical energy. Oxidation happens at the anode (negative electrode), reduction at the cathode (positive electrode); electrons flow through the external wire from anode to cathode, while a salt bridge maintains charge balance. The standard cell potential is $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ ; a positive value means the reaction is spontaneous.

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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
Exam-style questions

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

Anode, cathode and electron flow

A galvanic cell has two half-cells joined by an external wire and a salt bridge. At the anode (negative terminal) the more reactive metal is oxidised, releasing electrons. These electrons travel through the wire to the cathode (positive terminal), where reduction occurs. The cell harnesses a spontaneous redox reaction to push electrons through the circuit, doing electrical work.

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

Standard cell potential

E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}

Standard reduction potentials are measured against the standard hydrogen electrode (

E^\circ = 0.00\ \text{V}

) under standard conditions (

1\ \text{mol L}^{-1}

100\ \text{kPa}

25\ ^\circ\text{C}

). A more positive

E^\circ

means a stronger tendency to be reduced. Subtract the anode value from the cathode value; a positive result means a spontaneous cell.

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.

Adding potentials or flipping a half-cell sign

The most common SACE error is adding the two reduction potentials or changing the sign of a reduction potential when reversing a half-equation in the $E^\circ_\text{cell}$ formula. Use $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ directly with the reduction potentials as tabulated; do not flip signs. Another error is mislabelling the electrodes: the more negative potential is the anode (oxidised), and in a galvanic cell the anode is negative, the opposite of electrolysis.

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 questions

Questions in the style of SACE Board 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.

SACE 2022-style5 marks

A 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 2020-style4 marks

Using 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)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board