SCSA WACE 2021-style practice: A galvanic cell is built from a silver electrode in 1.0 mol L^-1\ AgNO_3 and a nickel electrode in 1.0 mol L^-1\ Ni(NO_3)_2. Standard reduction potentials: Ag^+/Ag = +0.80\ V; Ni^2+/Ni = -0.25\ V. (a) Identify the anode and cathode and write the half-equations. (b) Calculate the standard cell potential. (c) Write the cell using standard cell notation.

A 6 mark question rewards electrode assignment, the cell potential, and correct notation. **(a)** The half-cell with the more negative potential is oxidised, so nickel is the anode and silver the cathode. Anode: Ni(s) → Ni^2+(aq) + 2e^-. Cathode (note the electrons must balance, so multiply by 2): 2Ag^+(aq) + 2e^- → 2Ag(s). **(b)** $E^_cell = E^_cathode - E^_anode = (+0.80) - (-0.25) = +1.05\ V.$ The positive value confirms the reaction is spontaneous. **(c)** Anode on the left, cathode on the right: $Ni(s) Ni^2+(aq) Ag^+(aq) Ag(s).$ Markers reward correct anode/cathode with balanced electrons, E^_cell = +1.05\ V, and the notation with the salt bridge as the double line.

§-Syllabus dot point

WAChemistrySyllabus dot point

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

Describe the structure and operation of galvanic (voltaic) cells, including electrode reactions, electron and ion flow, and cell notation

A focused answer to the WACE Year 12 Chemistry dot point on galvanic cells, the roles of anode and cathode, electron and ion movement, the salt bridge, and cell notation, with a worked example and common exam mistakes.

Generated by Claude Opus 4.87 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

A galvanic cell (also called a voltaic cell) converts chemical energy from a spontaneous redox reaction into electrical energy. The trick is to separate the two halves of the reaction so that the electrons must travel through an external circuit, where they can do useful work.

The two electrodes

Each half-cell contains an electrode in a solution of its ions.

Anode: the electrode where oxidation occurs. In a galvanic cell it is the negative terminal because it releases electrons.
Cathode: the electrode where reduction occurs. In a galvanic cell it is the positive terminal because it draws electrons in.

Electron and ion movement

Electrons released by oxidation at the anode flow through the external wire to the cathode, where they are used in reduction. This electron flow is the current the cell delivers.

The salt bridge completes the circuit internally and maintains electrical neutrality. As oxidation produces positive ions in the anode compartment, negative ions migrate from the salt bridge into it; as reduction removes positive ions in the cathode compartment, positive ions flow in. Without the salt bridge, charge would build up and the cell would stop almost immediately.

Cell notation

A shorthand summarises the cell. For the zinc-copper cell:

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

The anode (oxidation) is written on the left, the cathode (reduction) on the right. A single vertical line is a phase boundary; the double line is the salt bridge.

Describing a zinc-copper cell

Question

A galvanic cell uses a zinc electrode in zinc sulfate solution and a copper electrode in copper(II) sulfate solution. Write the electrode half-equations, the overall equation, and state the direction of electron flow.

Solution

Zinc is more easily oxidised (more negative electrode potential), so it is the anode:

\text{Anode (oxidation): } \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2\text{e}^-

Copper ions are reduced at the cathode:

\text{Cathode (reduction): } \text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)

Overall: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$ .

Electrons flow through the external wire from the zinc anode (negative) to the copper cathode (positive). In the salt bridge, anions move towards the zinc compartment and cations towards the copper compartment.

Observations and the depleting cell

As the cell runs, the zinc anode loses mass (it dissolves) and the copper cathode gains mass (copper is deposited). The blue colour of the copper(II) solution fades as $\text{Cu}^{2+}$ is consumed. The cell stops when a reactant runs out or the electrode potentials equalise.

Why this matters

Galvanic cells are the basis of all batteries, from a simple cell to lithium-ion cells. Understanding electrode assignment, electron flow and the role of the salt bridge is essential for the redox section and for contrasting galvanic cells with electrolytic cells, which run the same chemistry in reverse.

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 20216 marksA galvanic cell is built from a silver electrode in

\text{1.0 mol L}^{-1}\ \text{AgNO}_3

and a nickel electrode in

\text{1.0 mol L}^{-1}\ \text{Ni(NO}_3)_2

. Standard reduction potentials:

\text{Ag}^+/\text{Ag} = +0.80\ \text{V}

;

\text{Ni}^{2+}/\text{Ni} = -0.25\ \text{V}

. (a) Identify the anode and cathode and write the half-equations. (b) Calculate the standard cell potential. (c) Write the cell using standard cell notation.

Show worked answer →

A 6 mark question rewards electrode assignment, the cell potential, and correct notation.

(a) The half-cell with the more negative potential is oxidised, so nickel is the anode and silver the cathode.
Anode: $\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2\text{e}^-$ .
Cathode (note the electrons must balance, so multiply by 2): $2\text{Ag}^+(aq) + 2\text{e}^- \rightarrow 2\text{Ag}(s)$ .

(b)

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.80) - (-0.25) = +1.05\ \text{V}.

The positive value confirms the reaction is spontaneous.

(c) Anode on the left, cathode on the right:

\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag}(s).

Markers reward correct anode/cathode with balanced electrons, $E^\circ_{\text{cell}} = +1.05\ \text{V}$ , and the notation with the salt bridge as the double line.

WACE 20234 marksExplain the function of the salt bridge in a galvanic cell, and predict what would happen to the current if the salt bridge were removed. Refer to charge balance in your answer.

Show worked answer →

A 4 mark answer needs the neutrality function, the ion movement, and the consequence of removal.

The salt bridge completes the internal circuit and maintains electrical neutrality in each half-cell. Oxidation at the anode produces positive ions (for example $\text{Zn}^{2+}$ ), so anions from the salt bridge migrate into the anode compartment to balance the building positive charge. Reduction at the cathode removes positive ions from solution, so cations migrate from the salt bridge into the cathode compartment.

If the salt bridge is removed, charge cannot be balanced: positive charge accumulates in the anode compartment and negative charge in the cathode compartment. This opposing charge quickly halts the electron flow, so the current falls to zero almost immediately.

Markers reward the neutrality function, the direction of anion and cation movement, and the prediction that the current stops.