Galvanic cells: WACE Year 12 Chemistry

Syllabus dot point

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

7 min readUpdated 2 Jun 2026

Quick answer

A galvanic (voltaic) cell turns a spontaneous redox reaction into electricity by physically separating the oxidation and reduction into two half-cells joined by a wire and a salt bridge. Oxidation occurs at the anode (negative terminal), reduction at the cathode (positive terminal). Electrons flow through the external wire from anode to cathode; ions move through the salt bridge to keep each half-cell neutral.

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.

Galvanic cell

An electrochemical cell in which a spontaneous redox reaction generates an electric current. It consists of two half-cells connected by an external conducting wire and a salt bridge.

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.

Anode oxidation, cathode reduction

In every electrochemical cell, oxidation happens at the anode and reduction at the cathode (a useful memory aid is that both vowels pair: anode/oxidation, and the consonants pair: cathode/reduction). What changes between galvanic and electrolytic cells is the sign of each electrode.

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.

Reversing electron flow or electrode signs

A frequent error is saying electrons flow through the salt bridge; electrons travel only through the external wire, while ions move through the salt bridge. Another is mislabelling the anode as positive; in a galvanic cell the anode is negative (the opposite of an electrolytic cell). Always anchor your answer on oxidation at the anode and reduction at the cathode.

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 2021-style6 marks

A 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 2023-style4 marks

Explain 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.

Practise this

Sources & how we know this

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