Electrochemistry: TCE Chemistry (Tasmania)

Syllabus dot point

“Describe galvanic and electrolytic cells and calculate cell potentials from standard electrode potentials”

10 min readUpdated 2 Jun 2026

Quick answer

Galvanic (voltaic) cells turn spontaneous redox reactions into electrical energy; electrolytic cells use electrical energy to drive non-spontaneous reactions. Oxidation always occurs at the anode and reduction at the cathode. Standard electrode potentials let you calculate the cell voltage as $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ and predict whether a reaction is spontaneous (positive $E^\circ_{cell}$ ).

What this dot point is asking

TASC expects you to describe both cell types, label the electrodes and their signs, explain the salt bridge, and use the standard electrode potential table to calculate cell EMF and predict the spontaneous direction.

The two cell types

A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy. An electrolytic cell does the opposite, using an external supply to force a non-spontaneous redox reaction. Both rely on separating oxidation and reduction so electrons travel through an external circuit.

In any cell, oxidation occurs at the anode and reduction at the cathode (a memory aid: anode and oxidation both begin with vowels). The signs differ by cell type: in a galvanic cell the anode is negative and the cathode positive; in an electrolytic cell the anode is positive and the cathode negative, because the power supply imposes the polarity.

Galvanic cell

An electrochemical cell that generates electrical energy from a spontaneous redox reaction. Oxidation occurs at the negative anode and reduction at the positive cathode, with electrons flowing through the external circuit from anode to cathode.

Salt bridge and electron flow

A galvanic cell has two half-cells, each an electrode in an electrolyte. Electrons flow from the anode, through the external wire, to the cathode. A salt bridge allows ions to move so electrical neutrality is maintained; without it, charge would build up and the reaction would stop almost immediately. Anions migrate towards the anode and cations towards the cathode to balance the charge produced by electron flow.

Standard electrode potentials

The driving force is the cell potential, or electromotive force (EMF), in volts. Each half-reaction has a standard electrode potential $E^\circ$ , measured relative to the standard hydrogen electrode defined as $0\ \text{V}$ . Standard conditions are $25\ ^\circ\text{C}$ , $1\ \text{mol L}^{-1}$ concentrations and $100\ \text{kPa}$ for gases. Potentials are tabulated as reduction potentials, with the strongest oxidants at the top.

Standard cell potential

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

A positive

E^\circ_{cell}

means the reaction is spontaneous. The species higher in the reduction-potential table is reduced (acts as the oxidant); the species lower is oxidised.

To predict which reaction occurs, locate both half-reactions on the table. The half-reaction with the more positive reduction potential proceeds as a reduction at the cathode, and the other reverses to become an oxidation at the anode.

Calculating the EMF of a copper-zinc cell

Identify cathode and anode

A cell uses $\text{Cu}^{2+}/\text{Cu}$ ( $E^\circ = +0.34\ \text{V}$ ) and $\text{Zn}^{2+}/\text{Zn}$ ( $E^\circ = -0.76\ \text{V}$ ). The more positive potential is copper, so $\text{Cu}^{2+}$ is reduced at the cathode and zinc is oxidised at the anode.

Apply the cell potential formula

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

Interpret the result

The value is positive, confirming the reaction is spontaneous, so this is a working galvanic cell with zinc as the negative anode and copper as the positive cathode.

In electrolysis, an external voltage drives reactions at inert or active electrodes. When several species could react, the one easiest to reduce (highest reduction potential) is favoured at the cathode and the one easiest to oxidise at the anode, though concentration and overpotential can change the actual product. Electrolysis extracts reactive metals such as aluminium and is used for electroplating.

Reversing the electrode signs between cell types

Students often memorise that the anode is negative, which is true only for galvanic cells. In an electrolytic cell the anode is positive because the power supply sets the polarity. Always identify the cell type first, then assign signs. Oxidation is always at the anode regardless of cell type. A second slip is reversing a tabulated $E^\circ$ value but forgetting it is still used as-written in $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ .

When answering, state the half-equations, identify anode and cathode with the correct signs for the cell type, and use the standard potential table to justify the spontaneous direction.

Exam-style questions

Questions in the style of TASC 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.

Practice5 marks

A cell is built from a platinum rod in

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

\text{FeCl}_2(aq)

and

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

\text{FeCl}_3(aq)

, and a copper rod in

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

\text{CuCl}_2(aq)

, joined by wires, a voltmeter and a salt bridge. Given

E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}

and

E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}

, write the anode, cathode and overall equations and calculate the maximum voltage.

Show worked answer →

The half-cell with the more positive reduction potential is reduced (cathode); the other is oxidised (anode). So the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple is the cathode and copper is the anode.

Cathode (reduction): $\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)$ . Anode (oxidation): $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$ . (3 marks)

Balancing electrons (multiply the iron half by $2$ ): $2\text{Fe}^{3+}(aq) + \text{Cu}(s) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq)$ .

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 - 0.34 = +0.43\ \text{V}$ . The platinum electrode is inert and simply carries electrons. (2 marks)

Practice4 marks

A cell uses a copper electrode in

\text{Cu}^{2+}

solution and a chromium electrode in

\text{Cr}^{3+}

solution. Given

E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\ \text{V}

and

E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}

, write the half-equations and describe what is observed at each electrode as the cell operates.

Show worked answer →

Chromium has the more negative reduction potential, so it is the stronger reductant: chromium is oxidised at the anode and copper(II) is reduced at the cathode.

Anode (chromium, negative terminal): $\text{Cr}(s) \rightarrow \text{Cr}^{3+}(aq) + 3e^-$ . Observation: the chromium electrode loses mass and dissolves, and $[\text{Cr}^{3+}]$ rises. (2 marks)

Cathode (copper, positive terminal): $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ . Observation: copper is deposited so the electrode gains mass, and the blue colour of the $\text{Cu}^{2+}$ solution fades. Electrons flow through the external wire from chromium to copper. (2 marks)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC