What are 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^, measured relative to the standard hydrogen electrode defined as 0\ V. Standard conditions are 25\ ^C, 1\ mol L^-1 concentrations and 100\ kPa for gases. Potentials are tabulated as reduction potentials, with the strongest oxidants at the top.

TASC TCE 2022-style practice: A cell is built from a platinum rod in 1.00\ mol L^-1 FeCl_2(aq) and 1.00\ mol L^-1 FeCl_3(aq), and a copper rod in 1.00\ mol L^-1 CuCl_2(aq), joined by wires, a voltmeter and a salt bridge. Given E^(Fe^3+/Fe^2+) = +0.77\ V and E^(Cu^2+/Cu) = +0.34\ V, write the anode, cathode and overall equations and calculate the maximum voltage.

The half-cell with the more positive reduction potential is reduced (cathode); the other is oxidised (anode). So the Fe^3+/Fe^2+ couple is the cathode and copper is the anode. Cathode (reduction): Fe^3+(aq) + e^- → Fe^2+(aq). Anode (oxidation): Cu(s) → Cu^2+(aq) + 2e^-. (3 marks) Balancing electrons (multiply the iron half by 2): 2Fe^3+(aq) + Cu(s) → 2Fe^2+(aq) + Cu^2+(aq). E^_cell = E^_cathode - E^_anode = 0.77 - 0.34 = +0.43\ V. The platinum electrode is inert and simply carries electrons. (2 marks)

TASC TCE 2023-style practice: A cell uses a copper electrode in Cu^2+ solution and a chromium electrode in Cr^3+ solution. Given E^(Cr^3+/Cr) = -0.74\ V and E^(Cu^2+/Cu) = +0.34\ V, write the half-equations and describe what is observed at each electrode as the cell operates.

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): Cr(s) → Cr^3+(aq) + 3e^-. Observation: the chromium electrode loses mass and dissolves, and [Cr^3+] rises. (2 marks) Cathode (copper, positive terminal): Cu^2+(aq) + 2e^- → Cu(s). Observation: copper is deposited so the electrode gains mass, and the blue colour of the Cu^2+ solution fades. Electrons flow through the external wire from chromium to copper. (2 marks)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we harness electron transfer to produce or use electrical energy?

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

Galvanic and electrolytic cells, electrodes and salt bridges, the standard electrode potential table, calculating cell EMF, and predicting spontaneity, with worked TASC-style examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

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.

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.

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.

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 practice questions

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

TCE 20225 marksA 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)

TCE 20234 marksA 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)