WJEC WJEC 2021-style practice: Use standard electrode potentials to determine whether acidified MnO_4^- (E = +1.51\ V) will oxidise Cl^- to Cl_2 (E for Cl_2/Cl^- = +1.36\ V). Explain your reasoning.

For a redox reaction to be feasible, the species being reduced must have the more positive electrode potential. MnO_4^- (+1.51 V) is more positive than the Cl_2/Cl^- couple (+1.36 V), so MnO_4^- is the stronger oxidising agent. Therefore MnO_4^- will oxidise Cl^- to Cl_2; the cell EMF is +1.51 - 1.36 = +0.15 V (positive, so feasible). Markers reward comparing the potentials, identifying the stronger oxidising agent, and the positive EMF showing feasibility.

WalesChemistrySyllabus dot point

How do electrode potentials predict redox behaviour?

Standard electrode potentials, the standard hydrogen electrode, electrochemical cells, calculating cell EMF, and predicting the feasibility of redox reactions.

A focused answer to WJEC A-Level Chemistry Unit 3, covering standard electrode potentials and the standard hydrogen electrode, electrochemical cells, calculating cell EMF, and using electrode potentials to predict whether a redox reaction is feasible.

Generated by Claude Opus 4.812 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
The answer
Examples in context
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What this dot point is asking

WJEC wants you to define standard electrode potentials relative to the standard hydrogen electrode, describe electrochemical cells, calculate cell EMF, and use electrode potentials to predict whether a redox reaction is feasible.

The answer

Standard electrode potentials

Cells and EMF

An electrochemical cell joins two half cells through a salt bridge. The electrode with the more positive potential is the positive terminal (reduction); the more negative is the negative terminal (oxidation).

Predicting feasibility

Will iron(III) oxidise iodide?

Given $E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V and $E^\ominus(\text{I}_2/\text{I}^-) = +0.54$ V, decide whether $\text{Fe}^{3+}$ oxidises $\text{I}^-$ .

step 1 Identify the stronger oxidising agent

The more positive potential ( $\text{Fe}^{3+}/\text{Fe}^{2+}$ at $+0.77$ ) is reduced, so $\text{Fe}^{3+}$ is the oxidising agent.

step 2 Assign terminals

Positive electrode $\text{Fe}^{3+}/\text{Fe}^{2+}$ , negative electrode $\text{I}_2/\text{I}^-$ .

step 3 Calculate EMF

$E^\ominus_{cell} = 0.77 - 0.54 = +0.23$ V.

step 4 Conclude

A positive EMF means the reaction is feasible: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ .

Setting up and measuring a half cell

A metal-ion half cell is a metal rod dipping into a $1$ mol dm $^{-3}$ solution of its ions; an ion-ion half cell (such as $\text{Fe}^{3+}/\text{Fe}^{2+}$ ) uses an inert platinum electrode in a solution containing both ions at $1$ mol dm $^{-3}$ . To measure a standard electrode potential, connect the half cell to the standard hydrogen electrode through a high-resistance voltmeter and a salt bridge, under standard conditions ( $298$ K, $100$ kPa, $1$ mol dm $^{-3}$ ). The high resistance ensures no current flows, so the measured voltage is the true potential difference rather than a value reduced by the cell doing work.

Limitations of predictions

Standard electrode potentials predict only the thermodynamic feasibility of a redox reaction, not its rate or its behaviour away from standard conditions. A reaction with a positive EMF may be too slow to observe because of a high activation energy. Changing concentrations shifts the actual electrode potential away from the standard value (by Le Chatelier reasoning at the electrode), so a reaction predicted to be just feasible under standard conditions may not occur if concentrations differ. These caveats are exactly what extended-answer questions ask you to discuss.

Examples in context

Batteries and fuel cells. The EMF of a cell is the difference of electrode potentials, the basis of designing batteries and the hydrogen fuel cell, whose driving force comes from a large potential difference. Corrosion protection. Sacrificial anodes work because a metal with a more negative electrode potential (like zinc or magnesium) oxidises preferentially, protecting iron from rusting.

Try this

Q1. State the value assigned to the standard hydrogen electrode. [1 mark]

Cue. $0$ V (it is the reference).

Q2. Calculate the EMF of a cell from electrodes at $+0.80$ V and $-0.44$ V. [1 mark]

Cue. $0.80 - (-0.44) = +1.24$ V.

Q3. State what a positive cell EMF indicates about a reaction. [1 mark]

Cue. The reaction is feasible (spontaneous) under standard conditions.

Q4. State why a high-resistance voltmeter is used to measure electrode potential. [1 mark]

Cue. So that no current flows and the true (maximum) potential difference is measured.

Q5. State one limitation of using standard electrode potentials to predict reactions. [1 mark]

Cue. They predict feasibility only, not rate (a high activation energy may make a feasible reaction slow), and they assume standard conditions.

Exam-style practice questions

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

WJEC 20194 marksA cell is set up from the

\text{Zn}^{2+}/\text{Zn}

electrode (

E = -0.76\ \text{V}

) and the

\text{Cu}^{2+}/\text{Cu}

electrode (

E = +0.34\ \text{V}

). Calculate the standard cell EMF and write the overall cell reaction.

Show worked answer →

The more positive electrode is the positive terminal (cathode, reduction); the more negative is the negative terminal (anode, oxidation).

$E_{cell} = E_{positive} - E_{negative} = (+0.34) - (-0.76) = +1.10$ V.

The copper half is reduced and the zinc half is oxidised, so overall: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ .

Markers reward identifying the positive electrode, the subtraction giving $+1.10$ V, and the correct overall equation.

WJEC 20213 marksUse standard electrode potentials to determine whether acidified

\text{MnO}_4^-

(

E = +1.51\ \text{V}

) will oxidise

\text{Cl}^-

\text{Cl}_2

(

E

for

\text{Cl}_2/\text{Cl}^- = +1.36\ \text{V}

). Explain your reasoning.

Show worked answer →

For a redox reaction to be feasible, the species being reduced must have the more positive electrode potential.

$\text{MnO}_4^-$ ( $+1.51$ V) is more positive than the $\text{Cl}_2/\text{Cl}^-$ couple ( $+1.36$ V), so $\text{MnO}_4^-$ is the stronger oxidising agent.

Therefore $\text{MnO}_4^-$ will oxidise $\text{Cl}^-$ to $\text{Cl}_2$ ; the cell EMF is $+1.51 - 1.36 = +0.15$ V (positive, so feasible).

Markers reward comparing the potentials, identifying the stronger oxidising agent, and the positive EMF showing feasibility.

Related dot points

Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)