What is feasible but slow?

A positive EMF means the reaction can happen, not that it does so quickly; kinetic factors can make a feasible reaction immeasurably slow.

What are non-standard conditions?

Changing concentration shifts the electrode potential (by Le Chatelier on the half-equation), so predictions from standard values may fail when conditions are far from standard.

Using Mg^2+/Mg (-2.37\ V) and Zn^2+/Zn (-0.76\ V), calculate the EMF of a cell made from these half-cells. [1 mark]

WJEC Eduqas Eduqas 2019-style practice: A cell is made from a Zn^2+/Zn half-cell (E^ = -0.76\ V) and a Cu^2+/Cu half-cell (E^ = +0.34\ V). (a) Calculate the standard EMF of the cell. (b) Write the overall equation and state which metal is the negative electrode.

(a) E^_cell = E^_positive - E^_negative = (+0.34) - (-0.76) = +1.10\ V (2). (b) Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s) (1). Zinc, with the more negative potential, is oxidised and is the negative electrode (1).

WJEC Eduqas Eduqas 2021-style practice: Using the standard electrode potentials Fe^3+/Fe^2+ (E^ = +0.77\ V) and I_2/I^- (E^ = +0.54\ V), predict whether iron(III) ions will oxidise iodide ions, and justify your answer.

Iron(III) will oxidise iodide ions (1). The Fe^3+/Fe^2+ system has the more positive electrode potential, so Fe^3+ is reduced (to Fe^2+) and I^- is oxidised (to I_2) (1). The cell EMF is 0.77 - 0.54 = +0.23\ V, which is positive, so the reaction is feasible (1). Overall: 2Fe^3+ + 2I^- → 2Fe^2+ + I_2 (1).

EnglandChemistrySyllabus dot point

How do we measure the tendency of a species to be reduced, and predict whether a redox reaction will occur?

Oxidation states and half-equations, the standard hydrogen electrode, standard electrode potentials, electrochemical cells and their EMF, and using electrode potentials to predict feasibility.

An Eduqas A-Level Chemistry PI1.1 answer on oxidation states and half-equations, the standard hydrogen electrode, standard electrode potentials, cell EMF and predicting feasibility.

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 topic is asking
Oxidation states and half-equations
The standard hydrogen electrode
Standard electrode potentials
Cells and EMF
Predicting feasibility
Examples in context
Try this

What this topic is asking

Eduqas topic PI1.1 covers oxidation states and half-equations, the standard hydrogen electrode as the reference, standard electrode potentials, the construction and EMF of electrochemical cells, and using electrode potentials to predict whether a redox reaction is feasible. It is the quantitative heart of electrochemistry.

Oxidation states and half-equations

A redox reaction can always be split into an oxidation half-equation (electrons lost) and a reduction half-equation (electrons gained). Balancing a half-equation involves balancing atoms, then adding $\text{H}^+$ , $\text{H}_2\text{O}$ and electrons to balance oxygen, hydrogen and charge; the two halves are then scaled so the electrons cancel.

The standard hydrogen electrode

Standard electrode potentials

Cells and EMF

An electrochemical cell connects two half-cells with a salt bridge; electrons flow through the external circuit from the more negative to the more positive electrode. The standard cell EMF is the difference between the two electrode potentials.

Building a cell and predicting its EMF

Combine the $\text{Ag}^+/\text{Ag}$ ( $E^{\circ} = +0.80\ \text{V}$ ) and $\text{Cu}^{2+}/\text{Cu}$ ( $E^{\circ} = +0.34\ \text{V}$ ) half-cells.

step 1: Identify the positive electrode

The silver half-cell has the more positive potential, so it is the positive electrode and silver ions are reduced.

step 2: Calculate the EMF

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{positive}} - E^{\circ}_{\text{negative}} = 0.80 - 0.34 = +0.46\ \text{V}$ .

step 3: Write the overall reaction

Copper is oxidised and silver ions reduced: $\text{Cu}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})$ . The positive EMF confirms it is feasible.

Predicting feasibility

A reaction is thermodynamically feasible if the cell EMF is positive. The species with the more positive $E^{\circ}$ acts as the oxidising agent. A note of caution: feasibility says nothing about rate (a reaction with a positive EMF may be very slow if the activation energy is high), and standard potentials only apply at standard conditions.

Examples in context

Example 1. Why iron rusts but gold does not. Iron has a negative electrode potential, so it is readily oxidised by oxygen and water; gold has a very positive potential and resists oxidation, which is why it stays untarnished and is used in connectors.

Example 2. Rechargeable cells. The lithium-ion cell relies on a large positive EMF between its electrodes; understanding electrode potentials explains both the voltage it delivers and why reversing the current recharges it.

Try this

Q1. Define the standard electrode potential of a half-cell. [2 marks]

Cue. The EMF of the half-cell connected to a standard hydrogen electrode under standard conditions ( $298\ \text{K}$ , $100\ \text{kPa}$ , $1\ \text{mol dm}^{-3}$ solutions).

Q2. Using $\text{Mg}^{2+}/\text{Mg}$ ( $-2.37\ \text{V}$ ) and $\text{Zn}^{2+}/\text{Zn}$ ( $-0.76\ \text{V}$ ), calculate the EMF of a cell made from these half-cells. [1 mark]

Cue. $E^{\circ}_{\text{cell}} = (-0.76) - (-2.37) = +1.61\ \text{V}$ .

Exam-style practice questions

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

Eduqas 20194 marksA cell is made from a

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

half-cell (

E^{\circ} = -0.76\ \text{V}

) and a

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

half-cell (

E^{\circ} = +0.34\ \text{V}

). (a) Calculate the standard EMF of the cell. (b) Write the overall equation and state which metal is the negative electrode.

Show worked answer →

(a) $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{positive}} - E^{\circ}_{\text{negative}} = (+0.34) - (-0.76) = +1.10\ \text{V}$ (2).

(b) $\text{Zn}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu}(\text{s})$ (1). Zinc, with the more negative potential, is oxidised and is the negative electrode (1).

Eduqas 20214 marksUsing the standard electrode potentials

\text{Fe}^{3+}/\text{Fe}^{2+}

(

E^{\circ} = +0.77\ \text{V}

) and

\text{I}_2/\text{I}^-

(

E^{\circ} = +0.54\ \text{V}

), predict whether iron(III) ions will oxidise iodide ions, and justify your answer.

Show worked answer →

Iron(III) will oxidise iodide ions (1). The $\text{Fe}^{3+}/\text{Fe}^{2+}$ system has the more positive electrode potential, so $\text{Fe}^{3+}$ is reduced (to $\text{Fe}^{2+}$ ) and $\text{I}^-$ is oxidised (to $\text{I}_2$ ) (1).

The cell EMF is $0.77 - 0.54 = +0.23\ \text{V}$ , which is positive, so the reaction is feasible (1). Overall: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ (1).

Related dot points

Sources & how we know this

WJEC Eduqas GCE A Level Chemistry specification (from 2015) — WJEC Eduqas (2015)