Two half-cells have E^ values of +0.34\ V and -0.44\ V. Calculate the standard cell potential. [1 mark]

OCR OCR 2019-style practice: A cell is made from a Zn^2+/Zn half-cell (E^ = -0.76\ V) and a Ag^+/Ag half-cell (E^ = +0.80\ V). (a) Calculate the standard cell potential. (b) Write the overall cell reaction. (c) State which electrode is the positive terminal.

(a) E^_cell = E^_positive - E^_negative = (+0.80) - (-0.76) = +1.56\ V (1). (b) The silver half-cell (more positive E^) is reduced and the zinc is oxidised: Zn + 2Ag^+ → Zn^2+ + 2Ag (1)(1). (c) The silver electrode is the positive terminal (more positive electrode potential) (1). Markers reward the cell potential, the balanced overall equation (silver reduced, zinc oxidised), and identifying the silver electrode as positive.

EnglandChemistrySyllabus dot point

How do standard electrode potentials let us predict which redox reactions happen?

Redox half-equations, standard electrode potentials measured against the standard hydrogen electrode, cell notation and standard cell potential, predicting feasibility, and storage and fuel cells.

An OCR H432 module 5 answer on electrochemistry: redox half-equations, standard electrode potentials and the standard hydrogen electrode, cell notation and standard cell potential, predicting feasibility, and fuel cells.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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What this topic is asking
Redox half-equations
Standard electrode potentials
Cell notation and cell potential
Storage and fuel cells
Examples in context
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What this topic is asking

OCR specification point 5.2.3 wants you to write and combine redox half-equations, define and use standard electrode potentials measured against the standard hydrogen electrode, represent cells with cell notation, calculate the standard cell potential, predict whether a redox reaction is feasible, and describe storage cells and fuel cells. This is the electrochemical extension of the redox of Module 2.

Redox half-equations

Standard electrode potentials

Cell notation and cell potential

Predicting a redox reaction from electrode potentials

Will acidified manganate(VII) ( $E^{\circ}(\text{MnO}_4^-/\text{Mn}^{2+}) = +1.51\ \text{V}$ ) oxidise iron(II) to iron(III) ( $E^{\circ}(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$ )?

step 1: Identify which is reduced

The half-cell with the more positive $E^{\circ}$ is reduced. Manganate(VII) at $+1.51\ \text{V}$ is more positive, so it is reduced (to $\text{Mn}^{2+}$ ); the iron(II) is oxidised to iron(III).

step 2: Calculate the cell potential

E^{\circ}_{\text{cell}} = E^{\circ}_{\text{positive}} - E^{\circ}_{\text{negative}} = (+1.51) - (+0.77) = +0.74\ \text{V}.

step 3: Judge feasibility

The cell potential is positive, so the reaction is feasible: manganate(VII) does oxidise iron(II) to iron(III). (This is the basis of the manganate(VII) titration.)

Storage and fuel cells

Examples in context

Example 1. The manganate(VII) titration. Because $E^{\circ}(\text{MnO}_4^-/\text{Mn}^{2+})$ is highly positive, manganate(VII) oxidises iron(II) quantitatively, and its own purple-to-colourless change acts as the indicator, a direct use of electrode-potential feasibility.

Example 2. Hydrogen fuel-cell vehicles. A fuel-cell car runs the hydrogen-oxygen reaction to generate electricity with only water as exhaust, illustrating why fuel cells are attractive where clean hydrogen is available.

Try this

Q1. State the defined value of the standard electrode potential of the standard hydrogen electrode. [1 mark]

Cue. $0.00\ \text{V}$ .

Q2. Two half-cells have $E^{\circ}$ values of $+0.34\ \text{V}$ and $-0.44\ \text{V}$ . Calculate the standard cell potential. [1 mark]

Cue. $E^{\circ}_{\text{cell}} = (+0.34) - (-0.44) = +0.78\ \text{V}$ .

Exam-style practice questions

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

OCR 20194 marksA cell is made from a

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

half-cell (

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

) and a

\text{Ag}^+/\text{Ag}

half-cell (

E^{\circ} = +0.80\ \text{V}

). (a) Calculate the standard cell potential. (b) Write the overall cell reaction. (c) State which electrode is the positive terminal.

Show worked answer →

(a) $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{positive}} - E^{\circ}_{\text{negative}} = (+0.80) - (-0.76) = +1.56\ \text{V}$ (1).

(b) The silver half-cell (more positive $E^{\circ}$ ) is reduced and the zinc is oxidised: $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$ (1)(1).

Markers reward the cell potential, the balanced overall equation (silver reduced, zinc oxidised), and identifying the silver electrode as positive.

OCR 20213 marksAn alkaline hydrogen-oxygen fuel cell operates with the half-reactions

\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2e^-

and

\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-

. (a) Write the overall equation. (b) Give one advantage of a fuel cell over a petrol engine.

Show worked answer →

(a) Balance electrons: multiply the hydrogen half-reaction by two, then add. The overall reaction is $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (1)(1).

(b) Any one: the only product is water (no carbon dioxide or pollutants), or it is more efficient than a combustion engine (1).

Markers reward correctly combining the half-reactions to give water, and a valid advantage such as no polluting products or higher efficiency.

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)