What are standard electrode potentials?

The standard hydrogen electrode bubbles H_2 at 100\ kPa over a platinised platinum electrode in 1.00\ mol dm^-3 H^+. A salt bridge (e.g. saturated KNO_3) completes the circuit while a high-resistance voltmeter measures the potential difference without drawing current.

What are redox titrations?

Manganate(VII) reacts with iron(II) in a 1:5 ratio because MnO_4^- gains 5 electrons while each Fe^2+ loses one: $MnO_4^- + 8H^+ + 5Fe^2+ → Mn^2+ + 4H_2O + 5Fe^3+Iodine-thiosulfate titrations (\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$) use starch indicator, with the blue-black colour vanishing at the end point.

Two half-cells have E^ values of +0.77\ V and -0.76\ V. Calculate the cell EMF. [1 mark]

Pearson Edexcel Edexcel 2020-style practice: 25.0\ cm^3 of a solution of iron(II) sulfate was acidified and titrated with 0.0200\ mol dm^-3 potassium manganate(VII); 24.0\ cm^3 was required. (a) Write the ionic equation for the reaction. (b) Calculate the concentration of Fe^2+. (c) State the colour change at the end point.

Use the 1:5 mole ratio from the combined half-equations. (a) MnO_4^- + 8H^+ + 5Fe^2+ → Mn^2+ + 4H_2O + 5Fe^3+ (2). (b) n(MnO_4^-) = 0.0200 × 24.0/1000 = 4.80 × 10^-4\ mol (1). Ratio MnO_4^- : Fe^2+ = 1:5, so n(Fe^2+) = 5 × 4.80 × 10^-4 = 2.40 × 10^-3\ mol (1). Concentration = 2.40 × 10^-3 / (25.0/1000) = 0.0960\ mol dm^-3 (1). (c) The end point is the first permanent pale pink colour (the first drop of excess MnO_4^-) (1).

Pearson Edexcel Edexcel 2018-style practice: An electrochemical cell is built 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 cell EMF. (b) Deduce which metal is oxidised and write the overall cell reaction.

Take the more positive electrode as the positive terminal. (a) E^_cell = E^(positive) - E^(negative) = (+0.34) - (-0.76) = +1.10\ V (1). (b) The more negative electrode (Zn^2+/Zn) is the negative terminal, so zinc is oxidised (1). Overall: Zn + Cu^2+ → Zn^2+ + Cu (1).

EnglandChemistrySyllabus dot point

How can we predict and quantify the direction of electron transfer?

Standard electrode potentials, the standard hydrogen electrode, electrochemical cells and cell EMF, using electrode potentials to predict feasibility, and redox titrations.

An Edexcel 9CH0 Topic 13 answer covering standard electrode potentials, the standard hydrogen electrode, cell EMF, predicting feasibility from electrode potentials, and redox titrations.

Generated by Claude Opus 4.89 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

Jump to a section

What this topic is asking
The answer
Examples in context
Try this

What this topic is asking

Edexcel Topic 13 wants you to define and use standard electrode potentials, describe the standard hydrogen electrode, calculate cell EMF, predict the feasibility of redox reactions, and carry out redox titration calculations using manganate(VII) and thiosulfate.

The answer

Standard electrode potentials

The standard hydrogen electrode bubbles $\text{H}_2$ at $100\ \text{kPa}$ over a platinised platinum electrode in $1.00\ \text{mol dm}^{-3}$ $\text{H}^+$ . A salt bridge (e.g. saturated $\text{KNO}_3$ ) completes the circuit while a high-resistance voltmeter measures the potential difference without drawing current.

Cell EMF and feasibility

A positive $E^{\ominus}_{\text{cell}}$ indicates the reaction is feasible (thermodynamically spontaneous). However, feasibility is thermodynamic only: a high activation energy can still make a feasible reaction immeasurably slow. Also, the prediction assumes standard conditions; changing concentrations shifts the electrode potentials (a Le Chatelier effect on the half-cell equilibrium) and can change whether a borderline reaction occurs.

Combining half-equations

To get the overall reaction, reverse the half-equation at the negative electrode (so it shows oxidation), balance the electrons between the two half-equations, then add them. The electrons must cancel, which fixes the mole ratio.

Redox titrations

Manganate(VII) reacts with iron(II) in a $1:5$ ratio because $\text{MnO}_4^-$ gains $5$ electrons while each $\text{Fe}^{2+}$ loses one:

\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}

Iodine-thiosulfate titrations (

\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}

) use starch indicator, with the blue-black colour vanishing at the end point.

A manganate(VII) titration calculation

$25.0\ \text{cm}^3$ of acidified iron(II) solution needs $22.5\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{KMnO}_4$ to reach the end point. Find the concentration of $\text{Fe}^{2+}$ .

step 1: Moles of manganate(VII)

$n(\text{MnO}_4^-) = 0.0200 \times \dfrac{22.5}{1000} = 4.50 \times 10^{-4}\ \text{mol}$ .

step 2: Use the 1:5 ratio

$n(\text{Fe}^{2+}) = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\ \text{mol}$ .

step 3: Convert to concentration

$c(\text{Fe}^{2+}) = \dfrac{2.25 \times 10^{-3}}{25.0/1000} = 0.0900\ \text{mol dm}^{-3}$ .

step 4: State the result

The iron(II) concentration is $0.0900\ \text{mol dm}^{-3}$ , and the end point is the first permanent pale pink, no indicator being needed.

Examples in context

Example 1. Rusting and sacrificial protection. Iron rusts because $\text{O}_2$ reduction has a more positive $E^{\ominus}$ than the $\text{Fe}^{2+}/\text{Fe}$ couple, making the corrosion feasible. Attaching blocks of zinc ( $E^{\ominus} = -0.76\ \text{V}$ , more negative than iron) makes the zinc oxidise preferentially, protecting the iron. This sacrificial protection on ships' hulls and pipelines is a direct application of comparing electrode potentials to predict which metal is oxidised.

Example 2. Estimating iron in an ore. Industrial laboratories determine the iron content of ores by dissolving the sample, reducing all the iron to $\text{Fe}^{2+}$ , then titrating with standard manganate(VII). The self-indicating $1:5$ titration gives the moles of iron, which is converted to a percentage by mass of the ore. This is exactly the redox-titration calculation Edexcel sets, applied to quality control in metal extraction.

Try this

Q1. Two half-cells have $E^{\ominus}$ values of $+0.77\ \text{V}$ and $-0.76\ \text{V}$ . Calculate the cell EMF. [1 mark]

Cue. $E_{cell} = (+0.77) - (-0.76) = +1.53\ \text{V}$ .

Q2. Explain why a reaction with a positive cell EMF might not occur at a noticeable rate. [2 marks]

Cue. Positive EMF shows feasibility only; a high activation energy can make the reaction very slow.

Exam-style practice questions

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

Edexcel 20206 marks

25.0\ \text{cm}^3

of a solution of iron(II) sulfate was acidified and titrated with

0.0200\ \text{mol dm}^{-3}

potassium manganate(VII);

24.0\ \text{cm}^3

was required. (a) Write the ionic equation for the reaction. (b) Calculate the concentration of

\text{Fe}^{2+}

. (c) State the colour change at the end point.

Show worked answer →

Use the $1:5$ mole ratio from the combined half-equations.

(a) $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ (2).

(b) $n(\text{MnO}_4^-) = 0.0200 \times 24.0/1000 = 4.80 \times 10^{-4}\ \text{mol}$ (1). Ratio $\text{MnO}_4^- : \text{Fe}^{2+} = 1:5$ , so $n(\text{Fe}^{2+}) = 5 \times 4.80 \times 10^{-4} = 2.40 \times 10^{-3}\ \text{mol}$ (1). Concentration $= 2.40 \times 10^{-3} / (25.0/1000) = 0.0960\ \text{mol dm}^{-3}$ (1).

Edexcel 20183 marksAn electrochemical cell is built from a

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

half-cell (

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

) and a

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

half-cell (

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

). (a) Calculate the standard cell EMF. (b) Deduce which metal is oxidised and write the overall cell reaction.

Show worked answer →

Take the more positive electrode as the positive terminal.

(a) $E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{positive}) - E^{\ominus}(\text{negative}) = (+0.34) - (-0.76) = +1.10\ \text{V}$ (1).

(b) The more negative electrode ( $\text{Zn}^{2+}/\text{Zn}$ ) is the negative terminal, so zinc is oxidised (1). Overall: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ (1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)