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Redox and Standard Electrode Potential

Cell EMF and the feasibility of reactions

  • 3 things to know
  • A2 Unit 3

Overview

Recall questions on electromotive force of a cell.

What you need to know

Answer each one before you reveal it.

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Numbers

Calculate E°cell for a cell made from Zn²⁺/Zn, E° = −0.76 V, and Cu²⁺/Cu, E° = +0.34 V, and state which metal is oxidised.

Numbers

Calculate E°cell for a cell made from Cu²⁺/Cu, E° = +0.34 V, and Ag⁺/Ag, E° = +0.80 V, and write the overall equation.

Numbers

Given E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V and E°(I₂/I⁻) = +0.54 V, calculate E°cell and state whether acidified dichromate(VI) will oxidise iodide ions.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What is the electromotive force of a cell?

    Model answer

    It is the maximum potential difference between the two electrodes of the cell, measured when no current is being drawn. It is the driving force pushing electrons round the external circuit.

  2. 2

    What relationship is used to calculate the standard EMF of a cell from two standard electrode potentials?

    Model answer

    E°cell = E°(half-cell at which reduction occurs) − E°(half-cell at which oxidation occurs). Written from a cell diagram this is E°cell = E°(right-hand electrode) − E°(left-hand electrode).

  3. 3

    Calculate E°cell for a cell made from Zn²⁺/Zn, E° = −0.76 V, and Cu²⁺/Cu, E° = +0.34 V, and state which metal is oxidised.

    Model answer

    Copper(II) is reduced because its value is the more positive, and zinc is oxidised. E°cell = (+0.34) − (−0.76) = +1.10 V. Zinc is the metal that is oxidised.

  4. 4

    Calculate E°cell for a cell made from Zn²⁺/Zn, E° = −0.76 V, and Ag⁺/Ag, E° = +0.80 V.

    Model answer

    Silver ions are reduced and zinc is oxidised, so E°cell = (+0.80) − (−0.76) = +1.56 V.

  5. 5

    Calculate E°cell for a cell made from Cu²⁺/Cu, E° = +0.34 V, and Ag⁺/Ag, E° = +0.80 V, and write the overall equation.

    Model answer

    Silver has the more positive value, so silver ions are reduced and copper is oxidised. E°cell = (+0.80) − (+0.34) = +0.46 V. The overall equation is Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).

  6. 6

    Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Cu²⁺/Cu) = +0.34 V, calculate E°cell and state what reaction occurs.

    Model answer

    Iron(III) has the more positive value, so it is reduced and copper metal is oxidised. E°cell = (+0.77) − (+0.34) = +0.43 V, and the reaction is Cu(s) + 2Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺(aq).

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