WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
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.
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.
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
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
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
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
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
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
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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