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 Reactions
Combining half-equations into full equations
- 3 things to know
- A2 Unit 3
Overview
Fifteen recall questions on combining half-equations into full equations.
What you need to know
Answer each one before you reveal it.
What is the overall equation for the reaction of acidified dichromate(VI) with iodide ions?
What is the overall equation for the reaction of acidified manganate(VII) with iodide ions?
Questions from this subtopic
Each one with the model answer underneath.
- 1
How are two half-equations combined into an overall redox equation?
Model answer
Multiply each half-equation by whatever whole number is needed so that the number of electrons released by the oxidation equals the number taken up by the reduction. Add the two equations together, cancel the electrons, and cancel any water molecules or hydrogen ions that appear on both sides.
- 2
Combine MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O with Fe²⁺ → Fe³⁺ + e⁻.
Model answer
Multiply the iron half-equation by five so five electrons are released, then add: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. The charge is −1 + 10 + 8 = +17 on the left and +2 + 15 = +17 on the right.
- 3
Combine the dichromate(VI) half-equation with the oxidation of iron(II).
Model answer
Six electrons are gained by each dichromate ion, so multiply the iron half-equation by six: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. The charge is −2 + 12 + 14 = +24 on the left and +6 + 18 = +24 on the right.
- 4
What is the overall equation for the oxidation of ethanedioate ions by acidified manganate(VII)?
Model answer
Multiply the manganate(VII) half-equation by two and the ethanedioate half-equation by five so that ten electrons are transferred: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. The charge is −2 − 10 + 16 = +4 on both sides.
- 5
What is the overall equation for the reaction of acidified manganate(VII) with hydrogen peroxide?
Model answer
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. Ten electrons are transferred, and the charge is −2 + 6 = +4 on the left and +4 on the right.
- 6
What is the overall equation for the reaction of acidified dichromate(VI) with iodide ions?
Model answer
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O. Six electrons are transferred, and the charge is −2 − 6 + 14 = +6 on the left and +6 on the right.
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