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
Building ion–electron half-equations
- 3 things to know
- A2 Unit 3
Overview
Nineteen recall questions on Building ion–electron half-equations.
What you need to know
Answer each one before you reveal it.
What is the half-equation for the reduction of the manganate(VII) ion to Mn²⁺ in acid solution?
What is the half-equation for the reduction of VO₂⁺ to VO²⁺ in acid solution?
Questions from this subtopic
Each one with the model answer underneath.
- 1
How do you construct a half-equation for a species reacting in acid solution?
Model answer
Write the species that changes on each side and balance the atoms of the element being oxidised or reduced. Add water molecules to balance the oxygen atoms, then add hydrogen ions to balance the hydrogen atoms. Finally add electrons to whichever side is needed so that the total charge is the same on both sides.
- 2
What is the half-equation for the reduction of the manganate(VII) ion to Mn²⁺ in acid solution?
Model answer
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Four water molecules balance the four oxygen atoms, eight hydrogen ions balance the hydrogen, and five electrons make the charge +2 on both sides, since −1 + 8 − 5 = +2.
- 3
Construct the half-equation for the reduction of the chlorate(I) ion, ClO⁻, to chloride ions in acid solution.
Model answer
The half-equation is ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O. The single oxygen atom leaves as one water molecule, which needs two hydrogen ions to supply its hydrogen. Two electrons are then added on the left so that the charge is −1 + 2 − 2 = −1 on each side, matching chlorine falling from +1 in ClO⁻ to −1 in Cl⁻.
- 4
What is the half-equation for the oxidation of thiosulfate ions to tetrathionate ions?
Model answer
2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻. Two thiosulfate ions are needed to supply the four sulfur atoms, and two electrons balance the charge, since −4 on the left equals −2 − 2 on the right.
- 5
Why do hydrogen ions appear in the half-equation for the reduction of manganate(VII)?
Model answer
The manganate(VII) ion carries four oxygen atoms which are not present in the Mn²⁺ product. Those oxygen atoms leave as water, and the hydrogen needed to form that water is supplied by the acid, so hydrogen ions must appear on the left of the equation.
- 6
Why must a solution of dichromate(VI) or manganate(VII) be acidified before it can act as an oxidising agent in these ways?
Model answer
Both half-equations consume a large number of hydrogen ions, eight for each manganate(VII) ion and fourteen for each dichromate(VI) ion. Without a plentiful supply of acid the reduction to Mn²⁺ or Cr³⁺ cannot take place.
Other ways to work this
The same content, four other ways.
What is inside Levo
Six ways to work the same material.
Every question in Redox Reactions, on a schedule.
Attempt a question from memory, write the correct version, then attempt it again. Levo records the jump between your first attempt and your last, and brings each question back on the day you are due to forget it.
Or £46 a year, about £3.83 a month. Cancel anytime.