Use copper(II)/iodide and supplied redox data
Balance the Cu²⁺/I⁻ reaction and transfer the half-equation method to other systems when the syllabus supplies data.
Section 28 · Chemistry of transition elements
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 28. Use the first-row transition elements titanium to copper and the species explicitly named in the syllabus. Where redox potentials or stability constants are needed, use the values supplied in the question. Colour descriptions can depend on conditions and concentration. Written practice is self-checked
Before you begin
- Use electron configurations, oxidation numbers, ionic equations, equilibrium expressions and the AS redox and bonding models.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What iodine product is formed from oxidation of iodide ions?
What you will learn
- Describe Cu²⁺/I⁻ redox products including CuI and iodine
- Balance electron and atom counts in the overall equation
- Apply supplied redox data to an unfamiliar transition-element system
Copper(II) oxidises iodide while forming copper(I) iodide
The net reaction is 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. Cu²⁺ is reduced to Cu⁺ in the solid CuI, and I⁻ is oxidised to I₂. The precipitate formation helps drive the reaction.
Check electron balance: two Cu²⁺ ions each gain one electron, while four I⁻ ions lose a total of two electrons to form one I₂ molecule. The equation is ionic and does not require spectator ions.
Use data rather than memorising every redox system
For other redox systems, use the half-equations and potential values supplied in the question. Identify oxidation and reduction, calculate the cell potential when requested, balance electrons, and then check atoms and charge.
The syllabus explicitly allows calculations involving other systems when suitable data are provided. Do not import an unprovided potential or assume a reaction is feasible without applying the given values.
Pause and explain
What is the oxidation state of copper in CuI(s)?
Worked example
Write the balanced ionic equation for Cu²⁺ reacting with excess I⁻ and identify what is oxidised and reduced.
Show the worked solution
- Cu²⁺ gains electrons to form Cu⁺ in CuI.
- I⁻ loses electrons to form I₂.
- Balance electrons and precipitate units.
Answer 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. Cu²⁺ is reduced to Cu⁺; I⁻ is oxidised to I₂.
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Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Identify the oxidising agent and reducing agent in 2Cu²⁺ + 4I⁻ → 2CuI + I₂.
Give me a hint
The oxidising agent is reduced; the reducing agent is oxidised.
Compare my reasoning
- Cu²⁺ becomes Cu⁺ in CuI.
- I⁻ becomes I₂.
- Assign agents from those oxidation-state changes.
Cu²⁺ is the oxidising agent and I⁻ is the reducing agent.
Look for these in your work
- Reduction identified
- Oxidation identified
- Agent names assigned correctly
Check the atom and charge balance of 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂.
Give me a hint
Count Cu and I, then sum total charge on each side.
Compare my reasoning
- Two Cu atoms and four I atoms appear on both sides.
- Left charge is +4 − 4 = 0.
- Products are neutral overall.
Atoms balance and both sides have total charge 0.
Look for these in your work
- Atoms counted
- Reactant charge calculated
- Product charge checked
An unfamiliar M³⁺/M²⁺ couple is paired with X₂/X⁻ and the question supplies E° values. Give a reliable method for deciding the spontaneous direction.
Give me a hint
Treat both supplied potentials as reductions before deciding which half-equation to reverse.
Compare my reasoning
- Compare the reduction potentials.
- Use the more positive reduction as the cathode process.
- Calculate E°cell and balance electrons in the equation.
Choose the more positive reduction as the cathode, reverse the other half-equation for oxidation, calculate E°cell = E°cathode − E°anode, then balance electrons and check atoms and charge. A positive E°cell supports that direction under standard conditions.
Look for these in your work
- Data used as reductions
- Cell direction selected
- Equation and feasibility checked
Common mistakes
- Writing CuI₂ instead of CuI
- Forgetting iodine molecules are I₂
- Balancing atoms but not electron transfer
- Using memorised potentials for an unfamiliar system when data are supplied
What can you explain now?
What iodine product is formed from oxidation of iodide ions?
Compare with the explanation
I₂
Two iodide ions lose two electrons to form one iodine molecule.
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Make it stick
How do oxidation states, half-equations and supplied potentials help solve an unfamiliar transition-metal redox problem?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.