Redox Chemistry of Transition elements
A concise revision guide to the redox chemistry of the transition elements: chromium(III), dichromate(VI) and chromate(VI), iron(II) and iron(III), copper(I) and copper(II), and the half-equations and colour changes.
- 5.3.1(k)(i)
- 5.3.1(k)(ii)
- 5.3.1(k)(iii)-a
- 5.3.1(k)(iii)-b
- 5.3.1(l)-iii
What these spec points say
- 5.3.1(k)(i) redox reactions and accompanying colour changes for interconversions between Fe2+ and Fe3+
- 5.3.1(k)(ii) redox reactions and accompanying colour changes for interconversions between Cr3+ and Cr2O72–
- 5.3.1(k)(iii)-a redox reactions and accompanying colour changes for reduction of Cu2+ to Cu+
- 5.3.1(k)(iii)-b redox reactions and accompanying colour changes for disproportionation of Cu+ to Cu2+ and Cu
- 5.3.1(l)-iii interpretation and prediction of unfamiliar reactions including redox
Before You Start
This page uses the ideas of 5.2.3 Redox and electrode potentials and the variable oxidation numbers introduced on Electron Configurations and Properties of Transition elements.
What you need to remember
- Oxidation is loss of electrons: the oxidation number of the element increases.
- Reduction is gain of electrons: the oxidation number decreases.
- An oxidising agent takes electrons and is itself reduced. A reducing agent gives electrons and is itself oxidised.
- Oxygen is usually −2 and hydrogen +1. The oxidation numbers in an ion add up to its charge.
- A reaction is feasible when E⦵cell is positive: E⦵cell = E⦵(species reduced) − E⦵(species oxidised).
Oxidation Numbers, Colour and Half-Equations
A transition element has several stable oxidation numbers, so its ions take part in many redox reactions. Each oxidation number has a different number of d electrons, so each has its own colour.
Working out an oxidation number
- In Cr₂O₇²⁻: 2Cr + 7 × (−2) = −2, so each chromium is +6.
- In MnO₄⁻: Mn + 4 × (−2) = −1, so manganese is +7.
- In a simple ion such as Fe³⁺, the oxidation number equals the charge: +3.
Building a half-equation in acid
Dichromate(VI) ions are reduced to chromium(III) ions in acidic solution.
- Balance the metal: Cr₂O₇²⁻ → 2Cr³⁺
- Balance oxygen with water: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
- Balance hydrogen with H⁺: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
- Balance charge with electrons: the left is 12+ and the right is 6+, so add 6e⁻ to the left.
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Check: each chromium falls from +6 to +3, a gain of three electrons, and there are two chromium atoms.
Combining two half-equations
Step 1. Write both half-equations: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and Fe²⁺ → Fe³⁺ + e⁻.
Step 2. Multiply the iron half-equation by 6 so that the electrons cancel.
Answer. Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
A change of oxidation number changes the colour of a transition element ion: iron(II) to iron(III), and manganate(VII) to manganese(II).
Two half-equations are combined by multiplying them so that the electrons lost equal the electrons gained.
General pattern: Ions with a high oxidation number, such as MnO₄⁻ and Cr₂O₇²⁻, are oxidising agents. Ions with a low oxidation number, such as Fe²⁺, are reducing agents.
Quick Check: Half-Equation for a New Ion
Apply the method to an ion that is new to this card.
Iron(II) and Iron(III)
Oxidation of Fe²⁺ to Fe³⁺
Acidified manganate(VII) ions oxidise iron(II) ions.
MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)
- The purple MnO₄⁻ solution is decolourised as it is added.
- The pale green Fe²⁺ solution becomes pale yellow as Fe³⁺ forms.
- Oxygen in the air also oxidises Fe²⁺(aq) to Fe³⁺(aq), slowly, on standing.
- Iron changes from +2 to +3 (oxidation). Manganese changes from +7 to +2 (reduction).
Reduction of Fe³⁺ to Fe²⁺
Iodide ions reduce iron(III) ions.
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)
- The orange-brown Fe³⁺ solution turns brown as iodine forms.
- The pale green colour of Fe²⁺ is hidden by the brown iodine.
- Iron changes from +3 to +2 (reduction). Iodine changes from −1 to 0 (oxidation).
Interconverting iron(II) and iron(III): acidified manganate(VII) oxidises Fe²⁺, and iodide ions reduce Fe³⁺.
Check: Oxidation Numbers and Colour Changes
Assign oxidation numbers in transition element species given in the question, and decide what has been oxidised and what has been reduced.
Chromium(III) and Dichromate(VI)
| Oxidation number | Ion | Colour |
|---|---|---|
| +6 | Cr₂O₇²⁻, dichromate(VI) | orange |
| +6 | CrO₄²⁻, chromate(VI) | yellow |
| +3 | Cr³⁺ | green |
Oxidation of Cr³⁺
Chromium(III) ions are oxidised by hot alkaline hydrogen peroxide.
2Cr³⁺ + 3H₂O₂ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O
- The green solution turns yellow as chromate(VI) ions form.
- Chromium changes from +3 to +6 (oxidation). Oxygen in H₂O₂ changes from −1 to −2 (reduction).
- Adding dilute acid then turns the yellow solution orange: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O.
- The last step is not a redox reaction: chromium is +6 in both ions.
Reduction of Cr₂O₇²⁻
Acidified dichromate(VI) ions are reduced by zinc.
Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺
- The orange solution turns green.
- Chromium changes from +6 to +3 (reduction). Zinc changes from 0 to +2 (oxidation).
The chromium cycle: chromium(III) is oxidised to chromate(VI) in alkali, acid converts chromate(VI) to dichromate(VI), and zinc in acid reduces it back.
Check: Half-Equations and Redox Equations
Combine half-equations given in the question into balanced redox equations, and identify the oxidising agent and the reducing agent.
Copper(II) and Copper(I)
Reduction of Cu²⁺ to Cu⁺
Iodide ions reduce aqueous copper(II) ions.
2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq)
- The blue solution gives a white precipitate of copper(I) iodide in a brown solution of iodine.
- Copper changes from +2 to +1 (reduction). Iodine changes from −1 to 0 (oxidation).
- Copper(I) iodide is white because Cu⁺ has a full 3d sub-shell.
Disproportionation of Cu⁺
Disproportionation is the oxidation and reduction of the same element in one reaction.
Cu₂O(s) + H₂SO₄(aq) → Cu(s) + CuSO₄(aq) + H₂O(l)
Ionic equation: 2Cu⁺ → Cu + Cu²⁺
- A brown solid (copper) and a blue solution (copper(II) sulfate) form.
- One Cu⁺ ion is reduced from +1 to 0. Another is oxidised from +1 to +2.
- E⦵ for Cu⁺ + e⁻ ⇌ Cu is +0.52 V and for Cu²⁺ + e⁻ ⇌ Cu⁺ is +0.15 V.
- E⦵cell = (+0.52) − (+0.15) = +0.37 V, so the disproportionation is feasible.
Copper(II) is reduced to copper(I) by iodide ions, and copper(I) disproportionates in dilute sulfuric acid to copper and copper(II).
Interpreting Unfamiliar Redox Reactions
Questions may give a redox reaction that is new. The same checks always apply.
- Assign oxidation numbers to the transition element before and after.
- An increase in oxidation number is oxidation. A decrease is reduction.
- Write or use the two half-equations, and multiply them so that the electrons cancel.
- Link each colour given in the question to a species.
Worked example
Manganese(IV) oxide reacts with concentrated hydrochloric acid: MnO₂ + 4H⁺ + 2Cl⁻ → Mn²⁺ + Cl₂ + 2H₂O. Identify what is oxidised and what is reduced.
Step 1. Manganese: +4 in MnO₂ and +2 in Mn²⁺, so manganese is reduced.
Common Exam Points
Say
- “The oxidation number of the metal decreases, so it is reduced.”
- “Converting chromate(VI) to dichromate(VI) is not redox: chromium stays at +6.”
- “Copper(I) is both oxidised and reduced, so the reaction is a disproportionation.”
Do not say
- “The colour changes” without naming both colours.
- “The solution turns green when Fe³⁺ reacts with iodide” (the brown iodine hides the pale green).
Watch for
- Balance the electrons before adding two half-equations.
- Balance oxygen with H₂O and hydrogen with H⁺ in acidic solution.
- Check the final equation: the atoms and the total charge must balance.
- Copper(I) iodide is a white solid: the brown colour belongs to the iodine.
Check: Redox Reactions of Transition elements
Interpret redox reactions that are new to you: oxidation numbers, electrons transferred, balanced equations and the colour changes described.
FAQs
Use these quick answers to check oxidation numbers, half-equations and the redox chemistry of iron, chromium and copper.
How do I work out the oxidation number of chromium in Cr₂O₇²⁻?
Oxygen is −2, so seven oxygen atoms total −14. The ion has a charge of 2−, so the two chromium atoms total +12. Each chromium atom is therefore +6. In the same way, manganese in MnO₄⁻ is +7.
How do I combine the dichromate(VI) and iron(II) half-equations?
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O needs six electrons, and each Fe²⁺ → Fe³⁺ + e⁻ supplies one. Multiply the iron half-equation by 6, then add: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺.
How are iron(II) and iron(III) ions interconverted?
Acidified manganate(VII) oxidises Fe²⁺ to Fe³⁺: pale green turns to yellow and the purple MnO₄⁻ is decolourised. Iodide ions reduce Fe³⁺ back to Fe²⁺: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, which gives a brown solution of iodine.
What happens when iodide ions are added to copper(II) ions?
The equation is 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The oxidation number of copper falls from +2 to +1 and iodide ions are oxidised to iodine. A white precipitate of copper(I) iodide forms in a brown solution.
What is the disproportionation of copper(I)?
Copper(I) is both oxidised and reduced: 2Cu⁺ → Cu + Cu²⁺. Copper(I) oxide reacts with dilute sulfuric acid to give a brown solid, copper, and a blue solution: Cu₂O + H₂SO₄ → Cu + CuSO₄ + H₂O.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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