Redox Chemistry of Transition elements
A concise revision guide to the redox chemistry of the transition elements: and using E⦵ values to predict which reactions are feasible.
- 24.2.7
- 28.2.8
What these spec points say
- 24.2.7 construct redox equations using the relevant half-equations
- 28.2.8 predict, using E⦵ values, the feasibility of redox reactions involving transition elements and their ions
Before You Start
This page uses the ideas of Topic 24 Electrochemistry and the variable oxidation states introduced on Electron Configurations and Properties of Transition elements.
What you need to remember
- Oxidation is loss of electrons: the oxidation state of the element increases.
- Reduction is gain of electrons: the oxidation state 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 states 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 States, Colour and Half-Equations
A transition element has several stable oxidation states, so its ions take part in many redox reactions. Each oxidation state has a different number of d electrons, so each has its own colour.
Working out an oxidation state
- 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 state 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 state 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 state, such as MnO₄⁻ and Cr₂O₇²⁻, are oxidising agents. Ions with a low oxidation state, 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.
Predicting Feasibility with E⦵ Values
Standard electrode potentials show whether one species can oxidise another.
| Half-equation | E⦵ / V | Colour change, left to right |
|---|---|---|
| MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O | +1.52 | purple → almost colourless |
| Cl₂ + 2e⁻ ⇌ 2Cl⁻ | +1.36 | pale green → colourless |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O | +1.33 | orange → green |
| Fe³⁺ + e⁻ ⇌ Fe²⁺ | +0.77 | yellow → pale green |
| I₂ + 2e⁻ ⇌ 2I⁻ | +0.54 | brown → colourless |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 | blue solution → pink-brown solid |
The method
- Identify the species that is reduced and the species that is oxidised in the proposed reaction.
- Find the two half-equations and their E⦵ values.
- Calculate E⦵cell = E⦵(half-equation of the species reduced) − E⦵(half-equation of the species oxidised).
- A positive E⦵cell means the reaction is feasible under standard conditions.
Worked example 1
Can acidified manganate(VII) ions oxidise iron(II) ions?
Step 1. MnO₄⁻ would be reduced (+1.52 V) and Fe²⁺ oxidised (+0.77 V).
Step 2. E⦵cell = (+1.52) − (+0.77) = +0.75 V
Answer. E⦵cell is positive, so the reaction is feasible: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
Worked example 2
Can iron(III) ions oxidise iodide ions? Can they oxidise copper metal?
Iodide. E⦵cell = (+0.77) − (+0.54) = +0.23 V, feasible: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
Copper. E⦵cell = (+0.77) − (+0.34) = +0.43 V, feasible: 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺.
An E⦵ ladder for transition element systems. A species on the left of a half-equation oxidises any species on the right of a half-equation below it.
Key idea: The oxidising agent is on the left of the more positive half-equation. It oxidises the species on the right of a less positive half-equation.
Check: Oxidation States and Colour Changes
Assign oxidation states in transition element species given in the question, and decide what has been oxidised and what has been reduced.
More Feasibility Examples
| Half-equation | E⦵ / V |
|---|---|
| Cu⁺ + e⁻ ⇌ Cu | +0.52 |
| Cu²⁺ + e⁻ ⇌ Cu⁺ | +0.15 |
| Cr³⁺ + e⁻ ⇌ Cr²⁺ | −0.41 |
| Zn²⁺ + 2e⁻ ⇌ Zn | −0.76 |
Worked example 3: choosing the acid
Can dichromate(VI) or manganate(VII) ions oxidise chloride ions to chlorine?
Cr₂O₇²⁻. E⦵cell = (+1.33) − (+1.36) = −0.03 V, not feasible under standard conditions.
MnO₄⁻. E⦵cell = (+1.52) − (+1.36) = +0.16 V, feasible.
Worked example 4: zinc and dichromate(VI)
+6 to +3. E⦵cell = (+1.33) − (−0.76) = +2.09 V, feasible: orange → green.
+3 to +2. E⦵cell = (−0.41) − (−0.76) = +0.35 V, feasible: green → blue Cr²⁺.
Worked example 5: copper(I) ions
Predict whether 2Cu⁺(aq) → Cu(s) + Cu²⁺(aq) is feasible.
Limits of the prediction
- E⦵ values say nothing about rate: a feasible reaction with a high activation energy may be too slow to see.
- E⦵ values apply to standard conditions. Other concentrations change the electrode potentials.
- Both points are explained in Topic 24 Electrochemistry.
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.
Common Exam Points
Say
- “The oxidation state of the metal decreases, so it is reduced.”
- “E⦵cell is positive, so the reaction is thermodynamically feasible.”
Do not say
- “The colour changes” without naming both colours.
- “The E⦵ value is doubled because the half-equation is doubled” (E⦵ values are never multiplied).
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.
Check: Redox Reactions of Transition elements
Interpret redox reactions that are new to you: oxidation states, electrons transferred, balanced equations and the colour changes described.
FAQs
Use these quick answers to check oxidation states, half-equations and the redox chemistry of E⦵ values.
How do I work out the oxidation state 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 do I predict whether a redox reaction is feasible?
Find the two half-equations. E⦵cell = E⦵ of the species reduced − E⦵ of the species oxidised. A positive value means the reaction is feasible. For MnO₄⁻ with Fe²⁺: (+1.52) − (+0.77) = +0.75 V.
Why is hydrochloric acid not used to acidify manganate(VII)?
MnO₄⁻ can oxidise chloride ions to chlorine: E⦵cell = (+1.52) − (+1.36) = +0.16 V, which is feasible. Dichromate(VI) cannot do so under standard conditions: (+1.33) − (+1.36) = −0.03 V.
Are copper(I) ions stable in aqueous solution?
No. Cu⁺ + e⁻ ⇌ Cu is +0.52 V and Cu²⁺ + e⁻ ⇌ Cu⁺ is +0.15 V, so E⦵cell = +0.37 V for 2Cu⁺ → Cu + Cu²⁺. The reaction is feasible: copper(I) ions disproportionate.
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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