Variable Oxidation States and Redox Systems
Learn and apply Variable Oxidation States and Redox Systems in the published Chemistry course sequence.
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The core idea
On this page
Variable Oxidation States and Redox Systems: Orientation
This lesson focuses on the redox systems that actually show up in structured and data questions: Fe³⁺/Fe²⁺, acidified permanganate, and acidified dichromate, including the half-equations and colour-change marks.
This chapter depends on periodic trends, so keep The Periodic Table (A Level) nearby and use the Transition Elements hub for the full sequence.
This page is really testing:
- Can you recall and apply the standard acidic half-equations without missing H⁺ and H₂O?
- Can you write colour-change observations precisely for Mn and Cr oxidising agents?
- Can you combine half-equations into a balanced ionic equation with correct electrons and charges?
Definitions (Must Know)
A. Variable oxidation states
Many transition elements show more than one stable oxidation state because the 3d and 4s electrons are similar in energy and can both be involved in bonding.
B. Key oxidising agents (acidic conditions)
- MnO₄⁻ (permanganate): purple → Mn²⁺ very pale (acidic)
- Cr₂O₇²⁻ (dichromate): orange → Cr³⁺ green (acidic)
Detailed Explanations
A. Why variable oxidation states happen (what to write)
Write something like:
- “The 3d and 4s orbitals are close in energy, so different numbers of electrons can be lost or shared, giving multiple oxidation states.”
Because 3d and 4s orbitals are close in energy, therefore more than one number of electrons can be involved in bonding, giving multiple stable oxidation states.
B. Core redox half-equations (acidic conditions)
Permanganate reduction (acidic): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Dichromate reduction (acidic): Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Iron redox: Fe³⁺ + e⁻ → Fe²⁺ Fe²⁺ → Fe³⁺ + e⁻
C. Workflow: combining half-equations in acidic solution (exam method)
- Write both half-equations.
- Multiply to balance electrons.
- Add and cancel electrons (and any common species).
- Check atoms and charge.
Mini example: To combine permanganate reduction (5e−) with Fe²⁺ → Fe³⁺ + e⁻, multiply the iron oxidation by 5 so 5 electrons cancel.
D. Worked method for redox-identification questions
When a question gives observations plus reagents, this route is reliable:
- Identify reagent conditions first (acidified permanganate/dichromate means acidic half-equations).
- Write the observation line (purple to colourless, orange to green).
- Name which species is reduced (oxidising agent) and which is oxidised.
- Add balanced half-equation(s) only if asked.
E. Predicting likely oxidation states from a configuration
Remove 4s electrons first, then consider how many 3d electrons can be removed to reach a relatively stable configuration. Treat the result as a prediction, not proof: supplied chemical evidence controls which oxidation states are actually stable. For example, [Ar]3d^54s^2 readily suggests +2 and +3 because both ions retain incomplete d subshells.
F. Using E⦵ values
Write both couples as reductions. The more positive reduction potential remains as reduction; reverse the other half-equation for oxidation. A positive E⦵_cell predicts feasibility for the combined equation as written under standard conditions. Kinetic barriers and non-standard conditions remain separate limitations.
Worked Examples
Modelled example 1
Permanganate Reduction in Acid
Problem
Study the worked solution
Balance oxygen and hydrogen
Method
Add four waters to products and eight protons to reactants.Reason
This balances four O and eight H atoms in acidic medium.Working
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O.Balance charge
Method
Add five electrons to the reactants.Reason
Left charge becomes -1 + 8-5 = +2, matching Mn²⁺.Working
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Common misconception 2
Dichromate Reduction Colour
Learner claim
Match colour to oxidation state
View solution step by step
Identify reactant colour
Method
Start with orange dichromate(VI).Reason
The species being reduced is Cr₂O₇²⁻.Working
Orange.Identify product colour
Method
End with green chromium(III).Reason
Reduction forms Cr³⁺.Working
Orange → green.
Challenge 3
Permanganate Oxidising Iron(II)
Equation transfer
Equalise and cancel electrons
Hints
Hint 1: oxidation
Hint 2: scale
View solution step by step
Scale iron oxidation
Method
Multiply Fe²⁺ → Fe³⁺ + e⁻ by five.Reason
This supplies five electrons.Working
5Fe²⁺ → 5Fe³⁺ + 5e⁻.Add and cancel
Method
Combine with permanganate reduction and remove electrons.Reason
Transferred electrons do not appear in the overall ionic equation.Working
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
Mind Stretchers
Mind stretcher 1Extension
In acidic solution, Cr₂O₇²⁻ oxidises I⁻ to I₂. Using the half-equations, write the balanced overall ionic equation.
Show Hint
Use the acidic dichromate reduction half-equation and scale the iodide oxidation half-equation to six electrons.
Show Answer
Mark scheme:
- Reduction (dichromate): Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
- Oxidation (iodide): 2I⁻ → I₂ + 2e⁻ (×3)
- Overall: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
Mind stretcher 2: Using electrode potentials with a transition-metal coupleExtension
Question. Standard reduction potentials are E⦵(Fe³⁺/Fe²⁺) = +0.77 V and E⦵(I₂/I⁻) = +0.54 V. Predict whether Fe³⁺ oxidises I⁻ under standard conditions and write the equation.
Show Hint
The more positive reduction potential stays as reduction; reverse the other half-equation and subtract.
Show Answer
The more positive iron couple is reduced, so Fe³⁺ gains electrons while I⁻ is oxidised. E⦵_cell = 0.77-0.54 = +0.23 V, so the reaction is feasible as written under standard conditions: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂