Use standard potentials to predict transition-metal redox feasibility
Combine supplied standard reduction potentials to infer cell direction, feasibility and balanced redox equations.
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
When using tabulated E° values, how are the half-equations written?
What you will learn
- Use standard reduction potentials to compare redox couples
- Calculate standard cell potential with a consistent sign convention
- Predict feasibility and construct an equation from half-equations
Compare reduction potentials as reductions
A standard electrode potential is tabulated as a reduction. For a proposed cell reaction, the couple with the more positive E° is more likely to undergo reduction; the other half-equation is reversed to describe oxidation.
Using reduction potentials, E°cell = E°(cathode, reduction) − E°(anode, reduction). A positive E°cell predicts thermodynamic feasibility under the stated standard conditions; a negative value predicts the reverse direction is favoured. This does not tell you the reaction rate.
Balance electrons before adding half-equations
Write each relevant half-equation with electrons, then multiply by integers so electrons cancel. Combine the species and check atoms and total charge. If the question supplies potential values, use them; do not rely on a memorised potential for an unspecified couple.
A feasible equation can still be slow. Standard potential predicts thermodynamic direction, not kinetic speed or whether an experiment will be safe or practical.
Pause and explain
What does a positive standard cell potential predict?
Worked example
Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Cu²⁺/Cu) = +0.34 V, predict the spontaneous direction and calculate E°cell.
Show the worked solution
- The more positive Fe³⁺/Fe²⁺ couple is reduced.
- Reverse the copper half-equation to show Cu oxidation.
- Subtract the anode reduction potential from the cathode reduction potential.
Answer E°cell = +0.77 − (+0.34) = +0.43 V. The predicted direction is 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺; the positive value indicates feasibility under standard conditions.
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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.
Use E°(Ag⁺/Ag) = +0.80 V and E°(Fe³⁺/Fe²⁺) = +0.77 V to calculate E°cell when Ag⁺ is reduced and Fe²⁺ is oxidised.
Give me a hint
Both values are reduction potentials; use E°cathode − E°anode.
Compare my reasoning
- Ag⁺/Ag is the reduction half-cell.
- Fe³⁺/Fe²⁺ is written as a reduction in the table, though Fe²⁺ is oxidised here.
- Subtract +0.77 V from +0.80 V.
E°cell = +0.80 − (+0.77) = +0.03 V; the stated direction is feasible under standard conditions.
Look for these in your work
- Cathode selected
- Anode potential interpreted as reduction value
- Sign and result correct
Write the overall ionic equation when Fe²⁺ is oxidised by acidified MnO₄⁻ to Fe³⁺ and Mn²⁺.
Give me a hint
Balance oxygen and hydrogen in acid, then cancel electrons.
Compare my reasoning
- MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
- Fe²⁺ → Fe³⁺ + e⁻; multiply by five.
- Add and cancel electrons.
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
Look for these in your work
- Acidic half-equation balanced
- Electron ratio correct
- Atoms and charge balanced
A learner says a redox reaction with E°cell = +0.12 V will always proceed rapidly. Correct the conclusion.
Give me a hint
Separate thermodynamic feasibility from kinetics.
Compare my reasoning
- A positive E°cell supports the forward direction under standard conditions.
- Rate depends on activation energy and conditions.
- The potential does not specify reaction time.
The positive E°cell predicts thermodynamic feasibility in the stated direction under standard conditions. It does not establish a fast rate; kinetics and experimental conditions matter.
Look for these in your work
- Direction interpreted
- Rate distinguished
- Limitation stated
Common mistakes
- Reversing a tabulated reduction potential without changing its role
- Adding both reduction potentials without identifying oxidation and reduction
- Using a positive E°cell to claim the reaction is fast
- Failing to balance electrons in the overall equation
What can you explain now?
When using tabulated E° values, how are the half-equations written?
Compare with the explanation
As reductions
Standard electrode potentials are tabulated for reduction directions.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do reduction potentials give E°cell and what does the sign predict about direction versus rate?
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.