Predicting redox reactions, limitations and disproportionation

Feasibility (spec 16.10)

A reaction is thermodynamically feasible if E°cell is positive. Method:

1. Find the two half-equations and their E° values.

2. The half-equation with the more positive E° goes forward (reduction); the other goes backward (oxidation).

3. E°cell = E°(reduced) - E°(oxidised). If the proposed reaction gives a positive value, it is feasible.

Example: Cl2 + 2Fe2+ -> 2Cl- + 2Fe3+: E°cell = +1.36 - (+0.77) = +0.59 V, feasible. I2 + 2Fe2+ -> 2I- + 2Fe3+: +0.54 - 0.77 = -0.23 V, not feasible (the reverse reaction is).

Links to entropy and K (spec 16.11)

E°cell is directly proportional to ΔStotal and to ln K. A large positive E°cell (more than about +0.6 V) means K is very large and the reaction goes essentially to completion; a small positive value means an equilibrium mixture.

Limitations (spec 16.12)

• Kinetics: a feasible reaction may have such a high activation energy that it is too slow to observe. Copper does not react with dilute sulfuric acid (not feasible), but many feasible reactions, such as acidified dichromate oxidising water, are also not seen because they are very slow.
• Non-standard conditions: real concentrations, temperatures and pressures differ from standard, which changes E. A reaction with a small negative E°cell may happen if conditions shift the equilibria, for example by using concentrated reagents.

Disproportionation (spec 16.14)

In disproportionation one species is simultaneously oxidised and reduced. Test it with the two half-equations involving that species:

Cu+ + e- -> Cu, E° = +0.52 V; Cu2+ + e- -> Cu+, E° = +0.15 V

2Cu+ -> Cu + Cu2+: E°cell = +0.52 - (+0.15) = +0.37 V, so copper(I) disproportionates in aqueous solution.

Worked example

Will acidified manganate(VII) oxidise chloride ions to chlorine? MnO4-/Mn2+ +1.51 V; Cl2/Cl- +1.36 V.

1. MnO4- is reduced, Cl- is oxidised.
2. E°cell = +1.51 - (+1.36) = +0.15 V

Answer: Yes, it is feasible, which is why HCl is not used to acidify manganate(VII) titrations

Check yourself

1. Calculate E°cell for Cl2 + 2Fe2+ -> 2Cl- + 2Fe3+. (Cl2/Cl- +1.36 V; Fe3+/Fe2+ +0.77 V)
   Show answer

   +0.59 V. E°cell = +1.36 - (+0.77) = +0.59 V, so the reaction is feasible.

2. Can iodine oxidise iron(II) ions to iron(III) under standard conditions? (I2/I- +0.54 V; Fe3+/Fe2+ +0.77 V)
   • No, because E°cell = -0.23 V
   • Yes, because E°cell = +0.23 V
   • Yes, because iodine is a halogen
   • No, because E°cell = -1.31 V
   Show answer

   No, because E°cell = -0.23 V. +0.54 - 0.77 = -0.23 V; in fact Fe3+ oxidises I-.

3. Calculate E°cell for 2Cu+(aq) -> Cu(s) + Cu2+(aq). (Cu+/Cu +0.52 V; Cu2+/Cu+ +0.15 V)
   Show answer

   +0.37 V. E°cell = +0.52 - (+0.15) = +0.37 V: disproportionation is feasible.

4. What is disproportionation?
   • A reaction in which two different elements are oxidised
   • A reaction in which one species is simultaneously oxidised and reduced
   • The breaking of a complex ion
   • A reaction that is not feasible
   Show answer

   A reaction in which one species is simultaneously oxidised and reduced. For example 2Cu+ -> Cu + Cu2+: Cu+ goes to 0 and to +2.

5. A redox reaction has E°cell = +0.80 V, but no reaction is seen when the solutions are mixed. What is the most likely reason?
   • E°cell must be negative for a reaction to occur
   • The reaction is endothermic
   • The equilibrium constant is very small
   • The activation energy is very high, so the reaction is too slow
   Show answer

   The activation energy is very high, so the reaction is too slow. E° values say whether a reaction can happen, not how fast.

6. How is E°cell related to ΔStotal?
   • E°cell is inversely proportional to ΔStotal
   • E°cell is directly proportional to ΔStotal
   • They are unrelated
   • E°cell equals ΔSsurroundings
   Show answer

   E°cell is directly proportional to ΔStotal. Both are proportional to ln K, so a positive E°cell matches a positive ΔStotal.

7. Why is hydrochloric acid not used to acidify potassium manganate(VII) in titrations? (MnO4-/Mn2+ +1.51 V; Cl2/Cl- +1.36 V)
   • HCl is a weak acid
   • Chloride ions catalyse the reaction
   • HCl reacts with Fe2+
   • Manganate(VII) can oxidise chloride ions to chlorine, using up extra manganate(VII)
   Show answer

   Manganate(VII) can oxidise chloride ions to chlorine, using up extra manganate(VII). E°cell = +0.15 V, so some MnO4- is used up oxidising Cl-, giving a titre that is too large.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.