Predict qualitative concentration effects on electrode potential
Use the redox couple and the reaction quotient to predict how changing aqueous ion concentrations shifts electrode potential.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 24. Treat all tabulated electrode potentials as reduction potentials; state conditions, phases, signs and units. Written practice is self-checked
Before you begin
- Use AS redox half-equations, oxidation states, mole ratios, current and charge.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For Ox + ne⁻ ⇌ Red, which concentration ratio controls the qualitative shift?
What you will learn
- Use the oxidised and reduced forms in a half-equation
- Predict how changing concentration shifts E
- Distinguish concentration effects from changing the standard potential
Concentration changes the actual potential, not E°
E° is defined for standard conditions. If aqueous concentrations change, the electrode potential E changes; the tabulated E° remains the standard reference value.
For Ox + ne⁻ ⇌ Red, increasing [Ox] relative to [Red] tends to make reduction more favourable and E more positive. Increasing [Red] relative to [Ox] tends to shift E in the negative direction, assuming other conditions are fixed.
Treat pure solids and liquids separately
Pure solids and liquids do not appear in the concentration ratio for the electrode potential. For Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), the ion concentration changes E while the pure copper solid has activity treated as constant.
For Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), compare [Fe³⁺]/[Fe²⁺]. Increasing the oxidised-to-reduced ratio makes the reduction potential more positive; changing both concentrations in the same proportion leaves the ratio unchanged.
Pause and explain
If [Ox]/[Red] increases, which direction does the reduction potential tend to shift?
Worked example
For Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), predict qualitatively how lowering [Cu²⁺] affects E.
Show the worked solution
- Cu²⁺ is the oxidised form in the reduction half-equation.
- Lowering [Cu²⁺] reduces the oxidised-to-reduced ratio.
- Reduction is less favoured and E becomes less positive.
Answer Lowering [Cu²⁺] makes the Cu²⁺/Cu electrode potential less positive; E° is unchanged because only the actual concentration has changed.
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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.
What happens to E for Fe³⁺/Fe²⁺ if [Fe³⁺] increases while [Fe²⁺] stays constant?
Give me a hint
Compare the oxidised-to-reduced ratio.
Compare my reasoning
- Fe³⁺ is the oxidised form.
- The ratio [Fe³⁺]/[Fe²⁺] increases.
- The reduction potential becomes more positive.
E becomes more positive; E° itself is unchanged.
Look for these in your work
- Oxidised form identified
- Ratio direction correct
- E and E° distinguished
For Cu²⁺/Cu, what happens to E if [Cu²⁺] is multiplied by ten at constant temperature?
Give me a hint
The pure Cu(s) term is constant; the oxidised ion concentration changes.
Compare my reasoning
- [Cu²⁺] increases tenfold.
- The oxidised-to-reduced ratio increases.
- E shifts in the positive direction.
E becomes more positive. Cu(s) does not enter the concentration ratio; only [Cu²⁺] changes.
Look for these in your work
- Aqueous species used
- Solid omitted
- Potential direction correct
A student reports that doubling both [Fe³⁺] and [Fe²⁺] doubles E. Explain why that conclusion is not supported by the qualitative ratio.
Give me a hint
Check what happens to [Fe³⁺]/[Fe²⁺].
Compare my reasoning
- Both numerator and denominator double.
- Their ratio is unchanged.
- The Nernst concentration term is therefore unchanged at fixed temperature.
The ratio [Fe³⁺]/[Fe²⁺] is unchanged when both concentrations are doubled, so the concentration-dependent potential term does not change; E° is also unchanged.
Look for these in your work
- Ratio evaluated
- Equal scaling recognised
- E and E° response distinguished
Common mistakes
- Changing E° when only the solution concentration changes
- Including a pure metal solid in the concentration ratio
- Reversing oxidised and reduced forms
- Predicting a shift without stating which concentration changed
What can you explain now?
For Ox + ne⁻ ⇌ Red, which concentration ratio controls the qualitative shift?
Compare with the explanation
[Ox]/[Red]
Ox is reduced and Red is the reduced form.
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Make it stick
How does the oxidised-to-reduced concentration ratio shift E while leaving the standard value E° fixed?
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