Topics

Predicting spontaneity from cell potential

The question this answers

Given any two half reactions, which way will the electrons go?
The answer comes from a single comparison and a single sign.

The rule

Combine the two half reactions so that one is reduced (the oxidant) and the other is oxidised (the reductant).
The cell potential is:

$E°_{cell} = E°(\text{species reduced}) - E°(\text{species oxidised})$

Some texts write this as $E°_{cathode} - E°_{anode}$ , which is the same statement, since reduction always happens at the cathode.

$E°_{cell}$	Meaning
Positive	The reaction is spontaneous as written — it happens on its own
Negative	Not spontaneous as written — the reverse reaction is spontaneous
Zero	The system is at equilibrium

The shortcut that never fails

On the table, a reaction is spontaneous when the oxidant is higher than the reductant.
Draw a line from the left-hand species higher up to the right-hand species lower down. If the arrow goes down-left to up-right, the reaction goes.
This is worth internalising because it removes sign errors entirely: anything on the left of the table will oxidise anything on the right of the table that sits below it.

Worked through: zinc and copper

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ , $E° = +0.34$ V
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ , $E° = -0.76$ V

Copper is higher, so Cu²⁺ is the oxidant and Zn is the reductant:

$E°_{cell} = (+0.34) - (-0.76) = +1.10 \text{ V}$

Positive, so the reaction is spontaneous:

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Testing the reverse confirms it: copper metal in zinc sulfate gives $E°_{cell} = (-0.76) - (+0.34) = -1.10$ V, so no reaction — which is exactly what is observed.

First half cell

Second half cell

Zn has the more negative E°, so it is oxidised at the anode: Zn → Zn²⁺ + 2e⁻

Cu²⁺ has the more positive E°, so it is reduced at the cathode: Cu²⁺ + 2e⁻ → Cu

E°cell = E°(reduced) − E°(oxidised) = +0.34 − (−0.76) = 1.10 V — positive, so the reaction is spontaneous.

Swapping the two dropdowns does not change which electrode is the anode: that is fixed by the E° values, not by which one you picked first.

What $E°_{cell}$ does NOT tell you

This distinction is worth marks at Excellence, and getting it wrong is worth losing them.

$E°_{cell}$ is a thermodynamic quantity. It tells you whether a reaction is energetically favourable.
It says nothing about rate. A reaction with a large positive $E°_{cell}$ may still be immeasurably slow if its activation energy is high.
- Hydrogen and oxygen have $E°_{cell} = +1.23$ V and will sit together indefinitely without a spark.
- Aluminium has $E° = -1.66$ V and should react vigorously with water; it does not, because a passivating oxide layer blocks the surface.
It applies only at standard conditions. Real concentrations differ, so a reaction predicted to be marginal may go either way in practice.
The honest summary: $E°_{cell}$ tells you what is possible, not what will happen.

Using it to explain observations

The criteria require you to link calculations to observations. The pattern is:

Calculate $E°_{cell}$ .
Predict whether a reaction occurs.
State what would be seen if it does — the colour change, the deposit, the gas.
If nothing is observed, say whether that is because $E°_{cell}$ is negative or because the rate is too slow, and justify which.

Worked ExamplePredicting whether a reaction occurs

Predict whether iodide ions will reduce iron(III) ions in aqueous solution. Calculate E°cell, state what would be observed, and comment on the reliability of the prediction.

$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ , $E° = +0.77$ V $\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$ , $E° = +0.54$ V

Step 1 — Decide which species is oxidised and which reduced

The question proposes that iodide reduces iron(III), so:

Fe³⁺ is reduced — it is the oxidant.
I⁻ is oxidised — it is the reductant, so the I₂/I⁻ couple runs in reverse.

Step 2 — Calculate $E°_{cell}$

$E°_{cell} = E°(\text{reduced}) - E°(\text{oxidised}) = (+0.77) - (+0.54) = +0.23 \text{ V}$

Step 3 — Interpret the sign

$E°_{cell}$ is positive, so the reaction is spontaneous as written.

This agrees with the table shortcut: Fe³⁺ sits above I⁻ on the table, and anything on the left will oxidise anything on the right below it.

Step 4 — Write the equations

Oxidation: $2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^-$

Reduction: $\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)$

Matching the electrons at 2 and adding:

$2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)$

Step 5 — State the observations

The yellow-brown Fe³⁺ solution changes towards the pale green of Fe²⁺.
A brown colour develops as I₂ forms, and this dominates visually.
Adding starch would give a blue-black colour, confirming the iodine.

Step 6 — Comment on reliability

$E°_{cell} = +0.23$ V is positive but small. Two cautions follow:

The value applies at standard conditions, 1 mol L⁻¹ in every species. At lower Fe³⁺ concentration or higher I⁻ concentration the actual cell potential shifts, and with a margin this small the reaction could be much less favourable than the standard value suggests.
A positive $E°_{cell}$ guarantees only that the reaction is thermodynamically favourable. In this case the reaction is also fast, because both species are ions in solution and no bonds need to be broken in a rate-limiting step — so the prediction is borne out in practice.

Answer: E°cell = +0.23 V, so iodide does reduce iron(III) spontaneously; the solution turns brown as I₂ forms, confirmed blue-black with starch.

Predicting spontaneity from cell potential

Covers AS91393

The question this answers

Given any two half reactions, which way will the electrons go?
The answer comes from a single comparison and a single sign.

The rule

Combine the two half reactions so that one is reduced (the oxidant) and the other is oxidised (the reductant).
The cell potential is:

$E°_{cell} = E°(\text{species reduced}) - E°(\text{species oxidised})$

Some texts write this as $E°_{cathode} - E°_{anode}$ , which is the same statement, since reduction always happens at the cathode.

$E°_{cell}$	Meaning
Positive	The reaction is spontaneous as written — it happens on its own
Negative	Not spontaneous as written — the reverse reaction is spontaneous
Zero	The system is at equilibrium

The shortcut that never fails

On the table, a reaction is spontaneous when the oxidant is higher than the reductant.
Draw a line from the left-hand species higher up to the right-hand species lower down. If the arrow goes down-left to up-right, the reaction goes.
This is worth internalising because it removes sign errors entirely: anything on the left of the table will oxidise anything on the right of the table that sits below it.

Worked through: zinc and copper

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ , $E° = +0.34$ V
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ , $E° = -0.76$ V

Copper is higher, so Cu²⁺ is the oxidant and Zn is the reductant:

$E°_{cell} = (+0.34) - (-0.76) = +1.10 \text{ V}$

Positive, so the reaction is spontaneous:

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Testing the reverse confirms it: copper metal in zinc sulfate gives $E°_{cell} = (-0.76) - (+0.34) = -1.10$ V, so no reaction — which is exactly what is observed.

First half cell

Second half cell

Zn has the more negative E°, so it is oxidised at the anode: Zn → Zn²⁺ + 2e⁻

Cu²⁺ has the more positive E°, so it is reduced at the cathode: Cu²⁺ + 2e⁻ → Cu

E°cell = E°(reduced) − E°(oxidised) = +0.34 − (−0.76) = 1.10 V — positive, so the reaction is spontaneous.

Swapping the two dropdowns does not change which electrode is the anode: that is fixed by the E° values, not by which one you picked first.

What $E°_{cell}$ does NOT tell you

This distinction is worth marks at Excellence, and getting it wrong is worth losing them.

$E°_{cell}$ is a thermodynamic quantity. It tells you whether a reaction is energetically favourable.
It says nothing about rate. A reaction with a large positive $E°_{cell}$ may still be immeasurably slow if its activation energy is high.
- Hydrogen and oxygen have $E°_{cell} = +1.23$ V and will sit together indefinitely without a spark.
- Aluminium has $E° = -1.66$ V and should react vigorously with water; it does not, because a passivating oxide layer blocks the surface.
It applies only at standard conditions. Real concentrations differ, so a reaction predicted to be marginal may go either way in practice.
The honest summary: $E°_{cell}$ tells you what is possible, not what will happen.

Using it to explain observations

The criteria require you to link calculations to observations. The pattern is:

Calculate $E°_{cell}$ .
Predict whether a reaction occurs.
State what would be seen if it does — the colour change, the deposit, the gas.
If nothing is observed, say whether that is because $E°_{cell}$ is negative or because the rate is too slow, and justify which.

Worked ExamplePredicting whether a reaction occurs

Predict whether iodide ions will reduce iron(III) ions in aqueous solution. Calculate E°cell, state what would be observed, and comment on the reliability of the prediction.

$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ , $E° = +0.77$ V $\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$ , $E° = +0.54$ V

Step 1 — Decide which species is oxidised and which reduced

The question proposes that iodide reduces iron(III), so:

Fe³⁺ is reduced — it is the oxidant.
I⁻ is oxidised — it is the reductant, so the I₂/I⁻ couple runs in reverse.

Step 2 — Calculate $E°_{cell}$

$E°_{cell} = E°(\text{reduced}) - E°(\text{oxidised}) = (+0.77) - (+0.54) = +0.23 \text{ V}$

Step 3 — Interpret the sign

$E°_{cell}$ is positive, so the reaction is spontaneous as written.

This agrees with the table shortcut: Fe³⁺ sits above I⁻ on the table, and anything on the left will oxidise anything on the right below it.

Step 4 — Write the equations

Oxidation: $2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^-$

Reduction: $\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)$

Matching the electrons at 2 and adding:

$2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)$

Step 5 — State the observations

The yellow-brown Fe³⁺ solution changes towards the pale green of Fe²⁺.
A brown colour develops as I₂ forms, and this dominates visually.
Adding starch would give a blue-black colour, confirming the iodine.

Step 6 — Comment on reliability

$E°_{cell} = +0.23$ V is positive but small. Two cautions follow:

The value applies at standard conditions, 1 mol L⁻¹ in every species. At lower Fe³⁺ concentration or higher I⁻ concentration the actual cell potential shifts, and with a margin this small the reaction could be much less favourable than the standard value suggests.
A positive $E°_{cell}$ guarantees only that the reaction is thermodynamically favourable. In this case the reaction is also fast, because both species are ions in solution and no bonds need to be broken in a rate-limiting step — so the prediction is borne out in practice.

Answer: E°cell = +0.23 V, so iodide does reduce iron(III) spontaneously; the solution turns brown as I₂ forms, confirmed blue-black with starch.

Predicting spontaneity from cell potential

The question this answers

The rule

The shortcut that never fails

Worked through: zinc and copper

What E°cellE°_{cell}E°cell​ does NOT tell you

Using it to explain observations

Step 1 — Decide which species is oxidised and which reduced

Step 2 — Calculate E°cellE°_{cell}E°cell​

Step 3 — Interpret the sign

Step 4 — Write the equations

Step 5 — State the observations

Step 6 — Comment on reliability

Predicting spontaneity from cell potential

The question this answers

The rule

The shortcut that never fails

Worked through: zinc and copper

What E°cellE°_{cell}E°cell​ does NOT tell you

Using it to explain observations

Step 1 — Decide which species is oxidised and which reduced

Step 2 — Calculate E°cellE°_{cell}E°cell​

Step 3 — Interpret the sign

Step 4 — Write the equations

Step 5 — State the observations

Step 6 — Comment on reliability

What $E°_{cell}$ does NOT tell you

Step 2 — Calculate $E°_{cell}$

What $E°_{cell}$ does NOT tell you

Step 2 — Calculate $E°_{cell}$