Cell Potentials and Spontaneity
Calculate a cell potential and use it to predict whether a reaction is feasible.
On this page
This lesson turns two E⦵ values into a decision: identify anode/cathode, calculate E⦵_cell, and link the sign to feasibility under standard conditions.
This page is really testing:
- Can you choose cathode/anode correctly from two reduction potentials?
- Can you compute E⦵_cell with the correct sign in one line?
- Can you translate the sign into exam language: “feasible under standard conditions”?
Definitions (Must Know)
A. Standard cell potential, E⦵_cell
The standard cell potential, E⦵_cell, is the potential difference between the two half-cells of a cell under standard conditions.
B. Anode and cathode (galvanic cells)
- Anode: oxidation occurs.
- Cathode: reduction occurs.
Key Ideas (What Earns Marks)
- Use reduction potentials from the data booklet.
- The half-cell with the more positive E⦵ is the cathode (reduction occurs there).
- Electrons flow externally from anode → cathode.
- Calculation: E⦵_cell = E⦵_cathode - E⦵_anode
- Under standard conditions, a positive E⦵_cell indicates the cell reaction is feasible (thermodynamically favourable).
Because the cathode is where reduction occurs, therefore the half-cell with the more positive E⦵ is the cathode.
Data table
| Cell / pair | E°cell |
|---|---|
| Zn/Cu (Daniell) | 1.1 |
| Ag/Cu | 0.46 |
| Fe3+/Fe2+ vs I2/I- | 0.23 |
| Cu/Zn (reverse) | -1.1 |
Detailed Explanations
A. Workflow: E⦵_cell, direction, and feasibility (standard conditions)
- Write down the two E⦵ values from the data booklet (both are reductions).
- The more positive E⦵ stays as reduction → cathode.
- The other half-equation is reversed → oxidation at anode.
- Calculate: E⦵_cell = E⦵_cathode - E⦵_anode
- Interpret the sign:
- E⦵_cell > 0: feasible under standard conditions
- E⦵_cell < 0: not feasible under standard conditions (reverse is feasible)
Mini example: If E⦵(Cu²⁺/Cu) = +0.34 and E⦵(Zn²⁺/Zn) = -0.76, then Cu is the cathode and E⦵_cell = 0.34-(-0.76) = +1.10 V.
B. What the sign means (and what it does not mean)
- E⦵_cell > 0: feasible under standard conditions.
- E⦵_cell < 0: not feasible under standard conditions (the reverse reaction would be feasible).
But:
- changing concentration / pressure / temperature can change feasibility (this is why cells can be rechargeable or vary in voltage).
Link to energetics wording (optional but useful): Δ G⦵ = -nFE⦵_cell.
C. Worked method you can reuse (30-second structure)
Write this sequence every time:
- “More positive reduction potential is the cathode.”
- “Other half-cell is the anode.”
- “E⦵_cell = E⦵_cathode - E⦵_anode = …”
- “Since E⦵_cell is [positive/negative], the forward reaction is [feasible/not feasible] under standard conditions.”
D. Limits of feasibility predictions
A positive E⦵_cell is a thermodynamic prediction for the reaction as written under standard conditions. It does not guarantee an observable rate: activation energy, passivation or another kinetic barrier may make a feasible reaction very slow. Actual concentration, gas pressure and temperature may also differ from standard conditions, so the actual cell potential need not equal E⦵_cell.
E. Qualitative concentration effects
For a reduction written as Ox + ne⁻ ⇌ Red, increasing the concentration of the oxidised aqueous species generally favours reduction and makes the electrode potential more positive. Increasing the concentration of the reduced aqueous species generally opposes reduction and makes it less positive. Use the actual half-equation to identify which side is changed; do not memorise “higher concentration means higher potential” without naming the species.
Choose two half-cells, read E_cell and the direction of electron flow, then dilute the right-hand ions under More settings and watch that electrode’s potential move.
Half-cells Zn²⁺/Zn (left) and Cu²⁺/Cu (right, 1.00 mol/dm³) joined by a salt bridge. The voltmeter reads +1.10 V. The right electrode is positive. Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
- Charge, Q = It
- 0 C
- n(e⁻) = Q/F
- 0 mol
- At the cathode
- —
- At the anode
- —
- Voltmeter
- 1.10 V
- Electrons flow
- left → right
- ΔG = −nFE
- −212 kJ/mol
Try this
0 of 4 doneSwitch on with a solid salt (heat off), then melt it and switch on again. (not done yet)
A solid ionic compound does not conduct: its ions are held in a lattice. Once molten, the ions are free to move to the electrodes.
Electrolyse dilute, then concentrated, sodium chloride. Compare the gas at the anode. (not done yet)
In dilute solution OH⁻ is discharged and oxygen forms (half the volume of hydrogen). In concentrated solution the many Cl⁻ ions are discharged instead, giving chlorine.
In copper(II) sulfate, record the cathode's gain in mass at three different charges. (not done yet)
The mass of copper is proportional to the charge: 2 mol of electrons (193 000 C) deposit 1 mol (63.5 g) of copper, whatever the current.
Build a simple cell that gives the largest voltage you can. (not done yet)
The further apart the metals are in the reactivity series, the larger the voltage. The more reactive metal is the negative electrode: it loses electrons.
Your readings
| # | I / A | t / s | Q / C | Δm(cathode) / g | Remove |
|---|---|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | |||||
Worked Examples
Modelled example 1
Zn/Cu Standard Cell Potential
Problem
Study the worked solution
Choose the cathode
Method
Use copper as the reduction half-cell.Reason
The more positive reduction potential runs as reduction.Working
Cathode: Cu²⁺/Cu, + 0.34 V.Choose the anode
Method
Use zinc as oxidation.Reason
The lower reduction potential runs in reverse at the anode.Working
Anode reduction value: -0.76 V.Subtract values
Method
Calculate cathode minus anode reduction potential.Reason
The negative anode value must remain inside the subtraction.Working
E⦵_cell = +0.34-(-0.76) = +1.10 V.
Guided practice 2
Electron Flow in the Zn/Cu Cell
Problem
Track where electrons are made and used
Hints
Hint 1: production
Hint 2: rule
View solution step by step
Locate electron production
Method
Oxidise zinc at the anode.Reason
Zn → Zn²⁺ + 2e⁻ releases electrons.Working
Zinc electrode is the electron source.State external flow
Method
Send electrons from zinc to copper.Reason
Copper(II) reduction consumes them at the cathode.Working
Electron flow: Zn anode → Cu cathode.
Common misconception 3
Agents in the Zn/Cu Cell
Learner claim
Name agents by what they do to the other species
View solution step by step
Identify the oxidising agent
Method
Name Cu²⁺.Reason
It gains electrons and thereby oxidises zinc.Working
Cu²⁺ + 2e⁻ → Cu.Identify the reducing agent
Method
Name Zn(s).Reason
It loses electrons and thereby reduces copper(II) ions.Working
Zn → Zn²⁺ + 2e⁻.
Challenge 4
Fe³⁺/Fe²⁺ and I₂/I⁻ Cell
Full-method transfer
Commit to direction before calculation
Hints
Hint 1: cathode
Hint 2: agents
View solution step by step
Assign directions
Method
Reduce Fe³⁺ at the cathode and oxidise I⁻ at the anode.Reason
+ 0.77 V is more positive than + 0.54 V.Working
2I⁻ → I₂ + 2e⁻ at the anode.Calculate and judge
Method
Calculate + 0.23 V and call the forward reaction feasible under standard conditions.Reason
E⦵_cell = 0.77-0.54 is positive.Working
E⦵_cell = +0.23 V.Name the agents
Method
Name Fe³⁺ as oxidising agent and I⁻ as reducing agent.Reason
The former is reduced; the latter is oxidised.Working
OA: Fe³⁺; RA: I⁻.
Common Mistakes
- Adding the two E⦵ values (use subtraction).
- Picking the anode as “more positive” (it’s the opposite: reduction is more positive).
- Forgetting that E⦵ values are for reductions, even if you’re describing oxidation at the anode.
- Multiplying E⦵ when balancing electrons (never do this; E⦵ is an intensive quantity).
- Writing “spontaneous” with no condition. In this context, state under standard conditions.
Exam Tips
- When asked for electron flow direction, name the electrodes and the species:
- “electrons flow from the Zn electrode to the Cu electrode”.
- If asked which is the oxidising agent: it is the species being reduced (at the cathode).
- If asked which is the reducing agent: it is the species being oxidised (at the anode).
- Prefer “feasible/thermodynamically favourable under standard conditions” over vague words like “reacts fast”.
A. Phrase bank for short conclusions
- “The more positive reduction potential is the cathode, so reduction occurs there.”
- “E⦵_cell is positive, therefore the reaction is feasible under standard conditions.”
- “A negative E⦵_cell means the reverse direction is feasible under standard conditions.”
- “This is a thermodynamic statement; it does not tell us the reaction rate.”
Mind Stretchers
Mind stretcher 1Extension
Explain why the oxidising agent is associated with the more positive E⦵ value.
Show Hint
Link a more positive reduction potential to electron gain, then use the definition of an oxidising agent.
Show Answer
Mark scheme:
- A more positive E⦵ means a stronger tendency to be reduced (gain electrons).
- The species that gains electrons oxidises the other species, so it acts as the oxidising agent.
Mind stretcher 2: Separating standard prediction from actual behaviourExtension
Question. A reaction has a small positive E⦵_cell, but no visible change occurs when the reagents are mixed. One reactant is then made much more concentrated and the cell voltage changes. Explain why neither observation contradicts the original data-booklet prediction.
Show Hint
Separate thermodynamics from kinetics, then identify which condition no longer matches the standard-state symbol.
Show Answer
A positive E⦵_cell predicts thermodynamic feasibility only under standard conditions; it does not predict a fast rate, so a high activation energy can prevent visible change. Changing concentration makes the conditions non-standard and changes the relevant electrode potential qualitatively. The actual cell voltage can therefore differ from E⦵_cell, and a sufficiently small standard driving force may even be overcome.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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