Cell Potentials and Spontaneity

Learn and apply Cell Potentials and Spontaneity in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Cell Potentials and Spontaneity: Orientation

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.

Detailed Explanations

A. Workflow: E⦵_cell, direction, and feasibility (standard conditions)

  1. Write down the two E⦵ values from the data booklet (both are reductions).
  2. The more positive E⦵ stays as reduction → cathode.
  3. The other half-equation is reversed → oxidation at anode.
  4. Calculate: E⦵_cell = E⦵_cathode - E⦵_anode
  5. 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:

  1. “More positive reduction potential is the cathode.”
  2. “Other half-cell is the anode.”
  3. “E⦵_cell = E⦵_cathode - E⦵_anode = …”
  4. “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.

Worked Examples

Modelled example 1

Zn/Cu Standard Cell Potential

Core

Problem

Given E⦵(Cu²⁺/Cu) = +0.34 V and E⦵(Zn²⁺/Zn) = -0.76 V, calculate E⦵_cell.
Study the worked solution
  1. 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.
  2. 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.
  3. 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

About 4 min

Problem

For the Zn/Cu cell in Example 1, state the direction of electron flow through the external circuit.

Track where electrons are made and used

Electron source
Flow

Hints

Hint 1: production
Electrons are released where oxidation occurs.
Hint 2: rule
External electron flow is anode to cathode.
View solution step by step
  1. Locate electron production

    Method

    Oxidise zinc at the anode.

    Reason

    Zn → Zn²⁺ + 2e⁻ releases electrons.

    Working

    Zinc electrode is the electron source.
  2. 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

Find and correct the mistake

Learner claim

A learner calls zinc the oxidising agent because zinc is oxidised. Correct both agent identities for the Zn/Cu cell.

Name agents by what they do to the other species

Oxidising agent
Reducing agent

View solution step by step
  1. Identify the oxidising agent

    Method

    Name Cu²⁺.

    Reason

    It gains electrons and thereby oxidises zinc.

    Working

    Cu²⁺ + 2e⁻ → Cu.
  2. 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

Minimal support

Full-method transfer

Given E⦵(Fe³⁺/Fe²⁺) = +0.77 V and E⦵(I₂/I⁻) = +0.54 V, state the cathode, calculate E⦵_cell, comment on forward feasibility under standard conditions, and identify both agents.

Commit to direction before calculation

Cathode couple
Cell potential
Forward reaction

Hints

Hint 1: cathode
The more positive reduction potential stays in the reduction direction.
Hint 2: agents
The reduced reactant is the oxidising agent; the oxidised reactant is the reducing agent.
View solution step by step
  1. 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.
  2. 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.
  3. 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⁻.

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.