Applications of electrode potentials

Chemistry· Unit 19: Further electrochemistry, Subtopic 3· 30 min read

1. Calculating Standard Cell Potential★★☆☆☆⏱ 10 min

📘 Definition

Standard Cell Potential

$E_{ce l l}^{\circ}$

The potential difference between the cathode (reduction) and anode (oxidation) of an electrochemical cell under standard conditions.

Example:

For a Zn/Cu cell, $E_{ce l l}^{\circ} = E^{\circ} (C u^{2 +} / C u) - E^{\circ} (Z n^{2 +} / Z n)$

To calculate $E_{ce l l}^{\circ}$ , follow the convention that reduction always occurs at the cathode, and oxidation always occurs at the anode. The core formula is:

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

📐 Worked Example

Given $E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77 V$ and $E^{\circ} (B r_{2} / B r^{-}) = + 1.07 V$ , calculate $E_{ce l l}^{\circ}$ for the reaction where $F e^{2 +}$ is oxidised by $B r_{2}$ .

1
Step 1: Identify oxidation and reduction half-equations. $F e^{2 +}$ is oxidised (anode) and $B r_{2}$ is reduced (cathode).
2
Step 2: Substitute values into the formula:
3
$E^\circ_{cell} = +1.07 - (+0.77) = +0.30\ V$

Exam tip:

Always confirm which half-cell is the anode and which is the cathode before subtracting, to avoid sign errors.

2. Predicting Redox Reaction Spontaneity★★☆☆☆⏱ 15 min

There is a direct relationship between $E_{ce l l}^{\circ}$ and the standard Gibbs free energy change $Δ G^{\circ}$ , which tells us if a reaction is spontaneous:

\Delta G^\circ = -nFE^\circ_{cell}

Where $n$ is the moles of electrons transferred and $F$ is Faraday's constant. From this relationship, we get two simple rules:

If $E_{ce l l}^{\circ} > 0$ , $Δ G^{\circ} < 0$ : the forward reaction is spontaneous under standard conditions
If $E_{ce l l}^{\circ} < 0$ , $Δ G^{\circ} > 0$ : the forward reaction is non-spontaneous, reverse reaction is spontaneous

📐 Worked Example

Predict if magnesium metal will displace lead ions from solution, given $E^{\circ} (M g^{2 +} / M g) = - 2.37 V$ and $E^{\circ} (P b^{2 +} / P b) = - 0.13 V$ .

1
Step 1: Write the expected reaction: $M g (s) + P b^{2 +} (a q) \to M g^{2 +} (a q) + P b (s)$
2
Step 2: Identify oxidation (Mg, anode: $E^{\circ} = - 2.37 V$ ) and reduction (Pb²+, cathode: $E^{\circ} = - 0.13 V$ )
3
Step 3: Calculate $E_{ce l l}^{\circ}$ :
4
$E^\circ_{cell} = -0.13 - (-2.37) = +2.24\ V$
5
Step 4: Conclusion: $E_{ce l l}^{\circ}$ is positive, so the reaction is spontaneous, magnesium displaces lead.

✓ Quick check

Test your understanding:

What does $E_{ce l l}^{\circ} = - 0.45 V$ mean for the forward reaction?
- Forward reaction is spontaneous under standard conditions
- Reverse reaction is spontaneous under standard conditions
- No reaction can ever occur
- Reaction will be spontaneous at higher temperature
Reveal answer
1 —
Negative $E_{ce l l}^{\circ}$ gives positive $Δ G^{\circ}$ , so forward is non-spontaneous, reverse is spontaneous.

3. Predicting Products of Electrolysis★★★☆☆⏱ 15 min

For electrolysis with inert electrodes, we can use $E^{\circ}$ values to predict which species is discharged at each electrode, for similar concentrations of ions:

Cathode (negative electrode, reduction): The species with the most positive $E^{\circ}$ is reduced preferentially
Anode (positive electrode, oxidation): The species with the most negative $E^{\circ}$ is oxidised preferentially

📐 Worked Example

Predict the product at the cathode during electrolysis of dilute aqueous silver nitrate, given $E^{\circ} (A g^{+} / A g) = + 0.80 V$ and $E^{\circ} (2 H^{+} / H_{2}) = 0.00 V$ .

1
Step 1: List all cations available for reduction at the cathode: $A g^{+} (a q)$ and $H^{+} (a q)$ from water.
2
Step 2: Compare $E^{\circ}$ values: $A g^{+} / A g$ (+0.80 V) is more positive than $H^{+} / H_{2}$ (0.00 V).
3
Step 3: Conclusion: Silver ions are preferentially reduced, so solid silver metal is deposited at the cathode.

4. Limitations of E° Predictions★★★☆☆⏱ 10 min

Predictions based on standard $E^{\circ}$ values can fail for three key reasons:

Non-standard conditions: $E^{\circ}$ is only valid for 1 mol dm⁻³ concentration, 298 K and 1 atm pressure. Deviations change the actual potential.
Kinetics: $E^{\circ}$ only describes thermodynamics (spontaneity), not rate. A spontaneous reaction may have very high activation energy and proceed too slowly to observe.
Overpotential: Extra voltage is required for gas discharge at electrodes, which can change the product of electrolysis.

📐 Worked Example

Explain why chlorine is produced at the anode during electrolysis of concentrated brine (NaCl), even though $E^{\circ} (O_{2} / H_{2} O / O H^{-}) = + 0.40 V$ and $E^{\circ} (C l_{2} / C l^{-}) = + 1.36 V$ .

1
Based on $E^{\circ}$ values, OH- (with more negative $E^{\circ}$ ) should be oxidised preferentially to oxygen.
2
In concentrated brine, chloride ion concentration is much higher than 1 mol dm⁻³, and there is a high overpotential for oxygen discharge.
3
These effects make chloride oxidation to chlorine favourable, despite the standard $E^{\circ}$ prediction.

5. Common Pitfalls

Wrong move:

Calculating $E_{ce l l}^{\circ}$ as $E_{an o d e}^{\circ} - E_{c a t h o d e}^{\circ}$ (reversed order of subtraction)

Why:

This gives the wrong sign for $E_{ce l l}^{\circ}$ , leading to incorrect spontaneity predictions

Correct move:

Always use $E_{ce l l}^{\circ} = E^{\circ} (cathode, reduction) - E^{\circ} (anode, oxidation)$

Wrong move:

Assuming a positive $E_{ce l l}^{\circ}$ means the reaction will occur at an observable rate

Why:

$E^{\circ}$ only describes thermodynamic spontaneity, not the kinetics (rate) of reaction

Correct move:

Remember that high activation energy can make a spontaneous reaction too slow to observe

Wrong move:

Forgetting to include water when predicting products of aqueous electrolysis

Why:

Water can be oxidised or reduced, so it must always be included in $E^{\circ}$ comparisons

Correct move:

Always add the $E^{\circ}$ values for reduction and oxidation of water to your comparison

Wrong move:

Picking the most positive $E^{\circ}$ species for oxidation at the anode

Why:

Oxidation is the reverse of reduction, so the opposite rule applies

Correct move:

For oxidation at the anode, the species with the most negative (least positive) $E^{\circ}$ is oxidised preferentially

Wrong move:

Assuming $E^{\circ}$ predictions work for all concentrations and temperatures

Why:

$E^{\circ}$ values are only valid under standard conditions

Correct move:

Check if the reaction is under standard conditions before making a prediction

6. Quick Reference Cheatsheet

Concept	Rule/Formula	Interpretation
Calculate E°cell	$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$	Units: volts (V)
Spontaneous forward reaction	$E_{ce l l}^{\circ} > 0$	ΔG° < 0, spontaneous under standard conditions
Non-spontaneous forward reaction	$E_{ce l l}^{\circ} < 0$	ΔG° > 0, reverse reaction is spontaneous
Cathode (reduction)	Most positive E°	Reduced preferentially for similar concentrations
Anode (oxidation)	Most negative E°	Oxidised preferentially for similar concentrations
Common limitations	Non-standard conditions, activation energy, overpotential	Predictions may not match actual outcome

7. Frequently Asked

Why don't E° values always correctly predict reaction outcomes?

Standard E° values are only valid under standard conditions. Non-standard concentrations, non-standard temperatures, high activation energy, and overpotential effects can all change the actual outcome of a reaction.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Predict reaction spontaneity from E° values
2023 · 13
Calculate E°cell and predict electrolysis product
2024 · 21
Discuss limitations of E° predictions

Going deeper

syllabusCIE 9701 2022-2024 SyllabusCheck Unit 19 for full learning outcomes

What's Next

This subtopic connects electrochemistry to thermodynamics, and is a core foundation for all further electrochemistry content in CIE A-level Chemistry. Mastery of this content is essential for both multiple-choice and extended response questions, which frequently ask for spontaneity predictions, electrolysis product identification, and discussions of prediction limitations. Understanding how electrode potentials relate to reaction spontaneity also helps you connect concepts from redox, energetics, and electrochemistry across the syllabus. Next, you can explore how non-standard conditions affect cell potential, and learn about industrial electrochemical processes.