AHL: Electrolytic cells and standard electrode potentials

IB Chemistry HL· IB Chemistry AHL Topic 19.1, 19.2· 5 min read

1. Key Differences Between Cell Types★★☆☆☆⏱ 15 min

Both electrolytic and galvanic cells are electrochemical cells that involve redox reactions, but they have opposite purposes and electrode polarity. Galvanic cells generate electricity from spontaneous reactions, while electrolytic cells use electricity to drive non-spontaneous reactions.

📘 Definition

Anode and Cathode Consistency

Across all electrochemical cells, oxidation always occurs at the anode, and reduction always occurs at the cathode. Only the polarity of the electrodes changes between cell types.

Example:

In electrolytic cells, the anode is positive (to attract anions) and the cathode is negative (to attract cations), opposite to galvanic cells.

Property	Electrolytic Cell	Galvanic Cell
Redox spontaneity	Non-spontaneous	Spontaneous
Energy flow	External input required	Energy output produced
Anode polarity	Positive (+)	Negative (-)
Cathode polarity	Negative (-)	Positive (+)

Exam tip:

IB examiners often test whether you can correctly state the polarity of electrodes in electrolytic cells, don't mix it up with galvanic cells.

2. Predicting Reactions with Standard E°★★★☆☆⏱ 20 min

Standard electrode potential ( $E^{\circ}$ ) measures the tendency of a half-reaction to undergo reduction. For electrolysis, we use these values to determine which species will react at each electrode when multiple options are present.

At the cathode (reduction): The species with the most positive $E^{\circ}$ will be reduced first
At the anode (oxidation): The species with the most negative $E^{\circ}$ will be oxidized first

📐 Worked Example

Given two possible cathode half-reactions for electrolysis of aqueous copper(II) sulfate: $C u^{2 +} (a q) + 2 e^{-} \to C u (s) E^{\circ} = + 0.34 V$ and $2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q) E^{\circ} = - 0.83 V$ . Predict which species is reduced.

1
Recall that the species with the most positive $E^{\circ}$ has the highest tendency to reduce.
2
Compare the two values:
3
$+0.34\ V > -0.83\ V$
4
Conclusion: Copper(II) ions are reduced to solid copper metal at the cathode.

✓ Quick check

Check your understanding of prediction rules

Which species oxidizes first at the anode, given $E_{C l_{2} /2 C l^{-}}^{\circ} = + 1.36 V$ and $E_{O_{2} / H_{2} O}^{\circ} = + 1.23 V$ ?
- Chloride ions
- Water
Reveal answer
Chloride ions —
Correct! Oxidizing the chloride gives a reversed E° of -1.36 V, which is more negative than water's -1.23 V, so chloride oxidizes first.

3. Products of Electrolysis★★★★☆HL only⏱ 25 min

The products of electrolysis depend on whether the electrolyte is molten or aqueous. Molten electrolytes only have ions from the salt, so products are straightforward. Aqueous electrolytes also have water, which can be oxidized or reduced, leading to different products.

📐 Worked Example

Predict the products of electrolysis of molten magnesium chloride.

1
1. Identify all species present: Only molten $M g^{2 +}$ and $C l^{-}$ (no water).
2
1. Assign reactions: $M g^{2 +}$ migrates to negative cathode for reduction; $C l^{-}$ migrates to positive anode for oxidation.
3
1. Write half-equations:
4
$\text{Cathode: } Mg^{2+}(l) + 2e^- \rightarrow Mg(l)$
5
$\text{Anode: } 2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$
6
Products are liquid magnesium metal and chlorine gas.

4. Calculating Minimum Required Voltage★★★☆☆⏱ 15 min

🚫 No Calculator

For electrolysis, the overall reaction is non-spontaneous, so $E_{ce l l}^{\circ}$ will be negative. The external power supply must provide a voltage greater than the absolute value of this negative $E_{ce l l}^{\circ}$ to drive the reaction. The calculation of $E_{ce l l}^{\circ}$ follows the same rule as for galvanic cells:

E^\circ_{cell} = E^\circ_{cathode\ (reduction)} - E^\circ_{anode\ (reduction)}

📐 Worked Example

Calculate the minimum voltage required to electrolyze molten potassium iodide, given $E_{K^{+} / K}^{\circ} = - 2.93 V$ and $E_{I_{2} /2 I^{-}}^{\circ} = + 0.54 V$ .

1
1. Identify cathode and anode reactions: $K^{+}$ reduces at cathode, $I^{-}$ oxidizes at anode.
2
1. Substitute into the $E_{ce l l}^{\circ}$ formula:
3
$E^\circ_{cell} = (-2.93\ V) - (+0.54\ V) = -3.47\ V$
4
1. Minimum voltage equals the absolute value of $E_{ce l l}^{\circ}$ :
5
$V_{min} = |-3.47\ V| = 3.47\ V$

5. Common Pitfalls

Wrong move:

Swapping anode and cathode polarity for electrolytic cells

Why:

Students memorize polarity from galvanic cells and forget it reverses

Correct move:

Anode = oxidation (always), so anode is positive in electrolytic cells, cathode negative

Wrong move:

Predicting group 1/2 metal is reduced in aqueous solution

Why:

Students forget water has a more favorable reduction potential than group 1/2 metal ions

Correct move:

For aqueous group 1/2 metal salts, water reduces to hydrogen at the cathode, not the metal ion

Wrong move:

Reporting positive $E_{ce l l}^{\circ}$ for electrolytic reactions

Why:

Students confuse calculation rules with spontaneous galvanic cells

Correct move:

Non-spontaneous electrolytic reactions have negative $E_{ce l l}^{\circ}$ , minimum voltage is its absolute value

Wrong move:

Relying only on $E^{\circ}$ for aqueous halide oxidation

Why:

Students forget overpotential changes the predicted product for chloride, bromide and iodide

Correct move:

Aqueous halides (Cl⁻, Br⁻, I⁻) oxidize to the halogen at the anode, not water

Wrong move:

Treating molten and aqueous electrolytes the same

Why:

Students forget to add water's possible half-reactions for aqueous electrolytes

Correct move:

Always list all species present (including water) before predicting electrolysis products

6. Quick Reference Cheatsheet

Concept	Key Rule for Electrolytic Cells
Electrode role	Anode = oxidation, Cathode = reduction (always)
Electrode polarity	Anode = +, Cathode = -
Reduction preference	Most positive $E^{\circ}$ reduces first
Oxidation preference	Most negative $E^{\circ}$ oxidizes first
Minimum voltage	$V_{min} = ∣ E_{ce l l}^{\circ} ∣$ , $E_{ce l l}^{\circ} < 0$
Aqueous halide anode	Halide oxidizes, not water (overpotential)
Molten salt product	Metal at cathode, non-metal at anode

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.

2023 · Paper 2
Predict electrolysis products
2022 · Paper 1
Calculate minimum required voltage
2021 · Paper 2
Compare cell types

Going deeper

syllabusIB Chemistry HL Syllabus Topic 19

What's Next

This subtopic is a core AHL electrochemistry topic that is regularly tested across both Paper 1 and Paper 2 of IB Chemistry HL exams. Mastery of prediction rules and voltage calculations will prepare you for both qualitative and quantitative questions on electrolysis. Next, you can explore Faraday's laws of electrolysis, which let you calculate the mass of product formed from current and time, or review galvanic cells to reinforce your understanding of the differences between the two electrochemical cell types.