19.7 Electrolytic Cells
PRE-HEALTH PREP
Electrolytic Cells
Molten vs. Aqueous Electrolysis
General Chemistry 19.7
Lesson Overview
Electrolytic cells drive non-spontaneous chemical reactions by supplying external voltage to force ion migration toward electrodes. In molten electrolysis, product prediction is straightforward — cations reduce at the cathode and anions oxidize at the anode — while aqueous electrolysis introduces competing half-reactions from water that must be resolved using standard reduction and oxidation potentials. The minimum voltage required equals the absolute value of the sum of the cathode reduction potential and the anode oxidation potential.
Table of Contents
What Are the 2 Types of Electrolysis?
Electrolytic cells use electrical energy to drive non-spontaneous chemical reactions — a process called electrolysis. The essential requirement is that the ionic compound's cations and anions must be mobile: they need to be free to migrate toward the electrodes. In a solid ionic compound, the ions are locked in a rigid crystal lattice and cannot move, so electrolysis is impossible.
Mobility is achieved in one of two ways: you can melt the compound into a liquid, or you can dissolve it in water. This gives rise to the 2 types of electrolysis:
- Molten electrolysis — the ionic compound is melted to a liquid. No water is present.
- Aqueous electrolysis — a soluble ionic compound is dissolved in water. Water is present.
That single difference — the presence or absence of water — has profound consequences for which products form at each electrode. This lesson focuses on how to predict those products qualitatively.
Two lessons, two goals: This lesson covers qualitative product prediction (which species form). The follow-up lesson covers quantitative electrolysis (how much product forms over time using Faraday's law).
How Does Molten Electrolysis Work?
Why High Temperatures Are Required
Ionic compounds have exceptionally high melting points — a direct consequence of the strong electrostatic forces holding the crystal lattice together. Melting an ionic compound requires enough thermal energy to overcome those forces entirely. For reference, this is the same phenomenon that makes molten lava so dramatic: rock is largely composed of ionic compounds, and it takes enormous temperatures to reduce it to liquid. The resulting liquid ionic compound is called a molten salt.
Predicting Products: Cation Goes to the Cathode, Anion Goes to the Anode
In molten electrolysis, the prediction is refreshingly straightforward. Consider the electrolysis of molten NaBr. The compound dissociates into Na⁺ and Br⁻.
- At the cathode (reduction): Na⁺ gains one electron and is reduced to elemental sodium metal, Na.
- At the anode (oxidation): Br⁻ loses electrons and is oxidized. Because bromine's elemental form is the diatomic molecule, two bromide ions combine: 2 Br⁻ → Br₂ + 2e⁻.
Example: Molten Electrolysis of NaBr
Cathode (reduction): Na⁺ + e⁻ → Na
Anode (oxidation): 2 Br⁻ → Br₂ + 2e⁻
Net result: Sodium metal forms at the cathode; diatomic bromine gas forms at the anode.
The rule for molten electrolysis of any binary salt is simply this: the cation is reduced to its elemental form at the cathode; the anion is oxidized to its elemental form at the anode. There is no ambiguity and no competition — only the two ions of the salt are present.
Memory aid: Cation reduced at the cathode. Both words start with "cat." The anion is oxidized at the anode — both start with "an."
Additional Example: Molten Electrolysis of AgCl
In the molten electrolysis of AgCl(l), the cation Ag+ is reduced to elemental silver at the cathode, and the anion Cl− is oxidized to elemental chlorine (Cl2) at the anode.
Cathode (reduction): Ag+ + e− → Ag
Anode (oxidation): 2Cl− → Cl2 + 2e−
Practice Problem (a): What are the products of electrolysis of NaBr(l)?
Answer: At the cathode, Na+ is reduced to elemental sodium metal (Na). At the anode, Br− is oxidized to diatomic bromine (Br2). Products: Na at the cathode and Br2 at the anode.
How Does Aqueous Electrolysis Work?
2 Competing Half-Reactions at Each Electrode
Aqueous electrolysis is far more convenient experimentally — dissolving an ionic compound in water at room temperature is much easier than heating it to hundreds or thousands of degrees. The tradeoff is that predicting the products becomes significantly more involved.
When water is present, it is not a passive bystander. Water can itself be reduced at the cathode and oxidized at the anode. This means both electrodes now face a competition between 2 possible half-reactions, and only one will actually occur at each electrode. Your task is to determine which one wins.
| Electrode | Option 1 (from dissolved salt) | Option 2 (from water) |
|---|---|---|
| Cathode (reduction) | Na⁺ + e⁻ → Na | 2H₂O + 2e⁻ → H₂ + 2OH⁻ |
| Anode (oxidation) | 2Br⁻ → Br₂ + 2e⁻ | 2H₂O → O₂ + 4H⁺ + 4e⁻ |
The selection principle is the same at both electrodes: whichever half-reaction requires the least energy — that is, whichever is the least non-spontaneous — is the one that occurs. In practical terms, you compare the relevant potentials and select the most favorable one.
| Na+ + 1e− → Na | E°red = −2.71 V |
|---|---|
| 2H2O + 2e− → H2 + 2OH− | E°red = −0.83 V |
| 2Br− → Br2 + 2e− | E°ox = −1.07 V |
| 2H2O → O2 + 4H+ + 4e− | E°ox = −1.23 V |
Practice Problem (b): What are the products of electrolysis of a solution of NaBr?
Cathode: Compare E°red for Na+ (−2.71 V) vs. water (−0.83 V). Water has the higher (less negative) reduction potential, so water is reduced, producing H2 gas and OH−.
Anode: Compare E°ox for Br− (−1.07 V) vs. water (−1.23 V). Bromide has the higher (less negative) oxidation potential, so bromide is oxidized, producing Br2. Products: H2 at the cathode and Br2 at the anode.
Cathode: Selecting the Reduction Half-Reaction
At the cathode, 2 reduction half-reactions compete. Using standard reduction potentials from a reference table:
- Na⁺ + e⁻ → Na E° = −2.71 V
- 2H₂O + 2e⁻ → H₂ + 2OH⁻ E° = −0.83 V
Because electrolytic cells are driven by an external power source, you want to accomplish the reaction with the minimum energy input — which means choosing the half-reaction with the highest (least negative) reduction potential. The cell will preferentially carry out whichever reduction is easier.
Cathode Selection for Aqueous NaBr
Comparing −2.71 V (Na⁺ reduction) and −0.83 V (water reduction): −0.83 V is less negative, so the reduction of water is easier. Water is reduced at the cathode, producing hydrogen gas and hydroxide ions.
Cathode product: H₂ gas (and OH⁻ in solution)
Key rule for the cathode: Select the reduction half-reaction with the highest reduction potential (most positive or least negative). This is the easiest reduction — it requires the least voltage from the external power source.
Anode: Selecting the Oxidation Half-Reaction
Converting Reduction Potentials to Oxidation Potentials
The anode requires oxidation, but standard reference tables list reduction potentials. To find the oxidation potential for any half-reaction, simply reverse the half-reaction and change the sign of its potential.
For the 2 oxidation candidates in aqueous NaBr:
- Bromide oxidation: The table gives Br₂ + 2e⁻ → 2Br⁻ at E° = +1.07 V. Reversed (oxidation direction): 2Br⁻ → Br₂ + 2e⁻, E°ox = −1.07 V.
- Water oxidation: The table gives O₂ + 4H⁺ + 4e⁻ → 2H₂O at E° = +1.23 V. Reversed: 2H₂O → O₂ + 4H⁺ + 4e⁻, E°ox = −1.23 V.
Selecting the Easier Oxidation
At the anode, the same principle applies: the half-reaction that is least non-spontaneous is the one that occurs. For oxidation potentials, select the most positive (or least negative) value.
Anode Selection for Aqueous NaBr
Comparing −1.07 V (Br⁻ oxidation) and −1.23 V (water oxidation): −1.07 V is less negative, so bromide oxidation is easier. Bromide is oxidized at the anode, producing diatomic bromine.
Anode product: Br₂
Molten vs. Aqueous Electrolysis: 2 Different Outcomes for NaBr
Notice how the presence of water changes the outcome at the cathode but not the anode in this case:
| Electrolysis Type | Cathode Product | Anode Product |
|---|---|---|
| Molten NaBr | Na (sodium metal) | Br₂ (bromine gas) |
| Aqueous NaBr | H₂ (hydrogen gas) | Br₂ (bromine gas) |
In aqueous solution, the reduction of water outcompetes sodium ion reduction, so no sodium metal is produced. The much lower reduction potential of Na⁺ (−2.71 V) means it is simply too difficult to reduce in the presence of an easier alternative. Bromine, however, is produced in both cases — its oxidation potential (−1.07 V) is less demanding than water oxidation (−1.23 V).
What Is the Minimum Voltage Required for Electrolysis?
Once you have identified the half-reactions at each electrode, you can calculate the overall cell potential and thereby determine how much voltage the external power source must supply.
For the aqueous electrolysis of NaBr:
- Cathode (reduction of water): E° = −0.83 V
- Anode (oxidation of Br⁻): E°ox = −1.07 V
- Overall E°cell = −0.83 V + (−1.07 V) = −1.90 V
A negative cell potential confirms that the reaction is non-spontaneous — it will not proceed on its own. The external power source must supply at least +1.90 V to overcome this energy barrier and drive the reaction forward.
Example: Battery Selection
A standard AA battery supplies 1.5 V. For this reaction (minimum 1.90 V required), a single AA battery is insufficient. However, two AA batteries wired in series supply 3.0 V — more than enough to drive the electrolysis. Even with this excess voltage, only the two selected half-reactions occur: water reduction at the cathode and bromide oxidation at the anode.
Overvoltage does not change the products: Supplying more voltage than the minimum does not cause additional or different half-reactions to occur. The electrode reactions remain fixed at the thermodynamically easiest options; excess voltage simply drives those same reactions faster.
7 Key Takeaways
- 2 types of electrolysis: Molten electrolysis (no water, high temperature) and aqueous electrolysis (water present, room temperature). The presence of water is the defining distinction.
- Molten electrolysis is easy to predict: The cation is always reduced to its elemental form at the cathode; the anion is always oxidized to its elemental form at the anode. No competition exists.
- Aqueous electrolysis involves competition: Water can be reduced at the cathode and oxidized at the anode, competing with the dissolved ions. Only one half-reaction occurs at each electrode.
- Cathode selection rule: Choose the reduction half-reaction with the highest (least negative) standard reduction potential — it is the easiest reduction and requires the least energy input.
- Anode selection rule: Convert reduction potentials to oxidation potentials by reversing the reaction and changing the sign, then choose the most positive (least negative) oxidation potential.
- Minimum voltage: Add the cathode reduction potential and the anode oxidation potential to find E°cell. The external power source must supply at least |E°cell| volts to drive the reaction. Supplying more voltage does not alter which products form.
- Practical tradeoff: Molten electrolysis is simple to predict but experimentally demanding (very high temperatures). Aqueous electrolysis is easy to perform but requires careful analysis to predict products correctly.