DAT General Chemistry: Predicting Electrolysis Products

DAT High-Yield Concept

Molten Electrolysis: Contains no water; cations reduce at the cathode and anions oxidize at the anode.
Aqueous Electrolysis: Water competes with solute ions at both electrodes.
Cathode Rule: The species with the more positive reduction potential ( $E^\circ_{\text{red}}$ ) is reduced. Highly active metals (e.g., Group 1 and 2 ions) are not reduced in water; $\ce{H2O}$ is reduced to $\ce{H2(g)}$ instead.
Anode Rule & Overpotential: Polyatomic anions (e.g., $\ce{SO4^{2-}}$ , $\ce{NO3^-}$ ) do not oxidize in water; $\ce{H2O}$ oxidizes to $\ce{O2(g)}$ . However, in concentrated brine, chloride overpotential favors $\ce{Cl2(g)}$ formation over $\ce{O2(g)}$ .

Molten vs. Aqueous Electrolysis

Electrolysis uses electrical energy to drive non-spontaneous redox reactions. The presence or absence of water determines which chemical species participate in electrode half-reactions.

Molten Electrolysis

In a molten salt, the ionic compound is melted without any solvent present. Because only the solute's constituent cations and anions exist, product prediction is direct:

Cathode (Reduction): Metal cation gains electrons: $\ce{Na^+(l) + e^- -> Na(l)}$ .
Anode (Oxidation): Nonmetal anion loses electrons: $\ce{2Cl^-(l) -> Cl2(g) + 2e^-}$ .

Worked Example: Molten Salt Electrolysis
Problem: Predict the products of electrolyzing molten

\ce{MgCl2}

.
Step 1: Identify present ions:

\ce{Mg^{2+}(l)}

and

\ce{Cl^-(l)}

.
Step 2: Cathode reduction:

\ce{Mg^{2+}(l) + 2e^- -> Mg(l)}

.
Step 3: Anode oxidation:

\ce{2Cl^-(l) -> Cl2(g) + 2e^-}

.
Answer:

\ce{Mg(l)}

forms at the cathode;

\ce{Cl2(g)}

forms at the anode.
Common Exam Trap: Do not predict

\ce{H2(g)}

\ce{O2(g)}

for molten salts; water is entirely absent.

Water Competition in Aqueous Solutions

In aqueous electrolysis, water molecules molecules can undergo oxidation or reduction in competition with dissolved ions.

Water Reduction (Cathode):

\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)} \quad (E^\circ_{\text{red}} = \pu{-0.83 V})

Water Oxidation (Anode):

\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad (E^\circ_{\text{red}} = \pu{+1.23 V}, \text{ so } E^\circ_{\text{ox}} = \pu{-1.23 V})

Predicting Products at Electrodes

To predict aqueous electrolysis products, evaluate competing half-reactions using standard reduction potentials and kinetic factors.

1. Cathode Selection (Reduction)

Compare the reduction potential of the cation against water ( $E^\circ_{\text{red}} = \pu{-0.83 V}$ ):

If cation $E^\circ_{\text{red}} > \pu{-0.83 V}$ (e.g., $\ce{Cu^{2+}}$ , $\ce{Ag^+}$ ), the metal cation reduces.
If cation $E^\circ_{\text{red}} < \pu{-0.83 V}$ (e.g., $\ce{Na^+}$ , $\ce{K^+}$ , $\ce{Mg^{2+}}$ ), $\ce{H2O}$ reduces to $\ce{H2(g)}$ and $\ce{OH^-}$ .

2. Anode Selection (Oxidation)

Compare the oxidation potential of the anion against water ( $E^\circ_{\text{ox}} = \pu{-1.23 V}$ ):

Polyatomic anions containing central atoms in high oxidation states (e.g., $\ce{SO4^{2-}}$ , $\ce{NO3^-}$ ) are resistant to oxidation, so $\ce{H2O}$ oxidizes to $\ce{O2(g)}$ and $\ce{H^+}$ .
Halide ions (e.g., $\ce{Cl^-}$ , $\ce{Br^-}$ , $\ce{I^-}$ ) can oxidize to halogen gases.

Concept Focus: Kinetic Overpotential
Thermodynamically, water oxidation (

E^\circ = \pu{+1.23 V}

) requires less voltage than chloride oxidation (

E^\circ = \pu{+1.36 V}

). However, forming

\ce{O2(g)}

involves high kinetic activation energy (overpotential). In concentrated

\ce{NaCl(aq)}

, this overpotential shifts the kinetic preference, causing

\ce{Cl2(g)}

to evolve at the anode.

Worked Example 1: Aqueous $\ce{CuSO4}$ Electrolysis
Problem: Predict products for aqueous

\ce{CuSO4}

with inert electrodes.
Reasoning: Ions present are

\ce{Cu^{2+}(aq)}

\ce{SO4^{2-}(aq)}

, and

\ce{H2O(l)}

.
• Cathode:

\ce{Cu^{2+}}

(

E^\circ_{\text{red}} = \pu{+0.34 V}

) vs

\ce{H2O}

(

E^\circ_{\text{red}} = \pu{-0.83 V}

\ce{Cu^{2+}}

is more easily reduced:

\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}

.
• Anode:

\ce{SO4^{2-}}

does not oxidize easily;

\ce{H2O}

oxidizes:

\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}

.
Answer:

\ce{Cu(s)}

deposits at cathode;

\ce{O2(g)}

evolves at anode.

Worked Example 2: Aqueous $\ce{NaCl}$ Electrolysis (Chlor-Alkali Process)
Problem: Predict products for concentrated aqueous

\ce{NaCl}

.
Reasoning:
• Cathode:

\ce{Na^+}

(

E^\circ_{\text{red}} = \pu{-2.71 V}

) vs

\ce{H2O}

(

E^\circ_{\text{red}} = \pu{-0.83 V}

\ce{H2O}

reduces:

\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)}

.
• Anode: Chloride overpotential leads to oxidation of

\ce{Cl^-}

\ce{2Cl^-(aq) -> Cl2(g) + 2e^-}

.
• Net Reaction:

\ce{2H2O(l) + 2Cl^-(aq) -> H2(g) + Cl2(g) + 2OH^-(aq)}

.
Answer:

\ce{H2(g)}

and

\ce{OH^-}

form at cathode;

\ce{Cl2(g)}

forms at anode.
Common Exam Trap: Selecting

\ce{Na(s)}

as the cathode product or

\ce{O2(g)}

as the primary anode product in concentrated brine.

Electrolysis of an aqueous solution of sodium chloride (Source: Slower · CC BY-SA 3.0 · Wikimedia Commons)

Feature	Molten Electrolysis	Aqueous Electrolysis
Electrolyte State	Liquid pure salt (high temp)	Salt dissolved in water (room temp)
Competing Species	None (solute ions only)	Water ( $\ce{H2O}$ ) competes with solute ions
Active Metal Cations (e.g., $\ce{Na^+}$ )	Reduced to metal ( $\ce{Na(l)}$ )	Water reduces to $\ce{H2(g)}$ instead
Anode Overpotential Effect	Not applicable	Controls $\ce{Cl2(g)}$ vs $\ce{O2(g)}$ formation

DAT Memory Aids:
• RED CAT / AN OX: REDuction at CAThode, ANode OXidation.
• Active Metals Stay Solution-Bound: Group 1 and Group 2 cations are weaker oxidizing agents than water, yielding

\ce{H2(g)}

Critical Thinking Questions

Why does electrolyzing aqueous $\ce{KNO3}$ yield $\ce{H2(g)}$ and $\ce{O2(g)}$ while leaving $\ce{K^+}$ and $\ce{NO3^-}$ ions unchanged in solution?
If aqueous $\ce{NaCl}$ is diluted significantly, how does the primary anode product change, and why?
How does the pH near the cathode electrode change during the active electrolysis of aqueous $\ce{NaCl}$ ?

References

Sources returned by Google Search grounding during lesson verification.

20.9: Electrolysis - Chemistry LibreTexts — chem.libretexts.org
11.6: Industrial Electrolysis Processes - Chemistry LibreTexts — chem.libretexts.org

Predicting Electrolysis Products

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