General Chemistry 2 • Electrochemistry

Electrolysis Products — Predict products at electrodes in aqueous electrolysis

Learn to predict which species are oxidized and reduced when aqueous solutions undergo electrolysis using electrode potential hierarchies.

An atom with laboratory flask and electron orbit for General Chemistry 2
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The ability to decompose chemical compounds using electrical current ranks among the most transformative discoveries in the history of chemistry. In the late eighteenth and early nineteenth centuries, scientists began to realize that passing a sustained current through solutions could drive reactions that would never occur spontaneously — reactions that separate compounds into their constituent elements. This technique, called electrolysis, not only revealed the elemental composition of water and countless salts but also established the intimate connection between electrical energy and chemical change that underpins all of electrochemistry.

1800 • Nicholson & Carlisle Decompose Water

William Nicholson and Anthony Carlisle used Volta's newly invented pile to pass current through water, observing the evolution of hydrogen and oxygen gas at separate electrodes — the first deliberate electrolysis experiment.

1807 • Davy Isolates Alkali Metals

Humphry Davy employed electrolysis of molten alkali hydroxides to isolate potassium and sodium for the first time, demonstrating that these caustic substances were compounds rather than elements.

1834 • Faraday's Laws of Electrolysis

Michael Faraday published quantitative laws relating the mass of substance deposited or liberated at an electrode to the total charge passed, introducing the terms anode, cathode, electrolyte, and ion to the scientific lexicon.

1886 • Hall–Héroult Process

Charles Martin Hall and Paul Héroult independently developed the electrolytic process for producing aluminum from alumina dissolved in cryolite, transforming aluminum from a precious metal into an industrial commodity.

1900s–Present • Industrial Chlor-Alkali & Electrorefining

Modern large-scale electrolysis of brine produces chlorine and sodium hydroxide, while electrorefining purifies metals such as copper to 99.99% purity — applications that demand precise prediction of electrode products.

The central challenge that emerged from these discoveries is deceptively simple: when an aqueous solution containing multiple ionic and molecular species is electrolyzed, which species will actually react at each electrode? Water itself can be oxidized or reduced, and the dissolved solute ions may compete with water for electron transfer. Predicting the products requires a systematic approach grounded in standard reduction potentials, overpotential considerations, and concentration effects — the framework we will develop in this lesson.

SECTION 2

Core Principles & Definitions

In an electrolytic cell, an external power supply forces a non-spontaneous redox reaction to proceed. The electrode connected to the positive terminal of the power supply is called the anode, and oxidation occurs there. The electrode connected to the negative terminal is the cathode, where reduction takes place. When the electrolyte is an aqueous solution rather than a pure molten salt, water molecules are always present as potential reactants, competing with dissolved ions for electron transfer at each electrode.

1

Competing Cathode Reactions

At the cathode, the species with the least negative (most positive) reduction potential is preferentially reduced. Water can be reduced to H₂ at E° = −0.83 V (pH 7). If a dissolved cation has E° more positive than −0.83 V, it is reduced instead.
2

Competing Anode Reactions

At the anode, the species with the least positive oxidation potential (i.e., easiest to oxidize) is preferentially oxidized. Water can be oxidized to O₂ at E° = +0.82 V (pH 7). If an anion is easier to oxidize, it reacts instead of water.
3

Overpotential

Overpotential is the additional voltage beyond the thermodynamic value required to drive a reaction at a practical rate. Gaseous products (especially O₂) often have significant overpotential, which can change the predicted product at the anode.
4

Concentration Effects

Actual electrode potentials depend on ion concentrations via the Nernst equation. Concentrated solutions (e.g., concentrated NaCl) can shift the effective potential enough to change the product identity compared with dilute solutions.
5

Electrode Material

When an active electrode (e.g., Cu, Zn) is used as the anode, the electrode metal itself may be oxidized in preference to water or the dissolved anion. Inert electrodes (Pt, graphite) do not participate in the reaction.
✦

KEY TAKEAWAY

Think of aqueous electrolysis as a competitive audition at each electrode. Multiple species — dissolved ions and water molecules — are 'auditioning' to gain or lose electrons. The winner at the cathode is the species easiest to reduce (most positive E°), and the winner at the anode is the species easiest to oxidize (least positive E° for oxidation). Overpotential acts like a handicap that can change the outcome, especially for gas-producing reactions.

SECTION 3

Visual Explanation — The Aqueous Electrolysis Cell

Aqueous Electrolysis Cell — NaCl(aq)Aqueous NaCl solutionANODE (+)CATHODE (−)Cl₂ bubblesH₂ bubblesDC Power Supply+−Anode: 2 Cl⁻ → Cl₂ + 2 e⁻Cathode: 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻Cl⁻ migratesNa⁺ migrates(Concentrated NaCl — Cl₂ produced at anode instead of O₂)
An aqueous electrolysis cell for concentrated NaCl. At the anode (left, purple), chloride ions are oxidized to Cl₂ gas. At the cathode (right, cyan), water is reduced to H₂ gas and OH⁻ ions. Na⁺ ions migrate toward the cathode but are not reduced because water has a less negative reduction potential.

The diagram above illustrates a key principle: sodium ions (Na⁺) remain as spectator ions in solution even though they migrate toward the cathode. The standard reduction potential for Na⁺/Na is −2.71 V, far more negative than that of water (−0.83 V at pH 7), so water wins the competition at the cathode and is reduced instead. At the anode, in concentrated NaCl, chloride ions are oxidized to Cl₂ rather than water being oxidized to O₂, largely because the high chloride concentration and the significant overpotential for O₂ evolution on many electrode surfaces shift the balance in favor of Cl₂ production. In dilute NaCl, however, O₂ would be the anode product — a critical concentration dependence that students must internalize.

SECTION 4

Mathematical Framework — Electrode Potentials & Decision Rules

The thermodynamic basis for predicting electrolysis products lies in comparing standard reduction potentials (E°) of all species that could be reduced at the cathode and all species that could be oxidized at the anode. The species requiring the least additional driving force will react first. When concentration deviates from standard conditions, the Nernst equation allows us to calculate the effective potential, and the concept of overpotential adds an empirical kinetic correction.

NERNST EQUATION

E=E°−(RT/nF)×lnQE = E° − (RT / nF) × ln Q

E = electrode potential under non-standard conditions; E° = standard reduction potential; R = 8.314 J mol⁻¹ K⁻¹; T = temperature in K; n = number of electrons transferred; F = 96 485 C mol⁻¹; Q = reaction quotient.

NERNST EQUATION (25 °C FORM)

E=E°−(0.0592/n)×logQE = E° − (0.0592 / n) × log Q

At 25 °C, the factor RT/F evaluates to 0.02569 V, and converting from ln to log₁₀ introduces the factor 2.303, yielding the convenient coefficient 0.0592 V.

WATER REDUCTION (CATHODE)

2H2O(l)+2e−→H2(g)+2OH−(aq)E°=−0.83V(pH7)2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻(aq) E° = −0.83 V (pH 7)

Under standard conditions (pH 0), E° = 0.00 V for the H⁺/H₂ couple. At pH 7, using the Nernst equation, the effective potential shifts to approximately −0.41 V, but the commonly cited value for neutral water is −0.83 V when OH⁻ is referenced at unit activity.

WATER OXIDATION (ANODE)

2H2O(l)→O2(g)+4H+(aq)+4e−E°=+1.23V(standard)/+0.82V(pH7)2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻ E° = +1.23 V (standard) / +0.82 V (pH 7)

At the anode, water can be oxidized to O₂. In practice, the overpotential for O₂ evolution on most electrode surfaces adds 0.4–0.8 V to the thermodynamic value, making the effective required potential significantly higher.

💡

Decision Algorithm

At the cathode: compare E° values of all reducible species (cations plus water). The species with the most positive (least negative) E° is reduced. At the anode: compare E° values of all oxidizable species (anions plus water). The species with the least positive E° (when written as a reduction) is oxidized first — equivalently, the one that requires the smallest reversal of its reduction half-reaction. Always check: (1) Is the electrode inert or active? (2) Is the solution dilute or concentrated? (3) Is overpotential significant for gas evolution?

SECTION 5

Detailed Breakdown — Product Predictions by Electrolyte Type

A practical way to master electrolysis product prediction is to classify electrolytes by the nature of their cation and anion, then apply the reduction-potential hierarchy systematically. The following table and diagram summarize the outcomes for the most common categories encountered in general chemistry.

Electrolyte TypeCathode ProductAnode ProductReasoning
Active-metal salt, dilute (e.g., dilute NaCl, Na₂SO₄)H₂ (from water)O₂ (from water)Na⁺ has E° = −2.71 V, far more negative than water (−0.83 V). SO₄²⁻ and dilute Cl⁻ are harder to oxidize than water.
Active-metal salt, concentrated halide (e.g., conc. NaCl, NaBr)H₂ (from water)X₂ (halogen gas)Cathode same as above. High halide concentration plus O₂ overpotential favors halide oxidation over water.
Less-active-metal salt (e.g., CuSO₄, AgNO₃)Metal deposited (Cu, Ag)O₂ (from water)Cu²⁺ (E° = +0.34 V) and Ag⁺ (E° = +0.80 V) are more easily reduced than water. SO₄²⁻ and NO₃⁻ are extremely difficult to oxidize.
Acid solution (e.g., dilute H₂SO₄)H₂ (from H⁺)O₂ (from water)H⁺ is the only cation; sulfate is very difficult to oxidize, so water is oxidized.
Less-active-metal halide (e.g., CuBr₂)Metal deposited (Cu)X₂ (halogen, e.g., Br₂)Cu²⁺ is reduced over water; Br⁻ is more easily oxidized than water.

Common aqueous electrolysis product predictions using inert electrodes

Decision Flowchart — Predicting Electrolysis ProductsIdentify all species in solutionCATHODE (Reduction)ANODE (Oxidation)Is E°(cation) more positivethan E°(water) ≈ −0.83 V?Is anion a halide (Cl⁻, Br⁻, I⁻)AND concentrated?YESNOMetal depositedH₂ gas(water reduced)YESNOX₂ (halogen)Is electrodeactive metal?(e.g., Cu, Zn)YESNOElectrode dissolvesO₂ gasNote: F⁻ is never oxidized in aqueous solution; oxoanions (SO₄²⁻, NO₃⁻, CO₃²⁻) are also not oxidized.Halide oxidation ease: I⁻ (easiest, E° = +0.54 V) > Br⁻ (+1.07 V) > Cl⁻ (+1.36 V) ≫ F⁻ (not oxidized)Lower E° for the X₂/X⁻ couple → easier to oxidize the halide
A decision flowchart for predicting electrolysis products. Follow the left branch for cathode products and the right branch for anode products. Note that oxoanions (SO₄²⁻, NO₃⁻) and F⁻ are never oxidized in aqueous electrolysis.

Several important caveats deserve emphasis. First, fluoride is never oxidized in aqueous solution because the F₂/F⁻ standard potential (+2.87 V) far exceeds that of water oxidation; attempting to oxidize F⁻ would decompose water first. Second, oxoanions such as sulfate, nitrate, and phosphate involve breaking strong covalent bonds and are thermodynamically and kinetically resistant to oxidation; water is always oxidized in their place. Third, concentration matters enormously for halides: in dilute HCl, the anode product is O₂, but in concentrated HCl, it is Cl₂. These nuances are best internalized through practice rather than memorization alone.

SECTION 6

Worked Example — Electrolysis of Aqueous CuSO₄

Predict the products formed at each electrode when an aqueous solution of CuSO₄ is electrolyzed using inert platinum electrodes.

Electrolysis of Aqueous CuSO₄ with Inert Electrodes

  1. 1

    Step 1 — Identify All Species in Solution

    CuSO₄ dissociates completely: Cu²⁺(aq) and SO₄²⁻(aq). Water (H₂O) is also present as the solvent. The electrode is platinum (inert), so it will not participate in any reaction. The species that could be reduced at the cathode are Cu²⁺ and H₂O. The species that could be oxidized at the anode are SO₄²⁻ and H₂O.
  2. 2

    Step 2 — Determine Cathode Product

    Compare reduction potentials: • Cu²⁺ + 2 e⁻ → Cu(s): E° = +0.34 V • 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻: E° ≈ −0.83 V (pH 7) Cu²⁺ has the more positive reduction potential, meaning it is more easily reduced than water.
    Cathode product: Cu(s) — copper metal is deposited on the electrode.
  3. 3

    Step 3 — Determine Anode Product

    Compare oxidation candidates. SO₄²⁻ is an oxoanion with very strong S–O bonds; it is essentially impossible to oxidize under normal electrolysis conditions. The only remaining candidate is water: 2 H₂O → O₂ + 4 H⁺ + 4 e⁻ E° = +1.23 V (written as reduction) Since sulfate cannot be oxidized, water is oxidized by default.
    Anode product: O₂(g) — oxygen gas bubbles at the anode.
  4. 4

    Step 4 — Write the Overall Cell Reaction

    Balance electrons: the Cu²⁺ reduction transfers 2 electrons, and the water oxidation transfers 4 electrons per O₂. Multiply the cathode half-reaction by 2: Cathode: 2 Cu²⁺ + 4 e⁻ → 2 Cu(s) Anode: 2 H₂O → O₂ + 4 H⁺ + 4 e⁻ Overall: 2 Cu²⁺(aq) + 2 H₂O(l) → 2 Cu(s) + O₂(g) + 4 H⁺(aq)
    The solution becomes increasingly acidic as electrolysis proceeds, and the blue color fades as Cu²⁺ is consumed.
  5. 5

    Step 5 — Verify and Note Observations

    Observable evidence: a reddish-brown coating of copper appears on the cathode, bubbles of oxygen gas form at the anode, the blue color of the solution diminishes over time, and a pH probe would show the solution becoming more acidic. If a copper anode were used instead of platinum, the anode reaction would change: Cu(s) → Cu²⁺ + 2 e⁻, dissolving the electrode and maintaining the blue color — this is the basis of copper electrorefining.
    With an active Cu anode: Cu dissolves at the anode, Cu deposits at the cathode — net transfer of copper with no gas evolution.

SECTION 7

Comparing Molten vs. Aqueous Electrolysis

A common source of confusion in electrochemistry courses is the difference in product outcomes between molten electrolysis and aqueous electrolysis of the same compound. In a molten salt, there is no water to compete, so the only species available for reduction and oxidation are the ions of the compound itself. In aqueous solution, water introduces additional competing reactions, fundamentally altering the product distribution in many cases.

FeatureMolten ElectrolysisAqueous Electrolysis
Competing speciesOnly the cation and anion of the saltCation, anion, AND water molecules
Cathode product (NaCl)Na(l) — sodium metal depositedH₂(g) — water reduced instead
Anode product (NaCl)Cl₂(g) — chlorine gasCl₂(g) if concentrated; O₂(g) if dilute
Temperature requiredVery high (801 °C for NaCl)Room temperature to moderate
Energy costHigh (heating + electrolysis)Lower (no melting step)
Industrial useProducing reactive metals (Na, Al, Mg)Chlor-alkali, electroplating, electrorefining

Molten vs. aqueous electrolysis — key differences

✦

KEY TAKEAWAY

Molten electrolysis is like a two-candidate election — only the cation and anion compete, so the products are straightforward. Aqueous electrolysis is a multi-candidate primary with water acting as a strong third-party contender. Whether water 'wins' at either electrode depends on the relative reduction potentials, concentration, and overpotential effects. This is why sodium metal can be produced by molten electrolysis of NaCl but never by aqueous electrolysis — water is always easier to reduce than Na⁺.

SECTION 8

Connection to Advanced Theory — Overpotential & Industrial Applications

The simplified decision rules presented so far work well for most general chemistry problems, but industrial electrolysis and research-level electrochemistry demand deeper consideration of overpotential (η). Overpotential is the extra voltage beyond the thermodynamic equilibrium value that must be applied to sustain a reaction at a given rate. It arises from kinetic barriers including charge-transfer activation energy, mass transport limitations, and the difficulty of nucleating gas bubbles on electrode surfaces. The Tafel equation (η = a + b × log j, where j is current density) provides a quantitative framework for analyzing overpotential, connecting the macroscopic observation to molecular-level electron-transfer kinetics described by Butler–Volmer theory.

ConceptGeneral Chemistry 2 LevelAdvanced / Physical Chemistry Level
Product predictionCompare E° values; apply rules for concentration and halidesUse Nernst-corrected potentials, include overpotential terms, consider electrode material catalysis
OverpotentialQualitatively acknowledged; O₂ has high overpotentialQuantitatively modeled via Tafel and Butler–Volmer equations; depends on electrode surface, T, and j
Faraday's lawsApplied to calculate mass deposited given charge and molar massExtended to include current efficiency, side reactions, and coulometric analysis
Industrial designConceptual understanding of chlor-alkali and electroplatingCell engineering: membrane vs. diaphragm cells, electrode materials, energy efficiency optimization

Progression from general to advanced electrochemistry

For students continuing to physical chemistry or chemical engineering, the take-home message is that the E° comparison method is a necessary first approximation, but the actual behavior of an electrolysis cell is governed by the interplay of thermodynamics (E°), kinetics (overpotential), and transport phenomena (mass transfer). Modern research in water splitting for hydrogen fuel exemplifies this: the thermodynamic cell voltage for water electrolysis is only 1.23 V, but practical cells require 1.8–2.0 V due to overpotential at both electrodes, and reducing that gap through better catalysts is a major area of materials science research.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why electrolysis of aqueous NaF produces the same products at both electrodes as electrolysis of pure water, despite the presence of Na⁺ and F⁻ ions in solution.
Show answer
Sodium has a standard reduction potential of −2.71 V, which is far more negative than water's reduction potential of approximately −0.83 V at pH 7. Therefore, water is reduced at the cathode to produce H₂ gas. At the anode, fluoride has the highest standard reduction potential of any halide (E° = +2.87 V for F₂/F⁻), meaning it is extraordinarily difficult to oxidize; water (E° = +1.23 V for the O₂/H₂O couple) is oxidized instead, producing O₂. Both Na⁺ and F⁻ act as spectator ions, and the overall result is identical to the electrolysis of water: 2 H₂O → 2 H₂ + O₂. The NaF merely increases the solution's conductivity.

Question 2: BASIC CALCULATION

An aqueous solution of AgNO₃ is electrolyzed with inert electrodes. (a) Predict the cathode product. (b) Predict the anode product. (c) Write the balanced overall cell reaction.
Show answer
(a) At the cathode, Ag⁺ (E° = +0.80 V) competes with water (E° ≈ −0.83 V at pH 7). Since +0.80 V >> −0.83 V, silver ions are preferentially reduced: Ag⁺ + e⁻ → Ag(s). Silver metal deposits on the cathode. (b) At the anode, NO₃⁻ is an oxoanion and is not oxidized under electrolysis conditions. Water is the only viable oxidation candidate: 2 H₂O → O₂ + 4 H⁺ + 4 e⁻. Oxygen gas is produced. (c) Balancing electrons (multiply the cathode reaction by 4): 4 Ag⁺(aq) + 2 H₂O(l) → 4 Ag(s) + O₂(g) + 4 H⁺(aq). The solution becomes acidic as electrolysis proceeds.

Question 3: INTERMEDIATE

Concentrated aqueous NaBr is electrolyzed with inert graphite electrodes. Predict the products at each electrode and explain why the anode product differs from that obtained with dilute NaBr.
Show answer
Cathode: Na⁺ (E° = −2.71 V) cannot compete with water (E° ≈ −0.83 V), so water is reduced to H₂ gas in both dilute and concentrated solutions. Anode: Br⁻ has a standard reduction potential of +1.07 V (for Br₂/Br⁻), while water oxidation is at E° = +1.23 V. Thermodynamically, Br⁻ is already easier to oxidize than water. However, at low Br⁻ concentration, the Nernst equation shifts the Br₂/Br⁻ potential to more positive values, and water oxidation becomes competitive. In concentrated NaBr, the high Br⁻ activity keeps the potential favorable, and the substantial overpotential for O₂ evolution on graphite further favors Br⁻ oxidation: 2 Br⁻ → Br₂ + 2 e⁻. In dilute NaBr, these factors reverse and O₂ becomes the primary anode product. This concentration dependence is a classic example of how kinetics and the Nernst equation jointly determine electrolysis outcomes.

Question 4: APPLIED

In copper electrorefining, impure copper serves as the anode and pure copper serves as the cathode in a CuSO₄/H₂SO₄ electrolyte. The impure copper contains traces of Ag, Au, Zn, and Fe. Using standard reduction potentials (Cu²⁺/Cu = +0.34 V, Ag⁺/Ag = +0.80 V, Au³⁺/Au = +1.50 V, Zn²⁺/Zn = −0.76 V, Fe²⁺/Fe = −0.45 V), explain the fate of each impurity during refining.
Show answer
At the anode, the impure copper dissolves: Cu → Cu²⁺ + 2 e⁻. Metals less noble than copper (Zn with E° = −0.76 V and Fe with E° = −0.45 V) are even more easily oxidized, so they dissolve into solution as Zn²⁺ and Fe²⁺ along with Cu²⁺. At the cathode, however, only Cu²⁺ is selectively reduced because the cell voltage is carefully controlled so that Cu²⁺ (E° = +0.34 V) is deposited but Zn²⁺ and Fe²⁺ (with their much more negative E° values) are not. Metals more noble than copper — Ag (E° = +0.80 V) and Au (E° = +1.50 V) — are not oxidized at the anode because they require more energy to dissolve than copper; instead, they fall to the bottom of the cell as an insoluble 'anode sludge' or 'slime,' which is collected and refined separately for its precious metal content. This process purifies copper from ~99% to 99.99%.

Question 5: CRITICAL THINKING

A student electrolyzes an aqueous solution of NiSO₄ (E° for Ni²⁺/Ni = −0.26 V) and observes metallic nickel depositing at the cathode, even though −0.26 V is more negative than water's reduction potential at standard conditions (E° = 0.00 V for 2H⁺/H₂ at pH 0). Should the student have expected this result? Construct a thermodynamic and kinetic argument to explain the observation.
Show answer
At first glance, the standard reduction potential of Ni²⁺/Ni (−0.26 V) seems to indicate that water (or H⁺) should be preferentially reduced. However, several factors favor nickel deposition. First, in neutral or slightly acidic NiSO₄ solution, the effective H⁺/H₂ reduction potential is shifted negative by the Nernst equation: at pH 4 (typical for NiSO₄ solutions), E = 0.00 − 0.0592 × 4 = −0.237 V, making it comparable to Ni²⁺/Ni. Second, hydrogen evolution on nickel surfaces has a significant overpotential (approximately 0.2–0.4 V), which effectively makes it harder to reduce H⁺ than the thermodynamic potential suggests. The actual potential required to evolve H₂ at a practical rate is therefore more negative than −0.237 V, while nickel deposition has a relatively low overpotential. The combination of the Nernst correction and kinetic overpotential means nickel is deposited preferentially, though some hydrogen co-evolution may occur — evidenced by occasional bubbles alongside the metal deposit. This result illustrates that E° tables alone are insufficient; a complete prediction requires considering actual pH, ion concentrations, and electrode-specific overpotentials.

SUMMARY

Lesson Summary

Summary

Predicting products in aqueous electrolysis requires systematically comparing the standard reduction potentials of all species that could react at each electrode, including water. At the cathode, the species with the most positive E° is reduced: metals with E° more positive than about −0.83 V (such as Cu²⁺ and Ag⁺) are deposited, while cations of active metals (Na⁺, K⁺, Mg²⁺) are not reduced in aqueous solution — water produces H₂ instead. At the anode, halide ions (Cl⁻, Br⁻, I⁻) are oxidized to their respective halogens when present at sufficient concentration, whereas oxoanions (SO₄²⁻, NO₃⁻) and F⁻ are never oxidized — water is oxidized to O₂ in their place.

Three modifying factors refine these predictions: overpotential (especially for O₂ evolution) can shift the effective potential enough to change the product, concentration effects governed by the Nernst equation alter equilibrium potentials (notably for halides), and electrode material matters because an active metal anode may dissolve in preference to oxidizing solution species. Mastering this decision framework — combining thermodynamic data with an awareness of kinetic and concentration corrections — provides a reliable method for predicting the outcome of any aqueous electrolysis experiment encountered in general chemistry and beyond.

Varsity Tutors • General Chemistry 2 • Electrolysis Products — Predict products at electrodes in aqueous electrolysis