General Chemistry 2 • Electrochemistry

Electron Transfer & Stoichiometry — Relate electron transfer to stoichiometry

Quantify the moles of substance produced or consumed in electrochemical cells by tracking transferred electrons.

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 relationship between electricity and chemical change was not always obvious. In the late eighteenth century, scientists observed that passing electric current through solutions could decompose compounds or deposit metals, but they lacked a quantitative framework to predict how much substance would be produced for a given amount of charge. The quest to unite the language of stoichiometry — the quantitative accounting of chemical reactions — with the physics of electron transfer drove some of the most elegant experiments in the history of chemistry and ultimately gave rise to the field of electrochemistry.

1800 • Voltaic Pile

Alessandro Volta constructs the first battery (voltaic pile), providing a continuous source of electric current and enabling systematic electrolysis experiments.

1807 • Electrolytic Isolation of Elements

Humphry Davy uses electrolysis to isolate sodium and potassium from their molten salts, demonstrating that electric current drives chemical decomposition in predictable ways.

1834 • Faraday's Laws of Electrolysis

Michael Faraday publishes his two laws of electrolysis, establishing that the mass of substance deposited is proportional to the charge passed and inversely proportional to the ion's valence — the birth of quantitative electrochemistry.

1874 • Stoney and the Electron

George Johnstone Stoney coins the term 'electron' and estimates the fundamental unit of electric charge from Faraday's electrochemical data, linking macroscopic measurements to atomic-scale particles.

1909 • Millikan's Oil-Drop Experiment

Robert Millikan precisely determines the charge on a single electron, allowing Faraday's constant to be expressed as Nₐ × e and firmly connecting mole-scale stoichiometry to electron-scale charge transfer.

Faraday's insight was profound: the number of moles of a substance produced at an electrode is not arbitrary — it is dictated by the total charge delivered and the number of electrons each ion must gain or lose. This principle remains the foundation for every quantitative calculation in electrochemistry, from industrial electroplating to the design of lithium-ion batteries. The central question this lesson addresses is: how do we translate a measured electric current and time into moles and grams of product using stoichiometric reasoning?

SECTION 2

Core Principles & Definitions

Bridging electron transfer to stoichiometry requires mastery of a small set of interconnected ideas. Every electrochemical calculation ultimately rests on the recognition that electrons are chemical reagents — they appear in balanced half-reactions with definite stoichiometric coefficients, just as ions, molecules, and atoms do. The concepts below form the logical scaffold for all quantitative work in this domain.

1

Half-Reaction Electron Count (n)

Every half-reaction specifies n, the number of moles of electrons transferred per mole of substance oxidized or reduced. For example, Cu²⁺ + 2e⁻ → Cu(s) indicates n = 2. This integer is the stoichiometric bridge between charge and moles.
2

Faraday's Constant (F)

One mole of electrons carries a total charge of F = 96 485 C mol⁻¹. Faraday's constant converts between moles of electrons and coulombs of charge, acting as the universal exchange rate in electrochemical stoichiometry.
3

Charge (Q = I × t)

Total charge Q (in coulombs) delivered to an electrode equals the current I (in amperes) multiplied by the elapsed time t (in seconds). This relationship connects the measurable electrical quantities of a circuit to the chemical quantities at the electrode surface.
4

Mole–Mass–Electron Chain

The calculation pathway is Q → mol e⁻ → mol substance → mass (or volume). Dividing Q by F yields moles of electrons; dividing that by n gives moles of substance; multiplying by molar mass gives grams. Each arrow uses a single conversion factor.
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KEY TAKEAWAY

Think of electrons as a currency in a chemical transaction. Faraday's constant is the exchange rate between 'coulombs' (dollars in the electrical economy) and 'moles of electrons' (items you can buy in the chemical economy). The half-reaction tells you the price per mole of product — how many moles of electrons you must spend for each mole of metal deposited or gas evolved. Just as you cannot buy half an indivisible item, you cannot deposit a fraction of the stoichiometric ratio; the mathematics enforces a strict, integer-ratio accounting.

SECTION 3

Visual Explanation — The Electron–Stoichiometry Bridge

The Mole–Electron–Charge Conversion ChainQ = I × tCoulombs (C)÷ Fmol e⁻96 485 C/mol÷ nmol substancefrom half-rxn× MMass (g)gramsExample: Cu²⁺ + 2e⁻ → Cu(s)I = 3.00 At = 3600 sQ = 10 800 C0.1119 mol e⁻10800 ÷ 964850.05596 mol Cu÷ 2 (n = 2)3.555 g Cu× 63.55 g/molMaster Formulam = (I × t × M) / (n × F)Each arrow represents one conversion factor in dimensional analysis.Reading left to right: charge → mol e⁻ → mol substance → mass
The conversion chain reads left to right. Starting from total charge Q, divide by Faraday's constant F to obtain moles of electrons, then divide by n (electrons per formula unit) to get moles of substance, and finally multiply by molar mass M to obtain the mass deposited or consumed.

The diagram above captures the essential logic of every Faraday's-law calculation. Notice that each arrow corresponds to a single conversion factor, making the entire problem amenable to clean dimensional analysis. The half-reaction determines n, and every other quantity is either measured (I, t) or looked up (M, F). In the numerical example shown, a current of 3.00 A flowing for one hour through a CuSO₄ solution deposits 3.56 g of metallic copper — a result that can be verified experimentally with a simple analytical balance, illustrating the remarkable predictive power of Faraday's laws.

SECTION 4

Mathematical Framework

The quantitative treatment of electron-transfer stoichiometry rests on two experimentally verified laws and their modern restatement in terms of the electron charge. Below we develop the key equations, define every variable, and show how they interconnect.

FARADAY'S FIRST LAW

m=(M×Q)/(n×F)m = (M × Q) / (n × F)

where m = mass deposited or dissolved (g), M = molar mass (g mol⁻¹), Q = total charge (C), n = moles of electrons per mole of substance (from balanced half-reaction), and F = 96 485 C mol⁻¹ (Faraday's constant).

CHARGE FROM CURRENT

Q=I×tQ = I × t

where I = current (A = C s⁻¹) and t = time (s). For non-constant currents, Q = ∫₀ᵗ I(τ) dτ.

MOLES OF ELECTRONS

mole−=Q/F=(I×t)/Fmol e⁻ = Q / F = (I × t) / F

This expression directly yields the number of moles of electrons delivered. Dividing by n converts to moles of the electroactive species.

COMBINED MASTER EQUATION

m=(I×t×M)/(n×F)m = (I × t × M) / (n × F)

Substituting Q = I × t into Faraday's first law yields the single master equation from which any unknown (m, I, t, or even n) can be isolated, given the remaining variables.

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Unit Consistency Check

Dimensional analysis of the master equation confirms the result is in grams: (A)(s)(g mol⁻¹) / (mol e⁻ mol⁻¹)(C mol⁻¹) = (C × g mol⁻¹) / (C mol⁻¹) = g. Always convert minutes or hours to seconds before substituting, and confirm that molar mass carries units of g mol⁻¹.

SECTION 5

Determining n from Half-Reactions

The integer n — the number of moles of electrons transferred per mole of substance — is arguably the most conceptually important variable in Faraday's law. It is not a constant for a given element; rather, it depends on the specific half-reaction occurring at the electrode. An element that can exist in multiple oxidation states (e.g., iron as Fe²⁺ or Fe³⁺) will have different values of n depending on the experimental conditions. The table below catalogues common electrochemical half-reactions and their corresponding n values.

Half-Reactionn (mol e⁻ per mol)Application
Ag⁺ + e⁻ → Ag(s)1Silver electroplating
Cu²⁺ + 2e⁻ → Cu(s)2Copper refining, PCB manufacturing
Al³⁺ + 3e⁻ → Al(s)3Hall–Héroult process (aluminum smelting)
2H₂O → O₂ + 4H⁺ + 4e⁻4Water electrolysis (anode)
2H⁺ + 2e⁻ → H₂(g)2Water electrolysis (cathode), hydrogen fuel
Cl₂ + 2e⁻ → 2Cl⁻2 per mol Cl₂Chlor-alkali process

Common electrochemical half-reactions with the number of electrons transferred per formula unit.

Electrolytic Cell: Cu ElectroplatingCuSO₄(aq) SolutionAnode (+)Cu → Cu²⁺ + 2e⁻oxidationCathode (−)Cu²⁺ + 2e⁻ → CureductionExternal Circuit (wire)e⁻ →Cu²⁺Cu²⁺Cu²⁺Stoichiometry SummaryHalf-reaction:Cu²⁺ + 2e⁻ → Cu(s)n = 2 mol e⁻ / mol CuF = 96 485 C/mol e⁻M(Cu) = 63.55 g/molGiven: I = 5.00 A, t = 1.00 hQ = 5.00 × 3600 = 18 000 Cmol e⁻ = 18000/96485 = 0.1865mol Cu = 0.1865/2 = 0.09327m = 0.09327 × 63.55m = 5.93 g CuEvery 2 mol e⁻ deposits 1 mol Cu
An electrolytic cell for copper electroplating. The anode (red) oxidizes Cu to Cu²⁺, while the cathode (cyan) reduces Cu²⁺ back to metallic copper. The right panel shows the complete stoichiometric calculation for 5.00 A over one hour.

Careful attention must be paid to how the half-reaction is written. For the hydrogen evolution reaction, the half-reaction 2H⁺ + 2e⁻ → H₂(g) indicates that n = 2 per mole of H₂ produced — but if you instead express it as 4H⁺ + 4e⁻ → 2H₂(g), the ratio n/mol product remains the same (4 electrons per 2 mol H₂ = 2 electrons per mol H₂). The stoichiometric ratio is what matters, not the absolute coefficients. This mirrors the logic used in conventional stoichiometry with balanced molecular equations, reinforcing the idea that electrons are simply another 'reagent' to be balanced.

SECTION 6

Worked Example — Aluminum Electrolysis

The Hall–Héroult process produces aluminum by passing enormous currents through molten cryolite containing dissolved Al₂O₃. Consider the following problem: How many grams of aluminum are deposited when a current of 150 000 A is passed through the electrolytic cell for exactly 24.0 hours?

Aluminum Deposition via Faraday's Law

  1. 1

    Step 1 — Write the Half-Reaction and Identify n

    The cathode half-reaction is Al³⁺ + 3e⁻ → Al(s). Each aluminum atom gains three electrons, so n = 3 mol e⁻ per mol Al.
    n = 3
  2. 2

    Step 2 — Calculate Total Charge Q

    Convert time to seconds: t = 24.0 h × 3600 s h⁻¹ = 86 400 s. Then Q = I × t = 150 000 A × 86 400 s = 1.296 × 10¹⁰ C.
    Q = 1.296 × 10¹⁰ C
  3. 3

    Step 3 — Calculate Moles of Electrons

    mol e⁻ = Q / F = (1.296 × 10¹⁰ C) / (96 485 C mol⁻¹) = 1.343 × 10⁵ mol e⁻.
    mol e⁻ = 1.343 × 10⁵
  4. 4

    Step 4 — Convert to Moles of Aluminum

    Using the stoichiometric ratio from the half-reaction: mol Al = mol e⁻ / n = (1.343 × 10⁵) / 3 = 4.477 × 10⁴ mol Al.
    mol Al = 4.477 × 10⁴
  5. 5

    Step 5 — Convert to Mass

    m = mol Al × M(Al) = 4.477 × 10⁴ mol × 26.98 g mol⁻¹ = 1.208 × 10⁶ g ≈ 1208 kg. In practice, the Hall–Héroult process operates at roughly 95% current efficiency, so the actual yield would be approximately 1148 kg.
    m ≈ 1208 kg Al (at 100% efficiency)
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Current Efficiency

Real electrolytic processes rarely achieve 100% current efficiency because some charge is consumed by side reactions (e.g., oxygen evolution, parasitic reduction of impurities). The current efficiency (η) is defined as the ratio of the actual mass deposited to the theoretical mass predicted by Faraday's law: η = (m_actual / m_theoretical) × 100%. When solving problems, always check whether the problem specifies an efficiency less than 100%.

SECTION 7

Galvanic vs. Electrolytic Cells — Stoichiometric Considerations

The master equation m = (I × t × M)/(n × F) is cell-type agnostic — it applies equally to galvanic (voltaic) cells that generate electricity spontaneously and to electrolytic cells that require an external power supply. However, the practical context differs, and understanding these differences prevents common errors in stoichiometric reasoning.

FeatureGalvanic CellElectrolytic Cell
Thermodynamic spontaneityΔG < 0; spontaneousΔG > 0; non-spontaneous
Driving forceChemical potential differenceExternal voltage source
Anode reactionOxidation (negative terminal)Oxidation (positive terminal)
Cathode reactionReduction (positive terminal)Reduction (negative terminal)
Stoichiometry applies?Yes — mass of anode lost = mass predicted by Faraday's lawYes — mass deposited at cathode = (I × t × M)/(n × F)
Limiting reagentThe electrode or electrolyte that is consumed firstCharge (I × t); electrolyte/electrode assumed in excess
Typical calculation questionHow long can a battery deliver current before the anode is exhausted?How much metal is plated for a given current and time?

Comparison of stoichiometric considerations in galvanic and electrolytic cells.

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KEY TAKEAWAY

Faraday's law functions like a universal translator between the electrical world and the chemical world: regardless of whether the cell produces current or consumes it, the same stoichiometric bookkeeping applies. The only difference is the direction of the energy flow — in a galvanic cell, the chemical reaction pays for the electrons; in an electrolytic cell, the external power supply foots the bill. In both cases, the electron-to-mole ratio is dictated exclusively by the balanced half-reaction.

SECTION 8

Connections to Thermodynamics & Advanced Electrochemistry

The stoichiometric framework developed so far connects seamlessly to the thermodynamic treatment of electrochemical cells. The relationship ΔG = −nFE° bridges Faraday's law to the Gibbs free energy of reaction, while the Nernst equation extends the analysis to non-standard conditions. Understanding these connections places electron-transfer stoichiometry within a broader theoretical context that you will encounter in advanced courses and in research applications such as battery design and corrosion engineering.

ConceptIntroductory Treatment (This Lesson)Advanced Extension
Electron count nRead directly from the balanced half-reactionConnects to ΔG° = −nFE° and to reaction quotient Q in the Nernst equation
Faraday's constant FConversion factor: 96 485 C per mol e⁻F = Nₐ × e; foundational in statistical thermodynamics of charged species
Constant current assumptionQ = I × t with constant IFor variable current, Q = ∫ I(t) dt; handled via coulometry or chronoamperometry
100% current efficiencyAssumed unless stated otherwiseSide reactions quantified via current efficiency η; parasitic losses modeled by Butler–Volmer kinetics
Single half-reactionOne product per electrodeCompeting cathode reactions (e.g., H₂ evolution vs. metal deposition) require overpotential analysis

How the stoichiometric treatment extends into thermodynamic and kinetic electrochemistry.

In particular, the equation ΔG° = −nFE° reveals that the same integer n that governs mass deposition also controls the energetics of the cell. A reaction with a larger n (more electrons transferred per formula unit) produces a proportionally larger magnitude of ΔG° for a given cell potential E°. This insight is central to battery engineering: the energy density of a cell depends not only on voltage but also on how many electrons each redox couple exchanges per unit mass of active material.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In the half-reaction Ni²⁺ + 2e⁻ → Ni(s), what does the coefficient '2' on the electrons tell you about the relationship between charge passed and nickel deposited? Explain in terms of Faraday's law.
Show answer
The coefficient 2 indicates that every mole of nickel deposited at the cathode requires exactly 2 moles of electrons (n = 2). According to Faraday's law, m = (I × t × M)/(n × F), a larger n means that more charge must be passed to deposit the same number of moles of metal. Specifically, depositing 1 mol of Ni (58.69 g) requires 2 × 96 485 = 192 970 C. If the half-reaction were instead Ni³⁺ + 3e⁻ → Ni, the charge requirement would increase by 50% for the same mass of nickel, illustrating how the electron coefficient directly controls the charge-to-mass ratio.

Question 2: BASIC CALCULATION

A current of 2.50 A is passed through a solution of AgNO₃ for 40.0 minutes. How many grams of silver are deposited at the cathode? (M(Ag) = 107.87 g mol⁻¹; the half-reaction is Ag⁺ + e⁻ → Ag.)
Show answer
First convert time: t = 40.0 min × 60 s/min = 2400 s. Calculate charge: Q = I × t = 2.50 × 2400 = 6000 C. Determine moles of electrons: mol e⁻ = 6000 / 96 485 = 0.06219 mol. Since n = 1, moles of Ag = moles of e⁻ = 0.06219 mol. Finally, m = 0.06219 × 107.87 = 6.71 g Ag. Because n = 1, the stoichiometry is the simplest possible case — each electron deposits one silver atom.

Question 3: INTERMEDIATE

During electrolysis of a CrCl₃ solution, 15.6 g of chromium are deposited at the cathode. If the current was held constant at 4.80 A, for how many minutes was the electrolysis conducted? (M(Cr) = 52.00 g mol⁻¹; the half-reaction is Cr³⁺ + 3e⁻ → Cr.)
Show answer
Rearrange the master equation to solve for time: t = (m × n × F)/(I × M). Substitute: t = (15.6 × 3 × 96 485)/(4.80 × 52.00). Numerator: 15.6 × 3 × 96 485 = 4 515 558 C. Denominator: 4.80 × 52.00 = 249.6 A·g mol⁻¹. Therefore t = 4 515 558 / 249.6 = 18 092 s ≈ 301.5 min ≈ 5.03 h. This problem illustrates the power of the master equation — any one of the five variables (m, I, t, M, n) can be isolated algebraically, and the same dimensional-analysis chain works in reverse.

Question 4: APPLIED

A car bumper is to be chrome-plated by depositing 250 g of chromium from a Cr₂O₇²⁻ solution in which chromium is reduced from Cr⁶⁺ to Cr⁰. The power supply delivers a constant 50.0 A. Calculate (a) the value of n, (b) the total charge required, and (c) the time in hours needed. State any assumptions.
Show answer
(a) Chromium goes from +6 to 0 oxidation state, so each Cr atom gains 6 electrons: n = 6. The half-reaction can be written Cr⁶⁺ + 6e⁻ → Cr(s). (b) Rearrange m = (I × t × M)/(n × F) to Q = I × t = (m × n × F)/M = (250 × 6 × 96 485)/52.00 = 2.785 × 10⁷ C. (c) t = Q / I = 2.785 × 10⁷ / 50.0 = 5.570 × 10⁵ s = 154.7 h ≈ 6.4 days. This result highlights why industrial chrome plating typically uses high currents and large surface areas. The assumption of 100% current efficiency was used; in practice hydrogen evolution at the cathode can reduce efficiency to 10–25%, dramatically increasing the required time.

Question 5: CRITICAL THINKING

Two electrolytic cells are connected in series so that the same current passes through both. Cell A contains CuSO₄(aq) and Cell B contains AgNO₃(aq). After electrolysis, 3.175 g of Cu have been deposited in Cell A. Without knowing the current or time, calculate the mass of Ag deposited in Cell B. Then explain why series-connected coulometers were historically used to measure charge.
Show answer
Because the cells are in series, the same total charge Q passes through both. In Cell A: mol Cu = 3.175 / 63.55 = 0.04996 mol, and since n(Cu) = 2, mol e⁻ = 0.04996 × 2 = 0.09992 mol. In Cell B: n(Ag) = 1, so mol Ag = mol e⁻ / n(Ag) = 0.09992 / 1 = 0.09992 mol. Mass of Ag = 0.09992 × 107.87 = 10.78 g. The ratio m(Ag)/m(Cu) = 10.78/3.175 = 3.395, which equals (M(Ag) × n(Cu))/(M(Cu) × n(Ag)) = (107.87 × 2)/(63.55 × 1) = 3.395 — confirming the stoichiometric consistency. Historically, a silver coulometer (weighing the silver deposited) was placed in series with an experimental cell precisely because the mass of silver provides a direct, precise measure of the total charge passed, exploiting the n = 1 simplicity of the Ag⁺/Ag couple.

SUMMARY

Lesson Summary

Summary

Electron-transfer stoichiometry provides a quantitative bridge between the electrical and chemical domains of electrochemistry. The central principle is Faraday's law, expressed as m = (I × t × M)/(n × F), which relates the mass of substance deposited or consumed to the current, time, molar mass, and the electron count n from the balanced half-reaction. Faraday's constant (F = 96 485 C mol⁻¹) serves as the universal conversion factor between coulombs and moles of electrons.

The calculation pathway — charge → mol e⁻ → mol substance → mass — applies to both galvanic and electrolytic cells. Mastery of this framework requires identifying the correct half-reaction, extracting n, ensuring consistent units (seconds for time, grams per mole for M), and accounting for current efficiency in real-world applications. This stoichiometric reasoning extends naturally to thermodynamic relationships (ΔG° = −nFE°) and to modern electrochemical engineering, including battery design and industrial electrolysis.

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