Electrochemistry · Lesson 13 of 13
Chapter Summary and Practice
“Connect redox direction, potential, equilibrium, ionic transport, quantitative electrolysis and electrochemical applications.”
• Select a consistent method for cell-potential, conductivity and electrolysis problems. • Connect energy, charge and reaction stoichiometry across the chapter. • Compare batteries, fuel cells and corrosion through their shared electrode principles. • Solve integrated problems while checking units, signs, conditions and assumptions.
Three connected questions organise the chapter
Electrochemistry connects chemical change to charge movement. The first question is thermodynamic: which reaction direction is favoured at the actual composition? Electrode potentials, the Nernst equation, Gibbs energy and equilibrium address it. The second is transport: how does charge cross the apparatus? Electrons move in conducting solids while ions move in electrolytes, so geometry and ionic mobility matter. The third is stoichiometric: how much reaction corresponds to a supplied charge? Faraday’s laws answer that question.
These questions are related but different. A cell may have a favourable emf yet supply little current because of resistance or slow electrode kinetics. A conductive solution can carry ionic current without identifying which electrode product will dominate. A charge measurement predicts an ideal chemical amount only after the correct electrode reaction and its efficiency have been specified.
| Situation | Start with | Then check |
|---|---|---|
| A galvanic cell | Identify oxidation and reduction; balance electrons | Electrode signs, electron direction, internal ionic path |
| A standard-potential problem | Use reduction potentials for both couples | Subtract right minus left; do not multiply voltage |
| A nonstandard cell | Write the full reaction and its activity quotient | Use n, temperature, pH and gas pressure consistently |
| An equilibrium problem | Set actual E = 0 and Q = K | Keep standard and actual quantities distinct |
| A conductivity measurement | Calibrate cell geometry and determine R | Use consistent length and concentration units |
| An electrolysis yield | Count charge and electrons per product | State solvent, electrode reactions and efficiency |
Core formula reference with conditions
| Relationship | Symbols and units | Condition or limitation |
|---|---|---|
| E°cell = E°cathode − E°anode | E in V; both entries are reduction potentials | Standard activities; same temperature |
| E = E° − RT ln Q/(nF) | R: J mol⁻¹ K⁻¹; T: K; F: C mol⁻¹ | Reversible emf; Q dimensionless |
| E ≈ E° − (0.05916 V/n)log₁₀Q | n: electrons per balanced reaction | 298.15 K; use another coefficient at another T |
| ΔrG = −nFE | ΔrG: J mol⁻¹ of reaction | Maximum reversible non-expansion electrical work |
| ln K = nFE°/(RT) | K dimensionless | At specified temperature; activities define K |
| G = 1/R; κ = Ccell/R | G: S; κ: S m⁻¹; Ccell: m⁻¹ | Calibrated cell and controlled temperature |
| Λm = κ/c | Λm: S m² mol⁻¹; c: mol m⁻³ | Match the SI concentration and conductivity units |
| Λm = 1000κ/c | Λm: S cm² mol⁻¹; κ: S cm⁻¹; c: mol L⁻¹ | The factor is a litre-to-cm³ conversion |
| Λm = Λm° − A√c | Slope −A and intercept Λm° | Appropriate dilute strong-electrolyte range |
| Λm° = Σνiλi° | νi: ions per formula unit | Infinite-dilution limit; same solvent and T |
| α ≈ Λm/Λm°; Ka,c ≈ cα²/(1 − α) | α dimensionless; c in stated concentration units | Dilute weak monobasic acid; mobility and ideality approximations |
| Qcharge = It; m = MIt/(zF) | I: A; t: s; M: g mol⁻¹; m: g | Constant current and ideal product efficiency |
The same letter can represent different quantities in different contexts: R is gas constant in the Nernst equation and electrical resistance in a measurement formula. Q is a reaction quotient in thermodynamics and charge in Faraday calculations; write Qcharge when ambiguity is possible. The electrochemical electron number n and the per-product electron count z both arise from balanced chemistry, not from a memorised electrode label.
Keep reaction direction and conditions explicit
Oxidation occurs at the anode and reduction at the cathode in every electrochemical cell. During galvanic discharge the anode is negative and the cathode positive; a driven electrolytic cell has a positive anode and negative cathode. Reversing the cell reaction changes E to −E and replaces Q and K with their reciprocals. Multiplying the reaction multiplies ΔrG and the electron count but leaves potential unchanged.
Cell notation describes a written direction; a negative E means the reverse direction is thermodynamically favoured at the specified conditions. Pure solids and pure liquids have unit activity in the usual reference treatment, so their masses do not appear in the quotient. Gas pressures and powers of dissolved activities do appear. At equilibrium E is zero; an open switch or equal opposing source can produce zero current without that equilibrium condition.
Problem
At 298.15 K a Daniell cell has [Zn²⁺] = 0.10 M, [Cu²⁺] = 0.010 M and E° = 1.10 V. Find E and ΔrG per mole of reaction.
- 1.Write Zn + Cu²⁺ → Zn²⁺ + Cu, giving n = 2 and Q ≈ 0.10/0.010 = 10.
- 2.E = 1.10 − (0.05915935/2)log₁₀(10) = 1.07042 V.
- 3.ΔrG = −2(96485.332)(1.07042) = −206.56 kJ mol⁻¹. The positive E and negative energy change agree, and the nonstandard potential is below E° because products are relatively abundant.
Distinguish amount of ions from their mobility
Conductivity removes the physical cell dimensions but still depends on how many ions are present per volume and how readily they move. Molar conductivity normalises this transport to the amount of electrolyte. Strong electrolytes mainly gain mobility on dilution; weak electrolytes can also gain a much larger ionised fraction. Independent ionic contributions supply weak-electrolyte limiting values when simple extrapolation is unsuitable.
Problem
A cell calibrated with κ = 0.0100 S cm⁻¹ has R = 100 Ω. A 0.00100 M weak-acid solution in the same cell gives R = 25000 Ω. Use Λm° = 400 S cm² mol⁻¹.
- 1.Ccell = κR = 0.0100 × 100 = 1.00 cm⁻¹. The acid conductivity is 1.00/25000 = 4.00 × 10⁻⁵ S cm⁻¹.
- 2.Λm = 1000κ/c = 40.0 S cm² mol⁻¹, so α ≈ 40.0/400 = 0.100.
- 3.Ka,c ≈ 0.00100(0.100)²/(0.900) = 1.11 × 10⁻⁵ mol L⁻¹, assuming background conductivity and non-ideal effects are negligible.
Electrolysis combines thermodynamics, kinetics and counting
Product identity must be chosen before product mass is calculated. In water, solvent reactions compete with dissolved ions; reactive anodes can dissolve; concentration and overpotential can alter which reaction dominates. Once the intended half-reaction is established, charge divided by F gives moles of electrons, division by z gives moles of product, and multiplication by molar mass gives ideal mass. In series cells, equal charge does not mean equal masses.
Problem
How much silver can 193 C ideally deposit from Ag⁺, and how does the answer change at 90% current efficiency? Use M(Ag) = 107.9 g mol⁻¹.
- 1.Ag⁺ + e⁻ → Ag uses one electron per atom. n(Ag) = 193/96485.332 = 0.00200030 mol.
- 2.Ideal mass = 0.00200030 × 107.9 = 0.21583 g.
- 3.At 90% current efficiency the desired deposit is 0.90 × 0.21583 = 0.19425 g. The missing fraction of charge does not produce the target silver deposit.
The same principles explain energy storage and material loss
| Application | Oxidation and reduction connection | Distinctive feature |
|---|---|---|
| Dry cell | Zn oxidation coupled to Mn(IV) reduction | Finite primary-cell material store; moist paste electrolyte |
| Mercury cell | Zn oxidation coupled to HgO reduction | Condensed-phase net chemistry helps voltage stability |
| Lead storage cell | Pb(0) oxidised and Pb(IV) reduced to Pb(II) during discharge | Controlled charging regenerates Pb, PbO₂ and acid |
| Nickel–cadmium cell | Cd oxidation coupled to Ni(III) reduction | Secondary-cell construction supports cycling |
| Hydrogen–oxygen fuel cell | Hydrogen oxidation coupled to oxygen reduction | Continuously supplied reactants and product removal |
| Iron corrosion | Local iron dissolution coupled to an environmental reduction | A wet layer and metal body can complete a local cell |
The practical consequence depends on the purpose. A battery is designed to route electrons through a useful load. Corrosion allows a local reaction to consume valuable metal, so protection aims to block contact, interrupt transport or shift the metal into a cathodic role. A fuel-cell system supplies useful current while requiring a continuing fuel pathway. In every case, matter, charge and energy remain conserved.
A valid numerical answer must have the correct sign, unit, reaction basis and physical range. Check the chemical equation before the calculator, then check whether the result matches the intended behaviour: α between 0 and 1, mass proportional to charge, and E = 0 at chemical equilibrium.
Quick Quiz
Quiz
Which statement connects thermodynamics and kinetics correctly?
A favourable chemical driving force does not remove kinetic barriers or transport losses.
A cell has E° = 0.50 V, n = 2 and Q = 100 at 298.15 K. What is E approximately?
E = 0.50 − (0.05916/2) × 2 = 0.44084 V.
Which calculation sequence correctly turns charge into deposited mass?
The steps count electron moles, product moles and finally grams.
What is common to both a working alkaline fuel cell and local iron corrosion?
Different applications share the same need for balanced electron transfer and continuous current paths.
Which value is physically inconsistent within the simple weak-acid dissociation model?
A fraction of added acid cannot be greater than one. Such a result requires checking data, units and model assumptions.
Practice Problems
- For Zn|Zn²⁺||Ag⁺|Ag, write the half-reactions, identify signs, and calculate E° from −0.76 V and +0.80 V.
- Find E, ΔrG and the standard equilibrium constant for Zn + Cu²⁺ → Zn²⁺ + Cu with E° = 1.10 V, [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.10 M at 298.15 K.
- A reaction is reversed and then multiplied by three. Describe E°, ΔrG°, electron count and K.
- Derive E for a hydrogen electrode at pH 7 and H₂ pressure 1 bar at 298.15 K.
- A cell has Ccell = 0.800 cm⁻¹ and solution resistance 400 Ω. For c = 0.0200 M find κ and Λm.
- Explain a falling κ but rising Λm during dilution, contrasting a strong salt and a weak acid.
- Using Λm°(HCl) = 425.9, Λm°(sodium ethanoate) = 91.0 and Λm°(NaCl) = 126.4, find the acid limit and α when Λm = 39.05 in the same units.
- Calculate ideal charge to reduce 0.0200 mol dichromate to Cr³⁺ in acid, and write the half-reaction.
- Cells with Ag⁺ and Cu²⁺ are connected in series. If 0.0100 mol silver deposits, how much copper deposits ideally?
- Explain products for AgNO₃(aq) with Ag electrodes, AgNO₃(aq) with Pt electrodes, dilute H₂SO₄ with Pt, and CuCl₂(aq) with Pt.
- Compare the effects of a scratch through zinc and tin coatings on iron in a conducting wet environment.
- Write lead-cell discharge and charging equations and describe the acid change.
- For the alkaline fuel cell, write both half-reactions and calculate water production for 0.200 mol electrons transferred.
- A metal has a favourable standard oxidation coupling but appears unreactive in practice. Give possible explanations.
- Design a reasoning checklist for any electrochemical calculation using an example of metal deposition.
1. Anode: Zn → Zn²⁺ + 2e⁻. Cathode: 2Ag⁺ + 2e⁻ → 2Ag. Zinc is negative and silver positive during galvanic discharge. 2. E° = 0.80 − (−0.76) = 1.56 V. Electrons move externally from zinc to silver; ions maintain the internal current path. 3. Do not double the silver potential when doubling its half-reaction.
Key Takeaways
• Electrode reactions must conserve atoms and charge before any calculation. • Potential determines reversible driving force per charge; current also depends on transport and kinetics. • Nernst, Gibbs-energy and equilibrium relationships describe the same thermodynamic system. • Conductivity and molar conductivity compare different aspects of ionic transport. • Independent-ion contributions connect limiting conductivities and weak-electrolyte behaviour. • Faraday’s laws connect measured charge to an ideal reaction amount. • Solvent, electrode material, concentration and overpotential influence electrolysis products. • Batteries, fuel cells and corrosion share coupled electrode processes but serve different practical roles. • Every result needs a sign, unit, stated reaction basis and appropriate assumptions.
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Corrosion as an Electrochemical Process
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