Electrochemistry | ISC Class 12 Chemistry Notes

This note covers electrochemical cells, electrode potentials, the electrochemical series, the Nernst equation, Gibbs energy, electrical conductance, conductivity, Kohlrausch’s law, electrolysis, batteries, fuel cells and corrosion.

How does an electrochemical cell produce electricity?

Electrochemistry connects chemical reactions with electrical energy. A galvanic cell, also called a voltaic cell, converts energy from a spontaneous chemical reaction into electrical work. An electrolytic cell uses an external electrical supply to bring about a non-spontaneous chemical reaction.

A redox reaction combines oxidation, the loss of electrons, with reduction, the gain of electrons. An electrode conducts electrons into or out of the reacting system. An electrolyte conducts electricity through mobile ions, which are charged particles.

What happens in the Daniell cell?

The Daniell cell combines a zinc electrode in zinc sulphate solution with a copper electrode in copper sulphate solution. Each electrode-solution pair forms a half-cell. A wire connects the electrodes; a salt bridge connects the solutions ionically.

1. At the anode, the electrode where oxidation occurs, zinc loses electrons: Zn(s) → Zn²⁺(aq) + 2e⁻.
2. The electrons pass through the external circuit from zinc to copper. Conventional current in that circuit flows in the opposite direction.
3. At the cathode, the electrode where reduction occurs, copper ions accept electrons: Cu²⁺(aq) + 2e⁻ → Cu(s).
4. The combined reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Zinc dissolves and copper is deposited.

Here Zn means zinc, Cu copper and e⁻ an electron; the superscript 2+ indicates a doubly positive ion. The state symbols (s), (aq), (l) and (g) mean solid, aqueous solution, liquid and gas respectively.

The drawing shows zinc and copper strips in separate solutions containing their salts, joined by a salt bridge. Arrows above the connecting wire show electron flow from zinc to copper and current in the opposite direction.

See Fig. 2.1 in your NCERT textbook

How are cells represented and compared?

In Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s), a single line marks a phase boundary, where physical phases meet; the double line represents the salt bridge. The anode is on the left and cathode on the right.

How are electrode potentials measured using the hydrogen electrode?

An electrode potential is the electrode-electrolyte potential difference. Metal atoms may dissolve as ions, leaving electrons; ions may deposit on the metal.

A measurement gives the difference between two half-cell potentials; a single potential cannot be measured independently. The standard hydrogen electrode, abbreviated SHE, provides the reference: its standard potential is assigned zero volts at all temperatures; V denotes volt.

What are the construction and standard conditions?

A platinum electrode coated with platinum black, finely divided platinum, is immersed in an acidic solution. Pure hydrogen gas is bubbled around it at 1 bar pressure, with hydrogen-ion concentration 1 mol L⁻¹. Here mol means mole, and L means litre.

Platinum, symbol Pt, provides a conducting surface without participating in the reaction. Hydrogen gas is H₂ and the hydrogen ion is H⁺. The reversible electrode reaction is 2H⁺(aq) + 2e⁻ ⇌ H₂(g), where ⇌ indicates a reversible reaction.

The figure shows hydrogen entering at 1 bar, bubbles around platinum foil coated with finely divided platinum, and a surrounding solution labelled 1.0 M H⁺. M denotes molarity, measured in mol L⁻¹.

See Fig. 2.3 in your NCERT textbook

How does the reference establish zinc and copper potentials?

Standard electrode potential, written E°, means the standard reduction potential at the stated temperature. The circle denotes standard conditions. Dissolved species have unit activity, approximated by unit molar concentration in this treatment; gases have pressure 1 bar. Activity means effective concentration relative to a standard state.

At 298 K, where K denotes kelvin, connecting copper to the SHE gives E°(Cu²⁺/Cu) = +0.34 V. For zinc, E°(Zn²⁺/Zn) = −0.76 V. Zinc is oxidised in its spontaneously operating cell with the SHE; copper ions are reduced in theirs.

Electromotive force, or emf, is cell potential with no current drawn. For right and left reduction potentials Eright and Eleft, the cell emf Ecell is:

Ecell = Eright − Eleft

The SHE requires pure hydrogen and controlled conditions, making routine use inconvenient. Its conventional reference value does not give an independently measurable absolute electrode potential.

How does the electrochemical series predict a reaction?

The electrochemical series arranges redox couples by standard reduction potential. A redox couple contains the oxidised and reduced forms of a species. A more positive potential indicates a greater reduction tendency under standard conditions.

An oxidising agent accepts electrons and is reduced. A reducing agent supplies electrons and is oxidised.

How is the sign of cell emf used?

For a proposed reaction, subtract the anode’s reduction potential from the cathode’s reduction potential. A positive emf corresponds to a spontaneous forward reaction under the specified conditions. A negative emf corresponds to a non-spontaneous forward reaction; at equilibrium the emf is zero.

Worked example 1. Calculate the standard Daniell-cell emf from E°(Cu²⁺/Cu) = 0.34 V and E°(Zn²⁺/Zn) = −0.76 V.

Formula: E°cell = E°cathode − E°anode.

Substitute: E°cell = 0.34 − (−0.76).

Answer: E°cell = 1.10 V. Zinc can reduce copper ions, so a zinc vessel is unsuitable for storing copper sulphate solution.

Worked example 2. For Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), calculate standard emf given E°(Ag⁺/Ag) = 0.80 V and E°(Cu²⁺/Cu) = 0.34 V. Ag denotes silver.

Formula: E°cell = E°cathode − E°anode.

Substitute: E°cell = 0.80 − 0.34.

Answer: E°cell = 0.46 V. Silver ions can oxidise copper under standard conditions.

Note: Multiplying a half-reaction to balance electrons does not multiply its electrode potential. Potential is an intensive quantity, meaning that it does not scale with the amount of reacting material.

Electrode potential depends on chemical nature, concentration and temperature. The Nernst equation handles concentration changes. Thermodynamic feasibility does not establish reaction speed.

How does the Nernst equation account for concentration?

The Nernst equation connects emf with composition and temperature. Let n count electrons transferred in the balanced reaction and Q be the reaction quotient, the product-to-reactant activity ratio with stoichiometric powers.

Let Rg denote the gas constant, 8.314 J K⁻¹ mol⁻¹; J denotes joule. Let T be absolute temperature in kelvin and F the Faraday constant, the charge per mole of electrons, here 96487 C mol⁻¹. C denotes coulomb, the unit of charge.

Ecell = E°cell − (RgT/nF) ln Q

Here ln denotes natural logarithm. At 298 K, the base-ten logarithm, log, gives Ecell = E°cell − (0.059 V/n) log Q. The numerical coefficient 0.059 V is specific to this temperature and rounding.

How is the reaction quotient assembled?

Pure solids and pure liquids have unit activity and do not appear as variable concentration terms. For dilute solutions, use the numerical concentrations in mol L⁻¹ relative to the standard concentration. Square brackets around a dissolved species denote its concentration.

For the Daniell reaction, Q = [Zn²⁺]/[Cu²⁺] and n = 2. Thus increasing copper-ion concentration increases emf, whereas increasing zinc-ion concentration decreases it. These comparisons hold at constant temperature with the other concentration fixed.

Worked example 3. For Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s) at 298 K, E°cell = 1.05 V, [Ni²⁺] = 0.160 mol L⁻¹ and [Ag⁺] = 0.002 mol L⁻¹. Ni means nickel. Find the emf using 0.059 V.

Formula: Q = [Ni²⁺]/[Ag⁺]²; Ecell = E°cell − (0.059 V/2) log Q.

Substitute: Q = 40000; Ecell = 1.05 − (0.059/2) log 40000.

Answer: Ecell = 0.91 V, rounded to two decimal places.

Worked example 4. At 298 K, the reaction Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s) has E°cell = 3.17 V. Magnesium, Mg, has ion concentration 0.130 mol L⁻¹; silver-ion concentration is 0.0001 mol L⁻¹. Find the emf using 0.059 V.

Formula: Q = [Mg²⁺]/[Ag⁺]²; Ecell = E°cell − (0.059 V/2) log Q.

Substitute: Q = 0.130/(0.0001)² = 1.30 × 10⁷. The concentration correction is approximately 0.21 V.

Answer: Ecell = 3.17 − 0.21 = 2.96 V. The quotient is greater than unity, so the emf is below its standard value.

How are cell emf, Gibbs energy and equilibrium connected?

Gibbs energy change, ΔG, describes the thermodynamic driving force at constant temperature and pressure; Δ means change. The maximum electrical work obtainable from a galvanic cell requires reversible operation, with charge transferred against an opposing potential differing infinitesimally from the cell emf.

ΔG = −nFEcell

Here ΔG is expressed per mole of reaction as written, n is its electron-transfer number, and F is the Faraday constant. Under standard conditions, ΔG° = −nFE°cell. Positive emf means negative Gibbs energy change and spontaneous forward reaction.

Derivation: How is the equilibrium constant obtained?

1. Write the Nernst relation, Ecell = E°cell − (RgT/nF) ln Q, using the balanced reaction.
2. At equilibrium, the cell emf becomes zero. The reaction quotient becomes K, the equilibrium constant for that reaction.
3. Substitute these conditions to obtain 0 = E°cell − (RgT/nF) ln K, then rearrange.

E°cell = (RgT/nF) ln K

Combining this result with the standard Gibbs energy relation gives ΔG° = −RgT ln K. At 298 K, E°cell = (0.059 V/n) log K. Thus a positive standard emf corresponds to an equilibrium constant greater than one.

What changes when a reaction is multiplied?

Doubling the Daniell reaction doubles the number of transferred electrons and doubles its Gibbs energy change. Its emf remains unchanged.

For Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°cell = 0.46 V and n = 2. At 298 K, log K = 2 × 0.46/0.059 = 15.6 after rounding; K is approximately 3.92 × 10¹⁵.

During Daniell-cell operation, zinc-ion concentration rises and copper-ion concentration falls, increasing Q and decreasing emf. Equilibrium means zero actual emf, not necessarily zero standard emf.

How are resistance, conductance and conductivity measured?

Metallic conductance results from electron movement. Electrolytic conductance results from ion movement through solution. Metal composition remains unchanged during electron conduction; prolonged direct current through an electrolyte can alter its composition through electrode reactions.

Solvation means interaction of ions with surrounding solvent molecules; viscosity describes resistance to flow. These factors affect how readily ions move.

What quantities and units must be distinguished?

Let R be resistance, l the conducting column’s length and A its cross-sectional area. This R is electrical resistance; Rg remains the gas constant. Let ρ, rho, be resistivity, also called specific resistance.

R = ρl/A

G = 1/R

G is conductance. Conductivity, κ, pronounced kappa, is the reciprocal of resistivity: κ = 1/ρ. It is also called specific conductance. Conductance depends on the dimensions of the conducting sample; conductivity characterises the material under stated conditions.

• The SI unit of resistance is ohm, symbol Ω; 1 Ω = 1 V A⁻¹, where A in this unit expression means ampere.
• The SI unit of conductance is siemens, symbol S; 1 S = 1 Ω⁻¹.
• The SI unit of resistivity is ohm metre, Ω m; m in a unit means metre.
• The SI unit of conductivity is siemens per metre, S m⁻¹. Also, 1 S cm⁻¹ = 100 S m⁻¹, where cm means centimetre.
• The SI unit of cell constant is reciprocal metre, m⁻¹.

How is a conductivity cell calibrated?

A conductivity cell contains two platinum electrodes coated with platinum black. Its cell constant, G*, equals l/A. The asterisk distinguishes this geometric quantity from conductance G. Since R = l/(κA), the calibration relation is G* = κR.

Calibrate G* using potassium chloride solution, KCl, of known conductivity. Measure the unknown solution’s resistance in the same cell: κ = G*/R. Alternating current avoids composition changes caused by direct current.

Worked example 5. A cell containing 0.1 mol L⁻¹ KCl has resistance 100 Ω and conductivity 1.29 S m⁻¹. In the same cell, 0.02 mol L⁻¹ KCl has resistance 520 Ω. Calculate the cell constant and the second solution’s conductivity.

Formula: G* = κR; κ = G*/R.

Substitute: G* = 1.29 × 100 = 129 m⁻¹; κ = 129/520.

Answer: The cell constant is 129 m⁻¹, equivalent to 1.29 cm⁻¹, and the second solution’s conductivity is 0.248 S m⁻¹.

What are molar and equivalent conductance?

Molar conductivity, Λm, is the conductance associated with the volume of solution containing one mole of electrolyte between electrodes one unit distance apart. The Greek capital lambda, Λ, is used here with m indicating the molar quantity.

If concentration c is in mol m⁻³ and κ is in S m⁻¹, then Λm = κ/c in S m² mol⁻¹. If c is in mol L⁻¹ and κ in S cm⁻¹, use Λm = 1000κ/c in S cm² mol⁻¹.

Derivation: Why does molar conductivity equal conductivity divided by concentration?

1. For the solution column, conductance is G = κA/l, with A its cross-sectional area and l its length.
2. Choose unit electrode separation. Let Vm be the volume containing one mole; the required area accommodates that volume.
3. Then Λm = κVm; with matching volume units, Vm = 1/c.

Λm = κ/c

The SI unit of molar conductivity is S m² mol⁻¹. For conversion, 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹. The factor 1000 converts litres to cubic centimetres and is absent from the coherent SI formula.

How does equivalent conductance differ?

Equivalent conductance, Λeq, refers to the volume containing one equivalent of electrolyte. An equivalent expresses reacting capacity, corresponding to one mole of electrons in a specified redox process. Let f be the number of equivalents per mole for the stated process.

Normality, N, is concentration in equivalents per litre. If c is in mol L⁻¹, N = fc. With κ in S cm⁻¹, Λeq = 1000κ/N in S cm² equiv⁻¹, where equiv means equivalent. Thus Λm = fΛeq when compatible units are used.

Worked example 6. A 0.05 mol L⁻¹ sodium hydroxide solution, NaOH, occupies a column 50 cm long and 1 cm in diameter. Its resistance is 5.55 × 10³ Ω. Find resistivity, conductivity and molar conductivity; use π = 3.14.

Formula: A = πr²; ρ = RA/l; κ = 1/ρ; Λm = 1000κ/c. Here r is the column radius and π is the circle constant.

Substitute: r = 0.5 cm; A = 0.785 cm²; ρ = 5550 × 0.785/50 = 87.135 Ω cm.

Answer: Resistivity is 87.135 Ω cm; κ = 0.01148 S cm⁻¹ and Λm = 229.6 S cm² mol⁻¹.

Why does dilution affect different conductivity measures differently?

Dilution lowers electrolyte concentration by adding solvent. Conductivity decreases on dilution for both strong and weak electrolytes because the number of current-carrying ions per unit volume decreases. Molar conductivity increases for the expanding volume containing one mole.

A strong electrolyte is extensively dissociated into ions in solution; a weak electrolyte is only partly dissociated. For strong electrolytes, molar conductivity increases slowly on dilution. For weak electrolytes, increasing dissociation causes a much steeper increase, especially at low concentrations.

How should the concentration graph be interpreted?

Limiting molar conductivity, Λ°m, is the value as concentration approaches zero, also called infinite dilution. Here the circle denotes the limiting value. For strong electrolytes in dilute solution, Λm = Λ°m − B√c, where B is a positive constant and √c is the square root of concentration.

At a given solvent and temperature, B depends on electrolyte type, meaning the ionic charges. A plot of Λm against √c has intercept Λ°m and slope −B. Intercept means the vertical-axis value; slope measures change in the vertical variable per change in the horizontal variable.

The horizontal axis shows √c and the vertical axis Λm. The potassium chloride line slopes gently downwards to the right. The acetic acid curve, labelled CH₃COOH, rises steeply towards the vertical axis as concentration approaches zero.

See Fig. 2.6 in your NCERT textbook

Acetic acid’s limiting value cannot be obtained by the same straight-line extrapolation. Extrapolation extends a measured trend beyond its observations. At very low weak-electrolyte concentrations, conductivity is too low for accurate direct measurement; Kohlrausch’s law supplies the limiting value indirectly.

Equivalent conductance also increases on dilution for a fixed electrolyte and equivalent convention because Λeq = Λm/f with constant f.

How is Kohlrausch’s law used for weak electrolytes?

Kohlrausch’s law of independent migration of ions states that limiting molar conductivity is the sum of the individual ionic contributions. Each contribution must be multiplied by the number of that ion produced from one formula unit of electrolyte.

Write Λ°m = ν₊λ°₊ + ν₋λ°₋. The symbols ν₊ and ν₋, nu, are the numbers of cations and anions. The symbols λ°₊ and λ°₋, lower-case lambda, are their limiting ionic molar conductivities. All contributions must refer to the same solvent and temperature.

Worked example 7. At 298 K, the limiting ionic molar conductivities of calcium ions, Ca²⁺, and chloride ions, Cl⁻, are 119.0 and 76.3 S cm² mol⁻¹. Find the limiting molar conductivity of calcium chloride, CaCl₂.

Formula: Λ°m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻).

Substitute: Λ°m = 119.0 + 2 × 76.3.

Answer: Λ°m(CaCl₂) = 271.6 S cm² mol⁻¹. The factor two counts the two chloride ions in each formula unit.

How can a weak acid’s limiting value be found?

For acetic acid, CH₃COOH, combine hydrochloric acid, HCl; sodium acetate, CH₃COONa; and sodium chloride, NaCl. Adding the first two limiting conductivities and subtracting the third leaves hydrogen and acetate contributions.

Worked example 8. The limiting molar conductivities of HCl, CH₃COONa and NaCl are respectively 425.9, 91.0 and 126.4 S cm² mol⁻¹ at the same temperature. Calculate the value for acetic acid.

Formula: Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl).

Substitute: Λ°m = 425.9 + 91.0 − 126.4.

Answer: Λ°m(CH₃COOH) = 390.5 S cm² mol⁻¹. Sodium-ion and chloride-ion contributions cancel in the combination.

The degree of dissociation, α, alpha, is the fraction dissociated into ions. For a weak electrolyte, it can be approximated by α = Λm/Λ°m. This approximation must not be presented as an exact identity under all conditions.

For a weak acid producing one hydrogen ion and one anion per molecule, its dissociation constant, Ka, is the equilibrium concentration quotient. It is given by Ka = cα²/(1 − α), or cΛm²/[Λ°m(Λ°m − Λm)], in the concentration treatment.

How do Faraday’s laws describe electrolysis?

Electrolysis uses electrical energy to drive chemical change at electrodes. In copper sulphate solution with copper electrodes, copper dissolves from the anode and deposits at the cathode. This principle purifies copper by making impure copper the anode and collecting copper at the cathode.

What are the two laws and their mathematical forms?

Faraday’s first law states that the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed. Let q be charge, I current, t time and m the mass deposited. Lower-case q is charge; upper-case Q remains the reaction quotient.

q = It

m = Zq

Z is the electrochemical equivalent, mass deposited per unit charge. For constant current, m = ZIt. Current is measured in amperes, A, time in seconds, s, and charge in coulombs, C. Thus 1 C = 1 A s.

Faraday’s second law states that equal charges passing through different electrolytes liberate masses proportional to their chemical equivalent masses. If Eeq is equivalent mass, then m₁/m₂ = Eeq₁/Eeq₂ for the same charge. Subscripts 1 and 2 identify the two substances.

For metal deposition, let Mmol be molar mass and z the electrons required per metal ion. Then Eeq = Mmol/z, Z = Mmol/(zF), and m = MmolIt/(zF).

F = Nₐe

Nₐ is Avogadro’s constant, the number of particles per mole; e is the magnitude of one electron’s charge. One mole of electrons carries one faraday of charge. Use F ≈ 96500 C mol⁻¹ for approximate calculations, or the specified value.

Worked example 9. Copper sulphate solution is electrolysed at 1.5 A for 10 minutes. Find the copper deposited, taking Mmol(Cu) = 63 g mol⁻¹, F = 96487 C mol⁻¹ and Cu²⁺ + 2e⁻ → Cu. Here g means gram.

Formula: q = It; m = Mmolq/(zF).

Substitute: t = 600 s; q = 1.5 × 600 = 900 C; z = 2.

Answer: m = 63 × 900/(2 × 96487) = 0.2938 g of copper.

What determines the products?

Products depend on electrolyte, concentration and electrodes. Reactive electrodes participate; inert electrodes conduct electrons. Overpotential is the extra potential required when a thermodynamically feasible electrode process is kinetically slow. Standard potentials alone may therefore be insufficient to predict the observed product.

How do primary cells, storage batteries and fuel cells work?

A primary cell becomes unusable after its reactants are consumed in operation. A secondary cell can be recharged by passing current in the opposite direction, reversing the discharge reaction. A fuel cell receives reactants continuously and directly converts their chemical energy into electricity.

What happens in the Leclanche dry cell?

The Leclanche cell has a zinc container as anode and a graphite, or carbon, rod surrounded by manganese dioxide and carbon as the cathode arrangement. A moist paste of ammonium chloride and zinc chloride fills the space between them. Its potential is nearly 1.5 V.

The formulas are MnO₂ for manganese dioxide, NH₄Cl for ammonium chloride and ZnCl₂ for zinc chloride. The electrode reactions are complex, but can be written approximately as follows:

• Anode: Zn(s) → Zn²⁺ + 2e⁻.
• Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃.

NH₄⁺ is the ammonium ion, NH₃ is ammonia, and MnO(OH) is manganese oxyhydroxide. Manganese is reduced from oxidation state +4 to +3. Oxidation state is the formal charge assigned by electron-counting rules. Ammonia combines with zinc ions to form [Zn(NH₃)₄]²⁺, a complex ion with four ammonia molecules coordinated to zinc. Dry cells are used in transistor radios and clocks.

Why does a mercury cell maintain its potential?

A mercury cell uses zinc-mercury amalgam, an alloy containing mercury, as anode. Mercury(II) oxide, HgO, mixed with carbon forms the cathode material. The electrolyte is a paste of potassium hydroxide, KOH, and zinc oxide, ZnO. Mercury is represented by Hg.

• Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻. Zn(Hg) denotes zinc in the amalgam; OH⁻ is hydroxide and H₂O is water.
• Cathode: HgO + H₂O + 2e⁻ → Hg(l) + 2OH⁻.
• Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l).

Its approximately 1.35 V potential remains constant during its life: the overall reaction contains no solution ion whose concentration can change. It is suitable for low-current devices such as hearing aids and watches.

How is a lead accumulator recharged?

The lead storage battery, or lead accumulator, uses a lead anode, a lead grid packed with lead dioxide as cathode, and 38% sulphuric acid solution. Pb means lead, PbO₂ lead dioxide, H₂SO₄ sulphuric acid, SO₄²⁻ sulphate ion and PbSO₄ lead sulphate.

• Discharge anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻.
• Discharge cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l).
• Overall discharge: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l).

Charging reverses the overall reaction: lead sulphate is converted back to lead and lead dioxide, and sulphuric acid is regenerated. The battery is commonly used in automobiles and inverters.

How does the hydrogen-oxygen fuel cell operate?

Hydrogen and oxygen gases pass through porous carbon electrodes into concentrated aqueous sodium hydroxide. Finely divided platinum or palladium catalysts increase electrode-reaction rates. A catalyst increases reaction rate without overall consumption.

• Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻.
• Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq), where O₂ denotes oxygen gas.
• Overall: 2H₂(g) + O₂(g) → 2H₂O(l).

The cell operates continuously as long as reactants are supplied. It provided electrical power in the Apollo space programme, with condensed product water added to the astronauts’ drinking-water supply.

How does electrochemical corrosion occur and how is it prevented?

Corrosion is the deterioration of a metal through reactions that form oxides or other salts. Rusting of iron, tarnishing of silver and the green coating on copper and bronze are examples. Iron rusts in the presence of water and air.

Rusting is complex but may be considered essentially an electrochemical phenomenon. Separate iron-surface spots act as anode and cathode. Metal conducts electrons between them; moisture supports ionic reactions.

What are the steps in rust formation?

1. At an anodic spot, iron is oxidised: 2Fe(s) → 2Fe²⁺(aq) + 4e⁻. Fe denotes iron; Fe²⁺ is the ferrous, or iron(II), ion.
2. Electrons travel through the metal to another spot, where oxygen is reduced in the presence of hydrogen ions.
3. The cathodic reaction is O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l).
4. Atmospheric oxygen further oxidises ferrous ions to ferric, or iron(III), ions. Rust forms as hydrated ferric oxide, Fe₂O₃·xH₂O; x indicates a variable amount of associated water.

The hydrogen ions are believed to be available from carbonic acid formed when atmospheric carbon dioxide dissolves in water. Dissolved acidic oxides may also supply hydrogen ions. Carbon dioxide is CO₂ and carbonic acid is H₂CO₃.

Which conditions and protective methods matter?

Water and oxygen enable rusting; acidity affects the cathodic reaction. Surface isolation interrupts contact with the atmosphere.

A sacrificial electrode is consumed preferentially so that the protected metal is preserved.

Glossary

• Galvanic cell — A device converting the energy of a spontaneous redox reaction into electrical work.
• Electrolytic cell — A device using externally supplied electrical energy to drive a non-spontaneous chemical reaction.
• Anode — The electrode at which oxidation occurs, with electrons released by the reacting species.
• Cathode — The electrode at which reduction occurs, with reacting species accepting electrons.
• Cell emf — The potential difference between cell electrodes measured when no current is drawn.
• Reaction quotient — The activity ratio of products to reactants, each raised to its stoichiometric power.
• Conductivity — The reciprocal of resistivity, describing electrical conduction under specified material and temperature conditions.
• Molar conductivity — Conductance associated with the solution containing one mole of electrolyte at unit electrode separation.
• Limiting molar conductivity — The molar conductivity approached as electrolyte concentration tends towards zero, or infinite dilution.
• Faraday constant — The magnitude of electric charge carried by one mole of electrons.
• Overpotential — Additional potential needed when an electrochemical process is thermodynamically feasible but kinetically slow.
• Corrosion — Deterioration of a metal through chemical reactions forming oxides or other salts.

Common errors and misconceptions

• Misconception: The anode is negative in every cell. Correct: The anode is defined by oxidation; it is negative in a galvanic cell and positive in an electrolytic cell.
• Misconception: A zinc reduction potential must be multiplied when balancing electrons. Correct: Stoichiometric multiplication changes reaction amounts and Gibbs energy, but electrode potential remains unchanged.
• Misconception: A positive standard emf guarantees the same emf at every concentration. Correct: Actual emf depends on concentration and temperature through the Nernst equation.
• Misconception: Conductivity and molar conductivity both fall on dilution. Correct: Conductivity decreases, but molar conductivity increases because it refers to the volume containing one mole.
• Misconception: The factor 1000 belongs in every molar-conductivity calculation. Correct: Use it with conductivity in S cm⁻¹ and concentration in mol L⁻¹; coherent SI units need no such factor.
• Misconception: A weak electrolyte’s limiting conductivity follows from straight-line extrapolation. Correct: Its concentration dependence is strongly curved, so use Kohlrausch’s law to obtain the limiting value.
• Misconception: One faraday deposits one mole of every metal. Correct: The required charge depends on electron stoichiometry; copper-ion reduction to copper requires two faradays per mole.
• Misconception: Rusting involves oxidation alone. Correct: Iron oxidation is coupled to oxygen reduction at another region, with electron movement through the metal.

Exam-style questions with model answers

Q1. Define the anode and cathode of an electrochemical cell in terms of electron transfer. [2 marks]

1. The anode is the electrode where oxidation occurs: a reacting species loses electrons to the electrode.
2. The cathode is the electrode where reduction occurs: a reacting species gains electrons supplied through the electrode.

Q2. At 298 K, E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) under standard conditions, identify the electrodes, calculate the emf and state whether the reaction is spontaneous. [3 marks]

1. Zinc is oxidised to zinc ions at the anode, while copper ions are reduced to copper at the cathode. The cell therefore uses the copper reduction potential as the cathode value.
2. Standard cell emf equals cathode reduction potential minus anode reduction potential: E°cell = 0.34 − (−0.76) = 1.10 V.
3. The positive standard emf shows that the reaction as written is spontaneous under standard conditions.

Q3. For Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s) at 298 K, E°cell = 3.17 V, [Mg²⁺] = 0.130 mol L⁻¹ and [Ag⁺] = 0.0001 mol L⁻¹. Calculate the emf using Ecell = E°cell − (0.059 V/n) log Q and log(1.30 × 10⁷) = 7.114. [4 marks]

1. Magnesium loses two electrons and two silver ions each gain one. The balanced reaction therefore transfers n = 2 electrons, fixing the denominator in the Nernst correction.
2. Omitting pure solids, Q = [Mg²⁺]/[Ag⁺]² = 0.130/(0.0001)² = 1.30 × 10⁷.
3. The concentration correction is (0.059/2) × 7.114 = 0.210 V, to three decimal places.
4. Subtracting this correction gives Ecell = 3.17 − 0.210 = 2.96 V. The stated composition lowers the emf relative to its standard value.

Q4. A conductivity cell containing 0.1 mol L⁻¹ KCl of conductivity 1.29 S m⁻¹ has resistance 100 Ω. The same cell gives resistance 520 Ω with 0.02 mol L⁻¹ KCl. Calculate its cell constant, the second conductivity and its molar conductivity. Use 1 mol L⁻¹ = 1000 mol m⁻³ and 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹. [5 marks]

1. The cell constant is the electrode separation divided by cross-sectional area. Obtain it from the calibration solution: G* = κR = 1.29 × 100 = 129 m⁻¹.
2. The same cell retains this geometric constant. For the second solution, κ = G*/R = 129/520 = 0.248 S m⁻¹ after rounding.
3. Convert the second concentration before using the SI conductivity: c = 0.02 × 1000 = 20 mol m⁻³.
4. Molar conductivity equals conductivity divided by concentration, so Λm = 0.248/20 = 0.0124 S m² mol⁻¹.
5. The unit conversion gives Λm = 124 S cm² mol⁻¹. This quantity refers to one mole of electrolyte, whereas the measured resistance also depends on cell geometry.

Q5. At the same temperature, Λ°m(HCl) = 425.9, Λ°m(CH₃COONa) = 91.0 and Λ°m(NaCl) = 126.4 S cm² mol⁻¹. Use Kohlrausch’s law to calculate Λ°m(CH₃COOH) and explain the ionic cancellation. [3 marks]

1. Kohlrausch’s law expresses limiting molar conductivity as the sum of ionic contributions. For acetic acid, the required contributions are those of hydrogen ions and acetate ions.
2. Add the values for hydrochloric acid and sodium acetate, then subtract sodium chloride. The sodium-ion and chloride-ion contributions cancel, leaving the desired pair.
3. Therefore Λ°m(CH₃COOH) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹ at the stated common temperature.

Q6. Copper sulphate solution is electrolysed for 10 minutes at 1.5 A, depositing copper by Cu²⁺ + 2e⁻ → Cu. Calculate the deposited mass using Mmol(Cu) = 63 g mol⁻¹, F = 96487 C mol⁻¹ and 1 minute = 60 seconds. [4 marks]

1. Convert the duration to seconds so that it matches the ampere definition: t = 10 × 60 = 600 s.
2. The charge passed is current multiplied by time: q = It = 1.5 × 600 = 900 C.
3. The stated half-reaction requires two moles of electrons per mole of copper. Thus depositing 63 g of copper requires 2 × 96487 C.
4. Mass deposited is m = 63 × 900/(2 × 96487) = 0.2938 g, assuming the charge produces the stated copper-deposition reaction.

Q7. A lead storage battery uses 38% sulphuric acid solution. Describe its construction, write its two discharge half-reactions and overall reaction, and explain recharging. [5 marks]

1. The battery has a lead anode and a lead grid packed with lead dioxide as cathode, immersed in 38% sulphuric acid solution as electrolyte.
2. During discharge, lead is oxidised at the anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻. Lead sulphate forms on that electrode.
3. At the cathode, lead dioxide is reduced: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l).
4. Adding the half-reactions gives Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l). Both electrodes form lead sulphate during discharge.
5. Passing current in the opposite direction reverses the discharge reaction. Lead sulphate is converted back into lead and lead dioxide, sulphuric acid is regenerated, and the cell can be used again.

Q8. Explain the electrochemical mechanism of rusting, giving the anodic reaction, cathodic reaction and subsequent formation of rust. [3 marks]

1. At anodic regions of the iron surface, iron loses electrons: 2Fe(s) → 2Fe²⁺(aq) + 4e⁻. These electrons pass through the metal to another region.
2. At cathodic regions, oxygen accepts electrons in the presence of hydrogen ions: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l).
3. Atmospheric oxygen further oxidises ferrous ions to ferric ions. Rust separates as hydrated ferric oxide, Fe₂O₃·xH₂O, where x represents its variable associated water.

Key takeaways

• Oxidation occurs at the anode and reduction at the cathode; electrode signs depend on whether the cell generates or consumes electricity.
• Calculate cell emf by subtracting the anode reduction potential from the cathode reduction potential, with both values referring to the stated conditions.
• The Nernst equation connects emf with concentration and temperature; construct its reaction quotient from the balanced cell reaction.
• Positive cell emf corresponds to negative Gibbs energy change, while equilibrium requires zero actual cell emf.
• Conductivity decreases on dilution, but molar conductivity increases; distinguish a unit volume of solution from the volume containing one mole.
• Kohlrausch’s law combines limiting ionic contributions and allows limiting molar conductivity of weak electrolytes to be found indirectly.
• Faraday’s laws relate deposited mass to charge and equivalent mass; electron stoichiometry determines how much charge deposits one mole.
• Primary cells, rechargeable storage batteries and continuously supplied fuel cells use different arrangements for converting chemical energy into electricity.
• Rusting couples iron oxidation to oxygen reduction, and protection can isolate the surface or supply a sacrificial metal.

Test yourself

What does the double vertical line in cell notation represent?

It represents the salt bridge joining the electrolyte solutions of the two half-cells.

What makes the standard hydrogen electrode a reference?

Its standard electrode potential is assigned zero at all temperatures, allowing other electrode potentials to be measured relative to it.

Why is the silver-ion concentration squared in the quotient for Ni + 2Ag⁺ → Ni²⁺ + 2Ag?

The balanced equation contains two silver ions, so their concentration is raised to the second power in the reaction quotient.

What happens to actual emf when the cell reaction reaches equilibrium?

The actual cell emf becomes zero, while its standard emf need not be zero.

Why is alternating current used in conductivity measurements?

Direct current can change solution composition through electrode reactions; alternating current resolves this difficulty in resistance measurement.

Why does weak-electrolyte molar conductivity rise steeply on dilution?

Dissociation increases, producing more ions within the total volume containing one mole of electrolyte.

Why can a lead storage battery be used again after discharge?

Charging reverses its discharge reaction, converting lead sulphate back into lead and lead dioxide and regenerating sulphuric acid.

How does a sacrificial electrode protect iron?

A connected metal such as magnesium or zinc corrodes preferentially, preserving the iron object.