Electrochemistry

An electrochemical cell is a device in which a redox reaction and an electric current are coupled. There are two kinds, and the whole chapter hangs on the difference.

A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy. The Daniell cell is the standard example: zinc dipped in ZnSO₄ and copper dipped in CuSO₄, the two solutions joined by a salt bridge. The reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) runs on its own and the cell delivers 1.1 V under standard conditions.

An electrolytic cell does the opposite: an external source drives a non-spontaneous reaction. NCERT explains this with the same Daniell cell connected to an external battery. If the external potential is less than 1.1 V, the cell works as a galvanic cell: electrons flow from Zn to Cu, Zn dissolves and Cu deposits. When the external potential equals exactly 1.1 V, the reaction stops and no current flows. When it exceeds 1.1 V, the reaction reverses: Cu dissolves, Zn deposits, and the cell is now electrolytic.

Two definitions the examiner expects word for word: oxidation is loss of electrons and takes place at the anode; reduction is gain of electrons and takes place at the cathode. That is true in both kinds of cell. What changes is the sign: in a galvanic cell the anode is negative and the cathode positive; in an electrolytic cell the anode is connected to the positive terminal of the battery and the cathode to the negative terminal.

In a galvanic cell the redox reaction is split into two half reactions that take place in two separate compartments called half-cells. Each half-cell is a metal electrode in contact with a solution of its ions. In the Daniell cell the oxidation half-cell is Zn(s) → Zn²⁺(aq) + 2e⁻ and the reduction half-cell is Cu²⁺(aq) + 2e⁻ → Cu(s).

How current flows. Electrons released at the zinc anode travel through the external wire to the copper cathode. Inside the cell, current is carried by ions: cations move towards the cathode and anions towards the anode. The salt bridge, a U-tube of agar gel containing an inert electrolyte such as KCl or NH₄NO₃, completes the circuit and keeps both solutions electrically neutral; without it, positive charge would build up in the anode compartment and the cell would stop within seconds.

Electrode potential and emf. A potential difference develops between each electrode and its electrolyte; this is the electrode potential. When the concentrations of all species are unity (1 M, gases at 1 bar) and the temperature is 298 K it is the standard electrode potential E°. The electrode where reduction occurs is at higher potential (positive), the electrode where oxidation occurs is at lower potential (negative). The potential difference between the two electrodes when no current is drawn is the cell potential, and it is called the electromotive force (emf). E_cell = E_right − E_left = E_cathode − E_anode, where both are reduction potentials.

Cell notation (IUPAC). Anode on the left, cathode on the right, a single vertical bar for a phase boundary and a double bar for the salt bridge: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Concentrations go in brackets after the ion. For a gas electrode you write the inert metal too: Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M).

The potential of a single electrode cannot be measured, because any measurement needs two electrodes and gives only the difference. So one electrode is chosen as reference and assigned zero. That reference is the standard hydrogen electrode (SHE): a platinum foil coated with platinum black (to adsorb hydrogen and speed up the reaction), dipped in a 1 M solution of H⁺ ions at 298 K, with pure hydrogen gas bubbled over it at 1 bar. Its half reaction is H⁺(aq) + e⁻ → ½H₂(g), and E° is defined as 0.00 V at all temperatures.

To find the standard potential of any electrode Mⁿ⁺/M, set up the cell Pt(s) | H₂(g, 1 bar) | H⁺(1 M) || Mⁿ⁺(1 M) | M and measure its emf. Since the left electrode is 0, E_cell = E°(Mⁿ⁺/M). If the metal is a stronger reducing agent than hydrogen, it acts as the anode and its E° comes out negative (Zn, −0.76 V); if weaker, it is the cathode and E° is positive (Cu, +0.34 V). Note that E° is the same whether the half reaction is written for one mole or for two: potential is intensive, unlike ΔG.

Arranging electrodes in order of their standard reduction potentials gives the electrochemical series. The rules for using it: the lower (more negative) the E°, the stronger the metal as a reducing agent and the more easily it is oxidised; the higher (more positive) the E°, the stronger the oxidising agent (F₂ at +2.87 V is the strongest). A metal higher in reducing power displaces any metal below it from a solution of its salt, which is why zinc pots cannot store CuSO₄ solution. Metals with negative E° liberate hydrogen from acids; Cu, Ag, Au do not. A reaction is feasible if E°_cell (cathode minus anode) comes out positive.

Standard potentials apply only at unit concentration. The Nernst equation tells you the electrode or cell potential at any concentration, and it is the single most examined formula in the chapter.

For the electrode reaction Mⁿ⁺(aq) + ne⁻ → M(s), NCERT writes E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − (RT/nF) ln ([M]/[Mⁿ⁺]). The concentration of a pure solid is taken as unity, so this becomes E = E° − (RT/nF) ln (1/[Mⁿ⁺]). Here R is the gas constant (8.314 J K⁻¹ mol⁻¹), T the temperature in kelvin, F the Faraday constant (96487 C mol⁻¹) and n the number of electrons in the half reaction.

For the whole cell. For the Daniell cell, subtracting the anode expression from the cathode expression gives E_cell = E°_cell − (RT/2F) ln ([Zn²⁺]/[Cu²⁺]). In general, for aA + bB → cC + dD with n electrons transferred, E_cell = E°_cell − (RT/nF) ln Q, where Q = [C]^c[D]^d/[A]^a[B]^b. Converting to log₁₀ and putting T = 298 K gives 2.303RT/F = 2.303 × 8.314 × 298/96487 = 0.0591 V, so E_cell = E°_cell − (0.0591/n) log Q.

What the equation tells you. Raising the concentration of a reactant (Cu²⁺ in the Daniell cell) or lowering that of a product (Zn²⁺) makes Q smaller and E_cell larger. For a hydrogen electrode in a solution of pH p, [H⁺] = 10^−p and E = 0 − 0.0591 log (1/[H⁺]) = −0.0591 × pH. A concentration cell has identical electrodes in solutions of different concentration; E° = 0 and the emf comes entirely from the log term.

Worked example (NCERT). Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s), E°_cell = 3.17 V, n = 2. E = 3.17 − (0.0591/2) log (0.130/(10⁻⁴)²) = 3.17 − 0.02955 × log (1.3 × 10⁷) = 3.17 − 0.02955 × 7.11 = 3.17 − 0.21 = 2.96 V.

As a galvanic cell works, reactants are used up, products accumulate and E_cell falls. When the reaction reaches equilibrium, the cell can do no more work: E_cell = 0 and Q becomes the equilibrium constant K_c. Substituting these two conditions into the Nernst equation is the derivation the board asks for.

E_cell = E°_cell − (2.303RT/nF) log Q. At equilibrium, 0 = E°_cell − (2.303RT/nF) log K_c, so E°_cell = (2.303RT/nF) log K_c, and at 298 K, E°_cell = (0.0591/n) log K_c, or log K_c = nE°_cell/0.0591.

Because E° sits in the exponent, a modest positive emf corresponds to an enormous equilibrium constant. For the Daniell cell, E° = 1.1 V and n = 2, so log K_c = 2 × 1.1/0.0591 = 37.2, K_c ≈ 2 × 10³⁷: the reaction goes essentially to completion. NCERT's example: for Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°_cell = 0.46 V, log K_c = 2 × 0.46/0.0591 = 15.6, K_c ≈ 3.7 × 10¹⁵.

The examiner wants three things in such a numerical: the correct n, the log form of the relation, and the antilog taken properly (write K_c = antilog(15.6) and then the value). If E° is negative, log K is negative and K is less than 1; the reaction as written does not proceed appreciably.

Electrical work done by a galvanic cell in one second equals the potential difference multiplied by the charge that flows. If the cell delivers its emf reversibly (no current drawn, so E_cell is the maximum potential), the maximum work is the product of the emf and the total charge nF for n moles of electrons. Maximum work available from a process at constant temperature and pressure is the decrease in Gibbs energy, so ΔG = −W_max = −nFE_cell. Under standard conditions, ΔG° = −nFE°_cell.

Three consequences worth stating in an answer. First, a positive E°_cell means a negative ΔG°, which is the thermodynamic condition for spontaneity: this is why 'E°_cell positive means feasible'. Second, ΔG is extensive while E is intensive. If you double the reaction (2Zn + 2Cu²⁺ → ...), n doubles and ΔG doubles, but E stays 1.1 V. Third, combining with ΔG° = −RT ln K_c gives back the relation E°_cell = (RT/nF) ln K_c of the previous section, so the three quantities E°, ΔG° and K_c are all interconvertible and a single 3-mark question often asks for all of them.

Worked example (NCERT intext 2.6). For 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, E°_cell = 0.236 V, n = 2. ΔG° = −2 × 96487 × 0.236 = −45,542 J mol⁻¹ = −45.54 kJ mol⁻¹. Then log K_c = −ΔG°/(2.303RT) = 45542/(2.303 × 8.314 × 298) = 45542/5705.8 = 7.98, so K_c = antilog(7.98) ≈ 9.6 × 10⁷. Using log K_c = nE°/0.0591 gives the same 7.99.

Units: with F in C mol⁻¹ and E in V, ΔG comes in J mol⁻¹; convert to kJ mol⁻¹ at the end and keep the negative sign.

Electrolytic solutions conduct because ions move; metals conduct because electrons move. NCERT lists what electronic conductance depends on (nature and structure of the metal, number of valence electrons, temperature, which decreases it) and what electrolytic conductance depends on: the nature of the electrolyte, the size of the ions and their solvation, the nature and viscosity of the solvent, the concentration, and temperature, which increases it. Electrolytic conductance is much smaller than metallic.

Resistance and conductance. Ohm's law: R = ρ (l/A), where ρ is resistivity (Ω m). Conductance G = 1/R (unit siemens, S = Ω⁻¹) and G = κ (A/l), where κ = 1/ρ is the conductivity (S m⁻¹; S cm⁻¹ is more common in numericals). Conductivity is the conductance of a solution of unit length and unit area of cross-section, that is, the conductance of 1 cm³ of solution held between electrodes 1 cm apart. It is the intrinsic property; conductance depends on the geometry of the cell.

Molar conductivity. Λ_m is the conductance of a volume of solution containing one mole of electrolyte kept between two electrodes 1 cm apart and of area large enough to hold that volume. Λ_m = κ/c in SI units (S m² mol⁻¹ when κ is in S m⁻¹ and c in mol m⁻³). In the mixed units actually used in exams, Λ_m (S cm² mol⁻¹) = κ (S cm⁻¹) × 1000/c (mol L⁻¹), the 1000 converting litres to cm³. Note 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.

Both κ and Λ_m change with concentration, in opposite directions; that is the subject of Section 2.4.2 and the most repeated theory question of the chapter.

Two problems stop you from measuring conductivity the way you measure the resistance of a wire. First, passing direct current through a solution changes its composition by electrolysis. Second, a solution cannot be connected to a bridge like a wire. Both are solved by using an alternating current source and a specially designed conductivity cell.

The cell has two platinum electrodes coated with platinum black, of area A and distance l apart, immersed in the solution. The ratio l/A is called the cell constant, G^*, with unit m⁻¹ or cm⁻¹. Measuring l and A directly is not accurate, so the cell constant is found by filling the cell with a KCl solution of known conductivity (tabulated at various concentrations and temperatures) and measuring its resistance: G^* = κ × R. Once the cell constant is known, the resistance of any solution in the same cell gives its conductivity, κ = G^*/R, and its molar conductivity follows from Λ_m = κ × 1000/c.

The resistance itself is measured with a Wheatstone bridge: two known resistances R₃ and R₄, a variable resistance R₁ and the conductivity cell of unknown resistance R₂. The bridge is balanced by adjusting R₁ until the detector shows zero current, at which point R₂ = R₁R₄/R₃. The source is an oscillator (AC of a few hundred to a few thousand hertz) and the detector is a suitable headphone or electronic device.

Worked example (NCERT). The resistance of a cell filled with 0.1 M KCl is 100 Ω; filled with 0.02 M KCl it is 520 Ω. Conductivity of 0.1 M KCl is 1.29 S m⁻¹. Cell constant = κR = 1.29 × 100 = 129 m⁻¹. For 0.02 M KCl, κ = 129/520 = 0.248 S m⁻¹ = 0.248 × 10⁻² S cm⁻¹. Λ_m = 0.248 × 10⁻² × 1000/0.02 = 124 S cm² mol⁻¹.

Conductivity κ always decreases on dilution, for strong and weak electrolytes alike, because κ is the conductance of a fixed volume (1 cm³) and dilution reduces the number of ions in that volume.

Molar conductivity Λ_m always increases on dilution, because the volume containing one mole of electrolyte increases, and the increase in volume more than compensates for the fall in κ. The way it increases is different for the two classes, and the Λ_m versus √c graph is a favourite 5-mark part.

Strong electrolytes (KCl, NaCl): Λ_m rises slowly and linearly with dilution, following the Debye–Hückel–Onsager equation Λ_m = Λ°_m − A√c, where A depends on the type of electrolyte (all 1:1 electrolytes in the same solvent at the same temperature share one A), the solvent and the temperature. Extrapolating the straight line to c = 0 gives the limiting molar conductivity Λ°_m, the molar conductivity at infinite dilution. The rise is small because the number of ions does not change; only interionic attractions weaken on dilution.

Weak electrolytes (CH₃COOH): at higher concentrations Λ_m is low, and near infinite dilution it rises steeply because the degree of dissociation α increases sharply and many more ions appear. The plot is a curve, not a line, so Λ°_m cannot be found by extrapolation. It is obtained from Kohlrausch's law.

Kohlrausch's law of independent migration of ions: the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion. For A_xB_y, Λ°_m = xλ°₊ + yλ°₋. Example: Λ°_m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.6 + 40.9 = 390.5 S cm² mol⁻¹. Equivalently, from salts: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 425.9 − 126.4 = 390.5.

Applications: (i) Λ°_m of a weak electrolyte; (ii) degree of dissociation, α = Λ_m/Λ°_m; (iii) dissociation constant of a weak acid, K_a = cα²/(1 − α), which becomes K_a = cΛ_m²/[Λ°_m(Λ°_m − Λ_m)]. Worked example (NCERT 2.11): κ of 0.00241 M acetic acid is 7.896 × 10⁻⁵ S cm⁻¹. Λ_m = 7.896 × 10⁻⁵ × 1000/0.00241 = 32.76 S cm² mol⁻¹; α = 32.76/390.5 = 0.0839; K_a = 0.00241 × (0.0839)²/(1 − 0.0839) = 1.85 × 10⁻⁵.

In an electrolytic cell an external source pushes a non-spontaneous reaction. Example: a copper rod dipped in CuSO₄ solution and connected to the negative terminal becomes the cathode; copper deposits on it (Cu²⁺ + 2e⁻ → Cu). Another copper rod at the positive terminal, the anode, dissolves (Cu → Cu²⁺ + 2e⁻). This is the principle of electrorefining of copper: impure copper as anode, pure copper as cathode, and the net result is transfer of copper. Sodium and magnesium are obtained by electrolysing their fused chlorides, and aluminium from Al₂O₃ dissolved in cryolite.

Faraday's first law: the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed. Faraday's second law: for the same quantity of electricity passed through different electrolytes, the amounts of substances liberated are proportional to their chemical equivalent weights (atomic mass divided by the number of electrons needed per atom).

The modern way to use both laws is through the Faraday constant: one mole of electrons carries F = N_Ae = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ = 96487 C mol⁻¹, usually rounded to 96500. Read the electrode reaction, note how many moles of electrons each mole of product needs, and convert: Q = It coulombs; moles of electrons = Q/F; moles of product = (Q/F)/n; mass = (Q/F)(M/n). So 1 mol Al³⁺ → Al needs 3F, 1 mol MnO₄⁻ → Mn²⁺ needs 5F, 1 mol Cr₂O₇²⁻ → 2Cr³⁺ needs 6F.

Worked example (NCERT). CuSO₄ electrolysed for 10 minutes with 1.5 A. Q = 1.5 × 600 = 900 C. Cu²⁺ + 2e⁻ → Cu, so 2 × 96487 C deposit 63.5 g. Mass = 63.5 × 900/(2 × 96487) = 0.296 g.

Cells in series carry the same charge. If AgNO₃, CuSO₄ and ZnSO₄ cells are in series, moles of electrons are equal in all three, so moles of Ag : Cu : Zn deposited = 1 : ½ : ½.

The products depend on the state of the electrolyte (molten or aqueous), on the electrodes (inert like Pt and graphite, or reactive like Cu and Ag), and on the standard electrode potentials of every species that could react. Water itself can be oxidised or reduced, so in aqueous solutions it competes at both electrodes.

Molten NaCl: cathode Na⁺ + e⁻ → Na, anode Cl⁻ → ½Cl₂ + e⁻. Aqueous NaCl: at the cathode two reductions compete, Na⁺ + e⁻ → Na (E° = −2.71 V) and H₂O + e⁻ → ½H₂ + OH⁻ (E° = −0.83 V). The reaction with the higher reduction potential occurs, so hydrogen is evolved and the solution becomes NaOH. At the anode, Cl⁻ → ½Cl₂ + e⁻ (E° = 1.36 V) competes with 2H₂O → O₂ + 4H⁺ + 4e⁻ (E° = 1.23 V). Oxidation of water is thermodynamically preferred, yet chlorine is obtained, because the evolution of oxygen needs an extra potential called overpotential. This is the point examiners test.

Dilute H₂SO₄ with Pt: cathode 2H⁺ + 2e⁻ → H₂; anode 2H₂O → O₂ + 4H⁺ + 4e⁻ (sulphate cannot be oxidised at this potential; concentrated H₂SO₄ gives peroxodisulphate instead). AgNO₃ with Pt: cathode Ag⁺ + e⁻ → Ag; anode water oxidised to O₂. AgNO₃ with silver electrodes: cathode Ag deposits, anode silver dissolves, Ag → Ag⁺ + e⁻, so the concentration of the solution stays constant. CuCl₂ with Pt: cathode Cu, anode Cl₂ (overpotential again). CuSO₄ with Cu electrodes: copper transfers from anode to cathode, the basis of electrorefining.

General rule to write in an answer: at the cathode, the species with the higher reduction potential is reduced; at the anode, the species with the lower reduction potential (easier oxidation) is oxidised, unless overpotential intervenes as it does for O₂.

A battery is one or more galvanic cells connected in series. A primary battery works until the reactants are exhausted and cannot be recharged. A secondary battery can be recharged by passing current in the opposite direction, so it works through many discharge–charge cycles.

Dry cell (Leclanché cell): the anode is the zinc container, the cathode a carbon (graphite) rod surrounded by powdered MnO₂ and carbon, and the space between is a moist paste of NH₄Cl and ZnCl₂. Anode: Zn(s) → Zn²⁺ + 2e⁻. Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃. The ammonia produced forms a complex with Zn²⁺, [Zn(NH₃)₄]²⁺. Potential about 1.5 V, which falls as the cell is used because the electrolyte composition changes.

Mercury cell (watches, hearing aids): anode zinc–mercury amalgam, cathode a paste of HgO and carbon, electrolyte a paste of KOH and ZnO. Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻. Cathode: HgO + H₂O + 2e⁻ → Hg(l) + 2OH⁻. Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l). Its potential of about 1.35 V stays constant during its life because the overall reaction involves no ions whose concentration could change.

Lead storage battery (cars, inverters): anode lead, cathode a grid of lead packed with lead dioxide, electrolyte 38% H₂SO₄ by mass. Discharging, anode: Pb(s) + SO₄²⁻ → PbSO₄(s) + 2e⁻; cathode: PbO₂(s) + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄(s) + 2H₂O. Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. On charging the reaction reverses: PbSO₄ on the anode is reduced back to Pb, PbSO₄ on the cathode is oxidised back to PbO₂, and H₂SO₄ is regenerated, which is why the density of the acid rises again.

Nickel–cadmium cell: longer life than the lead cell but costlier; overall discharge reaction Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l).

A galvanic cell in which the reactants are supplied continuously from outside and the products are removed continuously is a fuel cell. Fuels such as hydrogen, methane or methanol are oxidised at the anode and oxygen is reduced at the cathode. The point of a fuel cell is that it converts chemical energy to electrical energy directly, bypassing the thermal-power route (fuel → heat → steam → turbine → generator), which wastes most of the energy. Fuel cells run at about 70% efficiency compared with roughly 40% for thermal plants, and they are pollution-free, since the only product of the hydrogen–oxygen cell is water.

The H₂–O₂ fuel cell (used in the Apollo space programme, where the water produced was used for drinking): hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide. Catalysts such as finely divided platinum or palladium are incorporated in the electrodes to speed up the reactions. Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq). Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻. Overall: 2H₂(g) + O₂(g) → 2H₂O(l). The cell runs continuously as long as the reactants are supplied.

Points that turn up as sub-questions: why a fuel cell is not called a battery (it never runs down while fuel is fed), why it is more efficient (no intermediate heat step, so it is not bound by the thermodynamic limits of a heat engine), why NaOH rather than acid (the OH⁻ forms the cycle between the electrodes; the electrolyte is not consumed), and what other fuels can be used (methane, methanol).

Corrosion is the slow oxidation of a metal by its surroundings: rusting of iron, tarnishing of silver (Ag₂S), the green coating on copper and bronze. It is essentially an electrochemical process, and the board asks you to show rusting as the setting up of a galvanic cell on the metal surface.

Mechanism. At one spot on the iron surface, oxidation takes place; that spot acts as the anode: 2Fe(s) → 2Fe²⁺ + 4e⁻, E°(Fe²⁺/Fe) = −0.44 V. The electrons released move through the metal to another spot, which becomes the cathode, where oxygen is reduced in the presence of H⁺ ions: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l), E°(H⁺/O₂/H₂O) = 1.23 V. The H⁺ ions come from H₂CO₃ formed by dissolution of atmospheric CO₂ in water, or from other acids. Overall: 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l), E°_cell = 1.23 − (−0.44) = 1.67 V. The Fe²⁺ ions are then oxidised further by atmospheric oxygen to Fe³⁺, which comes out as hydrated ferric oxide, Fe₂O₃·xH₂O, the rust. Rusting therefore needs both water and oxygen, is faster in acidic conditions, and is faster in saline water because dissolved ions raise the conductivity of the electrolyte film on the surface.

Prevention. The simplest method is to keep the surface away from the atmosphere: paint, oil or grease, or a coating of a metal that resists corrosion (tin, chromium, nickel), or chemical treatment to form a protective layer (bisphenol). Galvanisation coats iron with zinc; zinc, being more electropositive, is oxidised in preference to iron even if the coating is scratched. Sacrificial protection attaches a block of Mg or Zn to a buried pipe or a ship's hull; the more active metal corrodes instead of the iron. Alloying (stainless steel with Cr and Ni) also stops rusting.