← Chemistry — Std 12
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Electrochemistry

Ch. 5Std 12

Easy Overview

Batteries power your phone, rust eats your car, electroplating puts gold on your jewelry — all electrochemistry. This chapter is about converting chemical energy into electrical energy (in batteries) and electrical energy into chemical energy (in electrolysis). It is all about electrons going on a road trip through wires and ions swimming through solutions. An electrochemical cell consists of two electrodes dipped in electrolyte solutions, connected by a wire and a salt bridge. Oxidation happens at the anode (electrons leave), reduction happens at the cathode (electrons arrive). The salt bridge maintains electrical neutrality. The Daniell cell is the classic example: zinc anode in ZnSO₄, copper cathode in CuSO₄. Zn gets oxidized (Zn → Zn²⁺ + 2e⁻), Cu²⁺ gets reduced (Cu²⁺ + 2e⁻ → Cu), and electrons flow from Zn to Cu through the wire. Every half-reaction has a standard electrode potential (E°) measured against the standard hydrogen electrode (SHE, E° = 0.00 V). The standard cell potential E°_cell = E°_cathode - E°_anode. For Daniell cell: E° = 0.34 - (-0.76) = 1.10 V. The Nernst equation tells you how cell potential changes with concentration: E = E° - (RT/nF) ln Q. At 298 K: E = E° - (0.059/n) log Q. At equilibrium, E = 0 and Q = K, giving log K = nE°/0.059. Electrolytic cells force non-spontaneous reactions by applying external voltage. Faraday's laws govern the amount of product: m = (M × I × t)/(n × F). The chapter also covers conductivity, Kohlrausch's Law, batteries (primary and secondary), fuel cells, corrosion, and the electrochemical series. Understanding electrochemistry means understanding how your phone battery works, why ships rust, and how aluminum is extracted from ore.

Galvanic vs electrolytic cells

Galvanic (voltaic) cells convert chemical energy to electrical energy — spontaneous reactions produce electricity. Electrolytic cells use electrical energy to drive non-spontaneous reactions. Both have two electrodes (anode and cathode) in electrolytes. In galvanic cells: anode is negative (-), oxidation occurs; cathode is positive (+), reduction occurs. Electrons flow spontaneously from anode to cathode through the external circuit. The salt bridge completes the circuit by allowing ion migration. In electrolytic cells: anode is positive (+) (connected to the positive terminal of the battery), cathode is negative (-). The external power source forces electrons to flow in the opposite direction of the spontaneous direction.

Daniell cell — the classic galvanic cell

Daniell cell: Zn electrode in ZnSO₄ solution | Cu electrode in CuSO₄ solution. The two compartments are connected by a wire and a salt bridge (KCl in agar jelly). Half-reactions: Zn(s) → Zn²⁺(aq) + 2e⁻ (oxidation at anode, E° = -0.76 V). Cu²⁺(aq) + 2e⁻ → Cu(s) (reduction at cathode, E° = +0.34 V). Overall: Zn + Cu²⁺ → Zn²⁺ + Cu, E°_cell = +1.10 V. The salt bridge allows K⁺ and Cl⁻ ions to migrate, maintaining electrical neutrality in both compartments. Without the salt bridge, the cell stops working because charge builds up in each compartment. Cell diagram notation: Zn(s)|ZnSO₄(aq)||CuSO₄(aq)|Cu(s). Single vertical line = phase boundary; double vertical line = salt bridge. Anode is written on the left, cathode on the right.

Standard electrode potential (E°) and SHE

Standard electrode potential (E°) is the potential of a half-cell measured against the Standard Hydrogen Electrode (SHE) under standard conditions (1 M concentration, 1 bar pressure, 298 K). SHE: 2H⁺(aq) + 2e⁻ → H₂(g), E° = 0.00 V. A platinum electrode is used (inert metal). E° can be positive (easier to reduce than H⁺) or negative (harder to reduce). More positive E° = stronger oxidizing agent (easily reduced). More negative E° = stronger reducing agent (easily oxidized). E°_cell = E°_cathode - E°_anode. If E°_cell > 0, the reaction is spontaneous (galvanic cell). If E°_cell < 0, it is non-spontaneous (electrolytic cell needed). The electrochemical series arranges elements by E° values — a metal will displace any metal below it from solution (Zn displaces Cu²⁺ but not the reverse). This is also used to predict whether a metal dissolves in acid: metals with E°(M²⁺/M) < 0 dissolve in dilute acids producing H₂.

Nernst equation — potential changes with concentration

For a reaction aA + bB → cC + dD: E = E° - (RT/nF) ln Q. At 298 K: E = E° - (0.059/n) log Q. For the Daniell cell: E = 1.10 - (0.059/2) log([Zn²⁺]/[Cu²⁺]). When [Zn²⁺] = [Cu²⁺] = 1 M, E = E° = 1.10 V. As the cell discharges, [Zn²⁺] increases and [Cu²⁺] decreases, so E drops. When E = 0, the cell is 'dead' — equilibrium is reached. The Nernst equation allows calculation of cell potential under non-standard conditions and is used in potentiometric titrations and pH measurements. Worked example: calculate E for Daniell cell with [Zn²⁺] = 0.1 M, [Cu²⁺] = 0.01 M at 298 K. E = 1.10 - (0.059/2) log(0.1/0.01) = 1.10 - 0.0295 log 10 = 1.10 - 0.0295 = 1.0705 V.

Equilibrium constant from E°

At equilibrium, E_cell = 0 and Q = K. From Nernst: 0 = E° - (0.059/n) log K, so log K = nE°/0.059 at 298 K. This is powerful — a single E°_cell measurement gives the equilibrium constant. Example: for Zn + Cu²⁺ ⇌ Zn²⁺ + Cu, E° = 1.10 V, n = 2. log K = (2 × 1.10)/0.059 = 37.29, K = 10³⁷·²⁹ = 1.95 × 10³⁷ — the reaction overwhelmingly favors products. The relationship ΔG° = -nFE°_cell connects thermodynamics to electrochemistry: if ΔG° < 0, E°_cell > 0 (spontaneous). Also ΔG° = -RT ln K, so all three are linked: E°_cell, K, and ΔG°. This three-way relationship is a favorite exam topic.

Electrolysis and Faraday's laws

Electrolysis uses electrical energy to drive non-spontaneous reactions. Faraday's First Law: mass of substance liberated (m) ∝ charge passed (Q = I × t). m = ZIt, where Z = electrochemical equivalent. Faraday's Second Law: for the same charge, masses of different substances liberated are proportional to their equivalent masses. Combined: m = (M × I × t)/(n × F), where M = molar mass, I = current (A), t = time (s), n = number of electrons involved, F = 96485 C/mol (Faraday constant = charge of 1 mole of electrons). Example: How much copper is deposited when 0.5 A is passed for 30 minutes through CuSO₄? For Cu²⁺ + 2e⁻ → Cu: n = 2, M = 63.5 g/mol, I = 0.5 A, t = 1800 s. m = (63.5 × 0.5 × 1800)/(2 × 96485) = 57150/192970 = 0.296 g. Common mistake: forgetting to convert time to seconds.

Applications of electrolysis

Electrolysis is used for: (1) Electroplating — coating a cheap metal with gold, silver, or chromium. The object to be plated is the cathode; the coating metal is the anode in a solution of its salt. (2) Electrorefining — purifying copper. Impure copper anode, pure copper cathode in CuSO₄ solution. Impurities either dissolve or fall as anode mud (which contains valuable metals like Au, Ag, Pt). (3) Extraction of reactive metals: aluminum from bauxite (Hall-Heroult process), sodium from molten NaCl (Down's process). (4) Electrolysis of water: 2H₂O → 2H₂ + O₂. (5) Production of chlorine and NaOH from brine (chlor-alkali industry). The amount of product is always predictable using Faraday's laws — a common board exam numerical problem.

Conductivity of electrolytic solutions

Electrolytic solutions conduct electricity due to mobile ions. Resistance R = ρ(l/A), where ρ = resistivity, l = distance between electrodes, A = area. Conductivity κ = 1/ρ = 1/R × (l/A). The cell constant = l/A (determined by measuring the resistance of a standard KCl solution of known κ). Molar conductivity Λ_m = κ/c, where c is concentration. Λ_m increases with dilution because: (1) more ions are produced (for weak electrolytes — Ostwald's dilution law), and (2) ion-ion interactions decrease, increasing ion mobility (for strong electrolytes). At infinite dilution, Λ_m approaches Λ_m° (limiting molar conductivity), where ions move independently without interionic attractions.

Kohlrausch's Law of Independent Migration

At infinite dilution, molar conductivity Λ_m° is the sum of individual ion contributions: Λ_m° = ν₊λ₊° + ν₋λ₋°. Here ν = stoichiometric coefficients, λ° = limiting molar conductivities of individual ions. Applications: (1) Calculate Λ_m° for weak electrolytes. Example: Λ_m°(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.8 + 40.9 = 390.7 S cm² mol⁻¹. This cannot be measured directly by extrapolation because weak electrolytes do not completely dissociate even at low concentrations. (2) Calculate degree of dissociation α = Λ_m/Λ_m° for weak electrolytes. (3) Calculate K_a for weak acids: K_a = Cα²/(1-α). (4) Calculate solubility of sparingly soluble salts: Λ_m° = κ/S (S = solubility in mol/L).

Batteries — primary and secondary

Primary batteries are non-rechargeable. Leclanché dry cell: Zn anode (-), graphite cathode (+) in NH₄Cl/MnO₂ paste. Voltage ≈ 1.5 V. The NH₄Cl is acidic (can corrode the Zn). Alkaline battery: similar but uses KOH instead of NH₄Cl — longer shelf life, higher energy density. Mercury battery: Zn anode, HgO cathode in KOH — used in small devices (hearing aids, watches). Secondary batteries are rechargeable. Lead-acid battery: Pb anode, PbO₂ cathode in H₂SO₄. Each cell gives 2 V; 6 cells in series = 12 V car battery. Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Recharge: reverse. Lithium-ion battery: LiCoO₂ cathode, carbon anode. High energy density, lightweight — used in phones, laptops, EVs. Li⁺ ions move between electrodes during charge/discharge.

Fuel cells — continuous clean power

Fuel cells convert chemical energy directly to electricity as long as fuel is supplied. Hydrogen-oxygen fuel cell: Anode (oxidation): 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. Cathode (reduction): O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall: 2H₂ + O₂ → 2H₂O. Voltage ≈ 1.23 V. Advantages: high efficiency (60-80% vs 30-40% for combustion engines), water as the only byproduct (clean), quiet operation. Used in spacecraft (Apollo, Space Shuttle), electric vehicles (Toyota Mirai), and stationary power generation. Challenges: H₂ production (currently from natural gas, needs to be green), storage, and fuel cell cost. Other types: phosphoric acid fuel cells, molten carbonate fuel cells, solid oxide fuel cells. Fuel cells are NOT batteries — they produce electricity continuously from supplied fuel, while batteries store energy chemically.

Corrosion — the electrochemical rusting of iron

Corrosion is the electrochemical deterioration of metals. Rusting of iron: at anodic sites on the surface, Fe → Fe²⁺ + 2e⁻. At cathodic sites (where O₂ and H₂O are available), O₂ + 2H₂O + 4e⁻ → 4OH⁻. Fe²⁺ + 2OH⁻ → Fe(OH)₂, which further oxidizes to Fe₂O₃·xH₂O (rust). Rust is porous and flaky — it does not protect the underlying metal; instead, it exposes fresh metal to further corrosion. Factors: presence of O₂ and H₂O, electrolytes (salt accelerates — that is why cars rust faster in coastal areas and where roads are salted), and the presence of a less active metal (galvanic corrosion). Prevention: (1) Surface coatings — paint, oil, grease, plastic coatings. (2) Galvanizing — coating with Zn (Zn protects sacrificially). (3) Tin plating — coating with Sn (if scratched, Fe corrodes faster because Sn is less active). (4) Alloying — stainless steel (Fe + Cr + Ni) forms a protective Cr₂O₃ layer. (5) Sacrificial anodes — attaching Mg or Zn blocks to ship hulls (Mg/Zn corrodes instead of Fe). (6) Cathodic protection — applying a small electric current to make the metal the cathode.

Electrochemical series and its applications

The electrochemical series arranges elements in order of increasing E° values. Key applications: (1) Predicting spontaneity of redox reactions — a metal with more negative E° can displace a metal with more positive E° from its salt solution. Zn (E° = -0.76 V) displaces Cu²⁺ (E° = +0.34 V), but Cu cannot displace Zn²⁺. (2) Choosing electrodes for cells — two half-cells with widely differing E° give higher cell voltage. (3) Products of electrolysis — at the cathode, the species with the more positive E° gets reduced first. At the anode, the species with the more negative E° gets oxidized first. (4) Predicting metal-acid reactions — metals with E°(M²⁺/M) < 0 (Zn, Fe, Mg) dissolve in dilute acids producing H₂. Metals with E° > 0 (Cu, Ag, Au) do not. The series: Li > K > Ca > Na > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag > Hg > Au (from most reactive to least).

Key Points

  • •Galvanic: chemical → electrical (spontaneous); Electrolytic: electrical → chemical (non-spontaneous)
  • •Anode: oxidation (loses electrons); Cathode: reduction (gains electrons)
  • •Daniell cell: Zn + Cu²⁺ → Zn²⁺ + Cu, E°_cell = 1.10 V
  • •SHE: Pt|H₂(1 bar)|H⁺(1 M), E° = 0.00 V (reference electrode)
  • •E°_cell = E°_cathode - E°_anode; positive E°_cell = spontaneous
  • •Nernst equation: E = E° - (0.059/n) log Q at 298 K
  • •ΔG° = -nFE°_cell; log K = nE°/0.059 at 298 K
  • •Faraday: m = (M × I × t)/(n × F); F = 96485 C/mol
  • •Molar conductivity Λ_m = κ/c; increases with dilution
  • •Kohlrausch's Law: Λ_m° = ν₊λ₊° + ν₋λ₋° — independent ion migration
  • •α = Λ_m/Λ_m° (degree of dissociation for weak electrolytes)
  • •Primary batteries: dry cell (1.5 V, non-rechargeable), alkaline (longer life)
  • •Secondary: lead-acid (12 V car battery), Li-ion (phones, laptops — rechargeable)
  • •Fuel cell: H₂ + O₂ → H₂O + electricity; clean, efficient, continuous
  • •Corrosion: Fe → Fe²⁺ (anode); O₂ + 2H₂O + 4e⁻ → 4OH⁻ (cathode); rust = Fe₂O₃·xH₂O
  • •Prevention: painting, galvanizing (Zn), sacrificial anodes (Mg), cathodic protection
  • •Electrochemical series: Li > K > Ca > Na > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag > Au

Practice Questions

  • Calculate EMF of a Daniell cell using Nernst equation if [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M. (E°_Zn²⁺/Zn = -0.76 V, E°_Cu²⁺/Cu = +0.34 V)
  • Explain the working of a hydrogen-oxygen fuel cell. Write electrode reactions and give two advantages.
  • How much copper is deposited when 0.5 A is passed for 30 minutes through CuSO₄? (At. mass Cu = 63.5, F = 96500 C/mol)
  • Differentiate electrolytic and galvanic cells with an example of each.
  • State and explain Faraday's laws. Calculate mass of silver deposited when 9650 C passes through AgNO₃. (At. mass Ag = 108)
  • What is Kohlrausch's Law? How can it determine the dissociation constant of a weak acid?
  • Explain the mechanism of rusting of iron. Describe methods to prevent corrosion.
  • Derive the Nernst equation. Calculate K for Zn + Cu²⁺ ⇌ Zn²⁺ + Cu at 298 K (E°_cell = 1.10 V).