Electrochemistry

🎯 Key Points

• Galvanic cell: chemical → electrical energy (spontaneous, ΔG < 0); Electrolytic cell: electrical → chemical energy (non-spontaneous, forced)
• Anode = oxidation always; Cathode = reduction always (true for BOTH cell types — only the polarity sign flips between them)
• EMF = E°cathode − E°anode; positive EMF means spontaneous as written
• Nernst equation: E = E° − (0.0592/n)log Q at 25°C — links cell potential to non-standard concentrations
• ΔG° = −nFE°cell; also ΔG° = −RT ln K, linking electrochemistry directly to thermodynamics and equilibrium
• Faraday's First Law: mass deposited ∝ charge passed; 1 Faraday = 96500 C = charge of 1 mole electrons
• Molar conductivity increases with dilution; Kohlrausch's law allows calculating Λ°m of weak electrolytes from strong electrolyte data

Types of Electrochemical Cells

• Galvanic (Voltaic) cell: Chemical energy to electrical energy (e.g., Daniell cell)
• Electrolytic cell: Electrical energy to chemical energy (e.g., electroplating)

Galvanic Cell

• Anode: Negative electrode; oxidation occurs here
• Cathode: Positive electrode; reduction occurs here
• EMF of cell = E°cathode - E°anode
• Daniell cell: Zn|ZnSO₄||CuSO₄|Cu; E° = 1.10 V

A Daniell (galvanic) cell: zinc is oxidised at the anode (Zn → Zn²⁺ + 2e⁻) and copper ions are reduced at the copper cathode (Cu²⁺ + 2e⁻ → Cu). The salt bridge maintains electrical neutrality and completes the circuit, while the voltmeter reads the cell EMF. Image: Gringer, CC BY-SA 3.0, via Wikimedia Commons.

Standard Electrode Potential

Measured against Standard Hydrogen Electrode (SHE = 0 V). Higher value = greater tendency to be reduced.

• Zn²⁺/Zn = -0.76 V
• Cu²⁺/Cu = +0.34 V
• F₂/F⁻ = +2.87 V (strongest oxidising agent)

Nernst Equation

E = E° - (0.0592/n) × log Q (at 25°C). This lets you calculate the actual cell potential under non-standard concentrations, and explains why concentration cells (same electrode, different ion concentrations on each side) can generate a voltage even with E° = 0.

Relation to thermodynamics: ΔG° = −nFE°cell, and since ΔG° = −RT ln K, we get E°cell = (RT/nF) ln K — a spontaneous cell (E° > 0) always corresponds to a reaction with K > 1 (products favoured at equilibrium).

Faraday's Laws

• First Law: Mass deposited is proportional to charge passed (m = ZQ = ZIt, where Z is the electrochemical equivalent)
• Second Law: For the same charge, mass deposited is proportional to equivalent weight
• 1 Faraday = 96500 C = charge of 1 mole of electrons

Conductance and Molar Conductivity

• Conductivity (κ) increases with dilution for both strong and weak electrolytes (more ions per unit volume become more mobile as concentration drops, even though total ion count decreases)
• Molar conductivity (Λm) = κ/c also increases on dilution; for strong electrolytes it increases slightly and linearly (Λm vs √c), reaching a limiting value Λ°m at infinite dilution; for weak electrolytes it increases sharply near infinite dilution since more of the weak electrolyte dissociates
• Kohlrausch's Law of independent migration of ions: Λ°m of an electrolyte = sum of the limiting molar conductivities of its individual ions. This lets you calculate Λ°m for a WEAK electrolyte (which can't be measured directly by extrapolation, since it keeps dissociating further even at very low concentration) using data from strong electrolytes that share its ions.

Batteries and Corrosion

• Primary batteries (non-rechargeable): dry cell (Zn-C), mercury cell
• Secondary batteries (rechargeable): lead-acid battery (car batteries), nickel-cadmium cell, lithium-ion battery
• Fuel cells: H₂-O₂ fuel cell directly converts chemical energy of a fuel into electricity continuously (as long as reactants are supplied), with water as the only by-product — used in spacecraft and increasingly in vehicles
• Corrosion (rusting) is an electrochemical process: iron acts as an anode (oxidised to Fe²⁺), atmospheric O₂ + moisture acts as the cathode region (reduced to OH⁻), and the Fe²⁺/OH⁻ combine and are further oxidised to hydrated Fe₂O₃ (rust). Prevention methods include galvanisation (Zn coating, sacrificial anode) and painting.

Quick Tips

• OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
• Cathode: always reduction; Anode: always oxidation — true for both galvanic and electrolytic cells
• Higher E° = better oxidising agent (greater tendency to be reduced)

Cell Representation (Cell Notation)

• Anode is written on the LEFT, cathode on the RIGHT: Anode | Anode solution || Cathode solution | Cathode
• A single vertical bar (|) marks a phase boundary (electrode/solution); the double bar (||) represents the salt bridge
• Example (Daniell cell): Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
• Concentrations/pressures are shown in brackets, e.g. Zn²⁺(1 M); for gas electrodes an inert electrode like Pt is included, e.g. Pt | H₂(1 bar) | H⁺(1 M)

Electrochemical Series and Its Applications

Arranging electrodes in order of increasing standard reduction potential (E°) gives the electrochemical (activity) series. Applications:

• Predicting displacement: A metal higher in the series (more negative E°) displaces a metal below it from its salt solution (e.g. Zn displaces Cu from CuSO₄)
• Reaction with acids: Metals with negative E° (above hydrogen, e.g. Zn, Fe) liberate H₂ from dilute acids; metals below hydrogen (Cu, Ag, Au) do not
• Oxidising/reducing strength: Higher E° = stronger oxidising agent (F₂ strongest); lower/more negative E° = stronger reducing agent (Li strongest)
• Predicting cell feasibility: A cell is spontaneous only if the calculated E°cell = E°cathode − E°anode is positive

Products of Electrolysis

• Molten electrolyte: Only the ions of the compound are present. Molten NaCl gives Na at the cathode and Cl₂ at the anode
• Aqueous electrolyte: Water can also be oxidised/reduced, so the product depends on electrode potentials and overpotential. Aqueous NaCl (brine) gives H₂ at the cathode (not Na) and Cl₂ at the anode (overpotential favours Cl₂ over O₂)
• Aqueous CuSO₄ with Pt electrodes: Cu deposits at the cathode; O₂ evolves at the anode (water oxidised)
• Aqueous CuSO₄ with Cu electrodes: Cu deposits at cathode while the Cu anode dissolves — the basis of electro-refining of copper

Electrolysis in a Hoffman apparatus: an external power supply drives a non-spontaneous redox reaction, forcing oxidation at the anode (+) and reduction at the cathode (−). Electrolysing acidified water gives hydrogen and oxygen in a 2:1 volume ratio. Image: Ivan Akira, CC BY-SA 3.0, via Wikimedia Commons.

Degree of Dissociation from Conductivity

• For a weak electrolyte, the degree of dissociation α = Λm / Λ°m (ratio of molar conductivity at that concentration to its value at infinite dilution)
• The dissociation constant then follows Ostwald's dilution law: Ka = cα² / (1 − α) = c Λm² / [Λ°m (Λ°m − Λm)]
• This gives an experimental route to Ka of weak acids/bases purely from conductance measurements

Electrode Reactions in Batteries and Fuel Cells

• Lead storage battery (secondary): Anode Pb, cathode PbO₂, electrolyte H₂SO₄. On discharge both electrodes form PbSO₄; charging reverses this
• H₂–O₂ fuel cell (alkaline): Anode H₂ + 2OH⁻ → 2H₂O + 2e⁻; Cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻; overall 2H₂ + O₂ → 2H₂O with high efficiency (~70%) and water as the only product
• Mercury cell (primary): Gives a steady voltage (~1.35 V) throughout its life because the overall reaction involves no ions whose concentration changes

🚀 JEE Advanced Edge

Concentration cells: Two half-cells of the SAME metal/ion but different concentrations generate an EMF purely from the concentration difference: E = (0.0592/n) log(C₂/C₁). The more concentrated side acts as the cathode (reduction, ions deposit) as the cell tries to equalise concentrations.

Electrolysis selectivity: When multiple species can be reduced/oxidised at an electrode, the one with the more favourable (less negative for reduction, less positive for oxidation) electrode potential is discharged preferentially — this is why electrolysis of brine gives H₂ at the cathode (not Na, despite Na⁺ being present) and Cl₂ at the anode under normal conditions, due to overpotential effects.

Worked problem: Calculate the EMF of the cell Zn|Zn²⁺(0.1M)||Cu²⁺(1M)|Cu given E°cell = 1.10 V. Approach: n=2 for this reaction. Q = [Zn²⁺]/[Cu²⁺] = 0.1/1 = 0.1. E = 1.10 − (0.0592/2)log(0.1) = 1.10 − (0.0296)(−1) = 1.10 + 0.0296 ≈ 1.13 V.