What this chapter is about
Electrochemistry studies the relationship between chemical energy and electrical energy. It explains how chemical reactions can produce electric current (as in a battery) and how electric current can drive non-spontaneous chemical reactions (as in electroplating or the extraction of metals). This chapter builds on your understanding of redox reactions from earlier classes and connects thermodynamic ideas like Gibbs energy to the measurable quantity called electromotive force (emf).
A Class 12 student meets this chapter because electrochemistry has wide applications in daily life and industry: batteries in phones and vehicles, corrosion of iron, refining of metals, and electroplating of jewellery. The chapter also introduces the Nernst equation, which relates the emf of a cell to the concentrations of the species involved, linking chemistry to quantitative predictions.
After studying this chapter, you should be able to describe galvanic and electrolytic cells, write cell notation, calculate electrode potentials and cell emf using standard tables and the Nernst equation, apply Faraday's laws to calculate masses of substances deposited or liberated, and explain conductance in electrolytic solutions.
Key ideas
- A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction; an electrolytic cell uses external electrical energy to drive a non-spontaneous reaction.
- The standard electrode potential (E°) of a half-cell is measured against the Standard Hydrogen Electrode (SHE), which is assigned a potential of exactly 0 V at 298 K, 1 bar H₂ and 1 M H⁺.
- The standard emf of a cell is E°(cell) = E°(cathode) − E°(anode); a positive value means the reaction is spontaneous under standard conditions.
- The Nernst equation relates cell emf to ion concentrations: E = E° − (RT / nF) ln Q, or at 298 K, E = E° − (0.0591 / n) log₁₀ Q, where Q is the reaction quotient and n is the number of electrons transferred.
- Gibbs energy change and emf are linked by ΔG = −nFE; at equilibrium, E = 0 and ΔG = 0.
- Faraday's first law states that the mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge) passed.
- Faraday's second law states that for the same quantity of electricity, the masses of different substances deposited are proportional to their equivalent masses (molar mass / number of electrons involved).
- Conductivity (κ) of an electrolyte solution and molar conductivity (Λₘ = κ / c, where c is concentration in mol/m³) describe how well a solution conducts; Λₘ increases on dilution for both strong and weak electrolytes.
Formulas and facts to remember
- Standard cell emf: E°(cell) = E°(cathode) − E°(anode).
- Nernst equation at 298 K: E = E° − (0.0591 V / n) log₁₀ Q.
- Relation of Gibbs energy to emf: ΔG° = −nFE°, where F = 96485 C mol⁻¹ (Faraday constant).
- At equilibrium: E = 0, so E° = (0.0591 / n) log₁₀ K (K is equilibrium constant).
- Faraday's first law: m = ZIt, where Z = M / nF (electrochemical equivalent), I is current (A), t is time (s).
- Molar conductivity: Λₘ = κ / c; unit is S cm² mol⁻¹ when κ is in S cm⁻¹ and c in mol cm⁻³.
- Kohlrausch's law: The limiting molar conductivity of an electrolyte equals the sum of limiting molar conductivities of its cation and anion: Λ°ₘ = λ°₊ + λ°₋.
- One faraday of electricity (96485 C) deposits one mole of a monovalent ion or half a mole of a divalent ion.
Worked examples
Example 1: Calculating the emf of a Daniell cell under non-standard conditions
A Daniell cell has the half-reactions Zn → Zn²⁺ + 2e⁻ (anode) and Cu²⁺ + 2e⁻ → Cu (cathode). Given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, find the cell emf when [Zn²⁺] = 0.10 M and [Cu²⁺] = 1.0 M at 298 K.
Step 1: Find standard emf. E°(cell) = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V.
Step 2: Write the overall reaction and reaction quotient. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 1.0 = 0.10.
Step 3: Apply the Nernst equation (n = 2 electrons). E = 1.10 − (0.0591 / 2) log₁₀(0.10) E = 1.10 − 0.02955 × (−1) = 1.10 + 0.03 ≈ 1.13 V.
The cell emf is approximately 1.13 V.
Example 2: Mass deposited using Faraday's law
A current of 3.0 A is passed through molten sodium chloride for 1.5 hours. Calculate the mass of sodium deposited at the cathode. (Molar mass of Na = 23 g mol⁻¹, F = 96485 C mol⁻¹)
Step 1: Calculate total charge. Q = I × t = 3.0 A × (1.5 × 3600 s) = 3.0 × 5400 = 16200 C.
Step 2: Find moles of electrons. Moles of electrons = 16200 / 96485 ≈ 0.168 mol.
Step 3: At the cathode, Na⁺ + e⁻ → Na, so 1 mole of electrons deposits 1 mole of Na. Mass of Na = 0.168 × 23 ≈ 3.86 g.
About 3.9 g of sodium is deposited.
Example 3: Using Kohlrausch's law
The limiting molar conductivities are λ°(K⁺) = 73.5 S cm² mol⁻¹ and λ°(Cl⁻) = 76.3 S cm² mol⁻¹. Calculate Λ°ₘ for KCl.
Step 1: Apply Kohlrausch's law. Λ°ₘ(KCl) = λ°(K⁺) + λ°(Cl⁻) = 73.5 + 76.3 = 149.8 S cm² mol⁻¹.
Common mistakes
- Adding electrode potentials instead of subtracting → always use E°(cell) = E°(cathode) − E°(anode).
- Forgetting to convert time to seconds when using Faraday's law → convert hours or minutes to seconds before calculating charge.
- Using concentration of solid or pure liquid in the reaction quotient Q → only aqueous or gaseous species appear; activities of pure solids and liquids are taken as 1.
- Confusing conductivity (κ) with molar conductivity (Λₘ) → κ is for the solution, Λₘ is per mole of electrolyte.
- Ignoring the sign of E° when identifying anode and cathode → the electrode with lower (more negative) E° acts as the anode in a galvanic cell.
Quick revision
- Galvanic cell: spontaneous reaction produces current; electrolytic cell: current drives non-spontaneous reaction.
- E°(cell) = E°(cathode) − E°(anode); positive E° means spontaneous.
- Nernst equation at 298 K: E = E° − (0.0591/n) log Q.
- ΔG° = −nFE° links thermodynamics to electrochemistry.
- Mass deposited = (M × I × t) / (n × F) (Faraday's first law rearranged).
- Kohlrausch's law: Λ°ₘ = λ°₊ + λ°₋; useful for weak electrolytes.