Electrochemistry

1 exercises11 questions solved

Exercise 2.1Galvanic Cells, EMF, Conductance and Electrolysis

How would you determine the standard electrode potential of the system Mg²⁺/Mg?

Solution

To determine the standard electrode potential of Mg²⁺/Mg: 1. Set up a galvanic cell with: • Mg²⁺/Mg half-cell (unknown electrode) • Standard Hydrogen Electrode (SHE, E° = 0.00 V) as reference 2. Connect through a salt bridge and voltmeter. 3. Measure the cell EMF under standard conditions: (1 M Mg²⁺ solution, 298 K, 1 bar pressure) 4. If Mg acts as anode (negative terminal): E°cell = E°cathode – E°anode E°cell = E°(SHE) – E°(Mg²⁺/Mg) 5. The measured cell EMF = +2.37 V So E°(Mg²⁺/Mg) = –2.37 V The negative value means Mg is a stronger reducing agent than H₂.

Can you store copper sulphate solutions in a zinc pot?

Solution

No, copper sulphate solution cannot be stored in a zinc pot. Reason: The standard reduction potentials are: Cu²⁺/Cu: E° = +0.34 V Zn²⁺/Zn: E° = –0.76 V Zn is above Cu in the electrochemical series, so Zn is a stronger reducing agent than Cu. The spontaneous reaction would be: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) The zinc pot would dissolve in the copper sulphate solution, depositing copper and ultimately destroying the container. This is not a suitable combination for storage.

Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions (Fe²⁺) under standard conditions.

Solution

Ferrous ion is oxidised to ferric ion: Fe²⁺ → Fe³⁺ + e⁻ (E° = –0.77 V for reverse) For Fe²⁺ to be oxidised, the oxidising agent must have E° > +0.77 V Substances with E° > +0.77 V (can oxidise Fe²⁺): 1. Cl₂ (E° = +1.36 V): Cl₂ + 2e⁻ → 2Cl⁻ 2. Cr₂O₇²⁻/Cr³⁺ (E° = +1.33 V in acidic medium) 3. MnO₄⁻/Mn²⁺ (E° = +1.51 V in acidic medium) 4. H₂O₂ (E° = +1.77 V): H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O Any three from: Cl₂, K₂Cr₂O₇, KMnO₄, H₂O₂, F₂, etc.

Calculate the standard cell potentials of galvanic cells in which the following reactions take place: (i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s) (ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s). Calculate the ΔrG° and equilibrium constant of the reactions.

Solution

Standard reduction potentials: Cr³⁺/Cr: E° = –0.74 V, Cd²⁺/Cd: E° = –0.40 V Fe³⁺/Fe²⁺: E° = +0.77 V, Ag⁺/Ag: E° = +0.80 V (i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s) Cr is oxidised (anode), Cd²⁺ is reduced (cathode) E°cell = E°cathode – E°anode = –0.40 – (–0.74) = +0.34 V n = 6 (each Cr loses 3e⁻, 2 Cr atoms → 6e⁻) ΔrG° = –nFE° = –6 × 96485 × 0.34 = –1.97 × 10⁵ J = –197 kJ log K = nE°/0.0592 = 6 × 0.34/0.0592 = 34.46 K = 10³⁴·⁴⁶ ≈ 2.9 × 10³⁴ (ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s) Fe²⁺ is oxidised (anode), Ag⁺ is reduced (cathode) E°cell = +0.80 – 0.77 = +0.03 V n = 1 ΔrG° = –nFE° = –1 × 96485 × 0.03 = –2894.6 J ≈ –2.9 kJ log K = 1 × 0.03/0.0592 = 0.507 K = 10⁰·⁵⁰⁷ ≈ 3.21

Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg²⁺(0.001M)||Cu²⁺(0.0001M)|Cu(s) (ii) Fe(s)|Fe²⁺(0.001M)||H⁺(1M)|H₂(g)(1bar)|Pt(s)

Solution

(i) Mg(s)|Mg²⁺(0.001M)||Cu²⁺(0.0001M)|Cu(s) Overall: Mg + Cu²⁺ → Mg²⁺ + Cu, n = 2 E°cell = E°(Cu²⁺/Cu) – E°(Mg²⁺/Mg) = 0.34 – (–2.37) = 2.71 V Nernst equation: E = E° – (0.0592/n) log Q Q = [Mg²⁺]/[Cu²⁺] = 0.001/0.0001 = 10 E = 2.71 – (0.0592/2) × log(10) = 2.71 – 0.0296 × 1 = 2.68 V (ii) Fe(s)|Fe²⁺(0.001M)||H⁺(1M)|H₂(1bar)|Pt(s) Overall: Fe + 2H⁺ → Fe²⁺ + H₂, n = 2 E°cell = 0.00 – (–0.44) = 0.44 V Q = [Fe²⁺] × P(H₂) / [H⁺]² = (0.001 × 1) / (1)² = 0.001 E = 0.44 – (0.0592/2) × log(0.001) = 0.44 – 0.0296 × (–3) = 0.44 + 0.0888 = 0.529 V

In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq). Determine ΔrG° and E° for the reaction.

Solution

Half reactions: Oxidation: Zn → Zn²⁺ + 2e⁻ (E° = +0.76 V, as anode) Reduction: Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻ (E° = +0.344 V) E°cell = E°cathode – E°anode = 0.344 – (–0.76) = 1.104 V n = 2 ΔrG° = –nFE° = –2 × 96485 × 1.104 = –213,103 J = –213.1 kJ E° ≈ 1.10 V, ΔrG° ≈ –213 kJ mol⁻¹

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Solution

Conductivity (κ or σ): The ability of a solution to conduct electricity. κ = 1/ρ (inverse of resistivity) Unit: S m⁻¹ or S cm⁻¹ Molar conductivity (Λm): Conductivity of a solution per unit molar concentration. Λm = κ/C (where C is in mol/m³ or mol/L × 1000) Unit: S m² mol⁻¹ or S cm² mol⁻¹ Variation with concentration: Conductivity (κ): • For both strong and weak electrolytes, κ decreases with dilution • Fewer ions per unit volume as dilution increases Molar conductivity (Λm): • For strong electrolytes (e.g., NaCl, KCl): – Λm increases gradually with dilution – Λm = Λ°m – A√C (Debye-Hückel-Onsager equation) – At infinite dilution, inter-ionic attractions become negligible → Λ°m • For weak electrolytes (e.g., CH₃COOH): – Λm increases steeply with dilution – At high dilution, degree of dissociation → 1, Λm → Λ°m – Λ°m cannot be determined by extrapolation; use Kohlrausch's law

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm⁻¹. Calculate its molar conductivity.

Solution

κ = 0.0248 S cm⁻¹, C = 0.20 M = 0.20 mol/L Molar conductivity: Λm = (κ × 1000) / C = (0.0248 × 1000) / 0.20 = 24.8 / 0.20 = 124 S cm² mol⁻¹ Molar conductivity of 0.20 M KCl = 124 S cm² mol⁻¹

The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10⁻³ S cm⁻¹?

Solution

R = 1500 Ω, κ = 0.146 × 10⁻³ S cm⁻¹ Conductance G = 1/R = 1/1500 S Cell constant = κ/G = κ × R = 0.146 × 10⁻³ × 1500 = 0.219 cm⁻¹ Cell constant = 0.219 cm⁻¹

Q10

A solution of CaCl₂ is electrolysed between inert electrodes using a current of 5 A for 30 min. Calculate the mass of Ca deposited at cathode and volume of Cl₂ evolved at anode (at STP).

Solution

I = 5 A, t = 30 min = 1800 s Charge passed Q = I × t = 5 × 1800 = 9000 C Faraday constant F = 96485 C/mol At cathode: Ca²⁺ + 2e⁻ → Ca (n = 2, M = 40 g/mol) Moles of e⁻ = Q/F = 9000/96485 = 0.09329 mol Moles of Ca = 0.09329/2 = 0.04665 mol Mass of Ca = 0.04665 × 40 = 1.866 g ≈ 1.87 g At anode: 2Cl⁻ → Cl₂ + 2e⁻ (n = 2, for each mol Cl₂) Moles of Cl₂ = 0.09329/2 = 0.04665 mol Volume at STP = 0.04665 × 22400 mL = 1044.9 mL ≈ 1.04 L

Q11

A steady current of 2 amperes was passed through two electrolytic cells in series, one containing AgNO₃ solution and the other containing CuSO₄ solution, for 10 minutes. Calculate the ratio of mass of silver to mass of copper deposited.

Solution

I = 2 A, t = 10 min = 600 s Charge Q = 2 × 600 = 1200 C At cathode of AgNO₃: Ag⁺ + e⁻ → Ag n = 1 (valency of Ag), M(Ag) = 108 g/mol Moles of Ag = Q/(n × F) = 1200/(1 × 96485) = 0.01244 mol Mass of Ag = 0.01244 × 108 = 1.343 g At cathode of CuSO₄: Cu²⁺ + 2e⁻ → Cu n = 2 (valency of Cu), M(Cu) = 63.5 g/mol Moles of Cu = Q/(n × F) = 1200/(2 × 96485) = 0.00622 mol Mass of Cu = 0.00622 × 63.5 = 0.395 g Ratio: Ag/Cu = 1.343/0.395 ≈ 3.40 ≈ 108:31.75 (ratio of equivalent masses)

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