Electrochemistry

36 solved questions from Class 12 Chemistry, each with the full working. Tap any question to see the step-by-step solution.

• 1 What is the primary chemical reaction or bonding mechanism highlighted in 'Electrochemistry' (Class 12)? medium MCQ Reaction Mechanism & Bonding
• 2 What factor affects the rate or equilibrium constant of reactions in 'Electrochemistry'? medium MCQ Chemical Kinetics & Factors
• 3 Which functional group or IUPAC nomenclature rule applies to compounds in 'Electrochemistry'? medium MCQ IUPAC Nomenclature & Structure
• 4 What is the major product obtained in the characteristic reaction of 'Electrochemistry'? medium MCQ Reaction Products & Selectivity
• 5 Calculate the cell potential of the galvanic cell: Zn(s) | Zn²⁺ (0.01 M) || Cu²⁺ (0.1 M) | Cu(s) at 25 °C. Use the standard electrode potentials: Zn²⁺/Zn = –0.76 V and Cu²⁺/Cu = +0.34 V. easy 2 marks Cell potential calculation using Nernst equation
• 6 Which of the following is the correct order of standard electrode potentials (in increasing order of reduction potential) for the couples Fe³⁺/Fe²⁺, Cu²⁺/Cu and Zn²⁺/Zn? medium MCQ 3 marks Comparison of standard electrode potentials
• 7 In an electrolytic cell, explain why metal ions are reduced at the cathode even though the cathode is the negative electrode. hard 3 marks Electrolysis: cathodic reduction
• 8 Using the Lippmann equation, calculate the surface tension of a mercury electrode when the potential changes from 0 V to –0.5 V. The capacitance per unit area is 0.20 mF cm⁻² and the surface tension at 0 V is 485 mN m⁻¹. hard 5 marks Surface tension variation with potential (Lippmann equation)
• 9 Consider the galvanic cell: Fe(s) | Fe^2+(1 M) || Cu^2+(1 M) | Cu(s). Which electrode acts as the anode? easy MCQ 1 mark Identifying anode in a galvanic cell
• 10 A copper plate of area 10 cm^2 is to be electroplated with nickel to a thickness of 20 μm. Density of nickel =8.90 g cm^-3, molar mass =58.69 g mol^-1. (a) Calculate the total charge required. (b) Find the time needed if a constant current of 2.0 A is passed. Assume 100% current efficiency. Reaction: Ni^2++2e^-→ Ni. Use F=96485 C mol^-1. hard 4 marks Application of Faraday’s laws of electrolysis
• 11 Calculate the emf of the cell Zn(s) | Zn^2+(0.001 M) || Ag^+(0.10 M) | Ag(s) at 25°C. Standard potentials: E°_(Zn^2+/Zn) = -0.76 V, E°_(Ag^+/Ag) = +0.80 V. medium 2 marks Using Nernst equation for cell emf
• 12 For the half‑reaction Fe^3++e^-→ Fe^2+, E° = +0.77 V at 25°C (n=1). Calculate the equilibrium constant K for this half‑reaction. (Take RT/F = 0.0257 V.) medium MCQ 2 marks Link between electrode potential and equilibrium constant
• 13 A copper strip of area 5\,cm^2 is to be electro‑plated with silver. A current of 0.85\,A is passed for 30\,minutes. The current efficiency is 95\%. Given: M_Ag=107.87\,g\,mol^-1, density of Ag=10.5\,g\,cm^-3, and Faraday constant F=96485\,C\,mol^-1. (a) Calculate the mass of silver actually deposited. (b) Determine the thickness of the silver layer formed on the copper strip. hard 5 marks Faraday’s law of electrolysis
• 14 State why the anode of an electrolytic cell is positive even though oxidation (loss of electrons) occurs there. medium 2 marks Electrode polarity in electrolytic cells
• 15 Which of the following statements best explains the corrosion of iron in moist air? hard MCQ 3 marks Mechanism of iron corrosion
• 16 A galvanic cell is constructed as: Cu(s) | Cu^2+(0.5\,M) || Ag^+(1.0\,M) | Ag(s). Standard reduction potentials are E°_(Cu^2+/Cu)=+0.34\,V and E°_(Ag^+/Ag)=+0.80\,V. Calculate the cell emf at 25°C. easy 2 marks Nernst equation for cell potential
• 17 Calculate the emf of the galvanic cell: Cu(s) | Cu^2+ (0.01,M) || Zn^2+ (0.1,M) | Zn(s) at 25°C. Given standard potentials E^circ_(Cu^2+/Cu)=+0.34,V and E^circ_(Zn^2+/Zn)=-0.76,V. easy 2 marks Electrochemical cell potential calculation
• 18 For the half‑reaction Fe³⁺ + e⁻ → Fe²⁺, the standard potential E° is +0.77 V at 25 °C (n = 1). Calculate the equilibrium constant K for this reaction. Use RT/F = 0.0257 V. hard 4 marks Relation between electrode potential and equilibrium constant
• 19 Calculate the emf of the galvanic cell: Fe(s) | Fe³⁺(0.02 M) || Cu²⁺(1 M) | Cu(s) at 25 °C. Standard potentials: E°_(Fe³⁺/Fe)=+0.04 V and E°_(Cu²⁺/Cu)=+0.34 V. easy 3 marks Cell emf using Nernst equation
• 20 Explain why the anode of an electrolytic cell is positive even though oxidation occurs there. Include in your answer the role of the external power source and the direction of electron flow. hard 5 marks Electrolytic cell operation
• 21 A nickel plate of area 8 cm² is to be electrolytically plated with palladium to a thickness of 5 µm. The density of palladium is 12.0 g cm⁻³, its molar mass is 106.42 g mol⁻¹, and the current efficiency is 70 %. If a constant current of 1.5 A is passed, how long (in seconds) will it take to deposit the desired layer? Reaction: Pd²⁺ + 2e⁻ → Pd. Use F = 96485 C mol⁻¹. medium 4 marks Faraday's laws of electrolysis
• 22 A watch case of area 15 cm^2 is to be electro‑plated with gold to a thickness of 2 µm. The density of gold is 19.3 g cm^-3, its molar mass is 197 g mol^-1. A constant current of 0.50 A is passed and the current efficiency is 85 %. For the reduction reaction Au^3+ + 3e^- → Au, calculate the time required (in seconds) to obtain the desired coating. Use F = 96485 C mol^-1. medium 3 marks Faraday’s law with current efficiency
• 23 Calculate the emf of the cell at 25 °C: Zn(s) | Zn^2+ (0.01 M) || Cu^2+ (1.0 M) | Cu(s). Standard potentials: E°_(Zn^2+/Zn)= -0.76 V, E°_(Cu^2+/Cu)= +0.34 V. Use the Nernst equation. easy MCQ 2 marks Nernst equation for cell potential
• 24 Explain why the anode of an electrolytic cell is labelled positive even though oxidation (loss of electrons) occurs there. easy 2 marks Polarity of electrodes in electrolytic cells
• 25 The half‑reaction MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O has E° = +1.51 V at 25 °C. The Fe^3+/Fe^2+ couple has E° = +0.77 V. Write the balanced overall redox reaction when MnO_4^- acts as oxidising agent and Fe^2+ as reducing agent. Then calculate the equilibrium constant K for this overall reaction. Use the relation log_10K = nE°_cell/0.0592. hard 5 marks Standard cell potential and equilibrium constant
• 26 Calculate the emf of the galvanic cell Fe(s) | Fe^3+(0.02 M) || Cu^2+(1 M) | Cu(s) at 25 °C. Standard potentials: E°_(Fe^3+/Fe)=+0.04 V and E°_(Cu^2+/Cu)=+0.34 V. easy 2 marks Nernst equation for non‑standard concentrations
• 27 In the galvanic cell Cu(s) | Cu^2+(0.5 M) || Fe^2+(1 M) | Fe(s), which electrode acts as the anode at 25 °C? Standard potentials: E°_(Cu^2+/Cu)=+0.34 V, E°_(Fe^2+/Fe)=−0.44 V. medium 3 marks Identifying anode/cathode from electrode potentials
• 28 A disc of copper with area 12 cm² is to be electro‑plated with gold to a thickness of 1.5 µm. The density of gold is 19.3 g cm⁻³, its molar mass is 197 g mol⁻¹. A constant current of 0.80 A is passed for 90 % current efficiency. Calculate the time required in seconds. medium 3 marks Faraday’s law and current efficiency in electro‑plating
• 29 The half‑reaction MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O has E°=+1.51 V at 25 °C. The Fe²⁺/Fe³⁺ couple has E°=+0.77 V. Write the balanced overall redox reaction when MnO_4^- is the oxidising agent and Fe²⁺ the reducing agent. Then calculate the equilibrium constant K for this reaction using log_10K=nE°_cell/0.0592. hard 4 marks Balancing redox equations and calculating equilibrium constants
• 30 In acidic solution, potassium permanganate oxidises ferrous ions. Given E°_(MnO_4^-/Mn^2+) = +1.51 V and E°_(Fe^3+/Fe^2+) = +0.77 V, which of the following statements is correct? medium MCQ 3 marks Overall redox and equilibrium constant
• 31 A stainless‑steel spoon of area 20 cm² is to be electro‑plated with silver to a thickness of 3 µm. Density of Ag = 10.5 g cm⁻³, molar mass = 107.87 g mol⁻¹. A constant current of 0.60 A is passed and the current efficiency is 92 %. For the reduction reaction Ag⁺ + e⁻ → Ag, calculate the time required (in seconds) to obtain the desired coating. Use F = 96485 C mol⁻¹. hard 5 marks Electroplating charge‑time calculation
• 32 Explain why the anode of an electrolytic cell is labelled positive even though oxidation (loss of electrons) occurs there. easy 2 marks Electrode sign in electrolytic cell
• 33 Calculate the emf at 25 °C of the cell Zn(s) | Zn^2+(0.005 M) || Cu^2+(0.5 M) | Cu(s). Standard potentials: E°_(Zn^2+/Zn)= -0.76 V, E°_(Cu^2+/Cu)= +0.34 V. Use the Nernst equation. medium 3 marks Nernst equation emf calculation
• 34 A stainless‑steel spoon of area 25 cm² is to be electro‑plated with silver to a thickness of 2 µm. Density of Ag = 10.5 g cm⁻³ and molar mass = 107.87 g mol⁻¹. A constant current of 0.55 A is passed and the current efficiency is 90 %. For the reduction reaction Ag⁺ + e⁻ → Ag, calculate the time required (in seconds) to obtain the desired coating. medium 3 marks Faraday's laws of electroplating
• 35 When ceMnO4^- oxidises ceFe^2+ in acidic solution, the balanced overall reaction is ceMnO4^- + 5Fe^2+ -> Mn^2+ + 5Fe^3+. Given the standard potentials E^circ_(ceMnO4^- / Mn^2+)=+1.51 V and E^circ_(ceFe^3+ / Fe^2+)=+0.77 V, calculate the equilibrium constant K for this reaction. hard 4 marks Equilibrium constant from cell potential
• 36 A disc of copper of area 12 cm² is to be electro‑plated with gold to a thickness of 1.5 µm. Density of gold = 19.3 g cm⁻³, molar mass = 197 g mol⁻¹. A constant current of 0.80 A is passed with a current efficiency of 90 %. For the reduction reaction ceAu^3+ + 3e- -> Au, calculate the time required (in seconds) to obtain the desired coating. hard 4 marks Faraday’s law for multi‑electron deposition