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Electrochemistry

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Chapter Notes

Overview
This chapter explores the interconversion of chemical and electrical energy. It covers galvanic cells that produce electricity from spontaneous reactions and electrolytic cells that use electricity to drive non-spontaneous reactions. The central idea is that the Gibbs energy of a spontaneous redox reaction can be converted into electrical work, quantified by the cell potential, and that this potential depends on concentration via the Nernst equation. A student must understand the relationship between cell potential, Gibbs energy, and the equilibrium constant, as well as the principles of ionic conductance and electrolysis.
Key Concepts (18)
Electrochemical cell
A device that converts chemical energy into electrical energy (galvanic cell) or uses electrical energy to carry out non-spontaneous chemical reactions (electrolytic cell).
e.g. A Daniell cell converts the chemical energy of the Zn-Cu redox reaction into electrical energy to light a small bulb.
Galvanic cell (voltaic cell)
An electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy.
e.g. The Daniell cell, where Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) produces a potential of 1.1 V.
Electrolytic cell
A device that uses electrical energy to carry out non-spontaneous chemical reactions.
e.g. Electrolysis of water to produce hydrogen and oxygen gas.
Electrode potential
The potential difference developed between the electrode and the electrolyte at equilibrium.
e.g. For a copper electrode in CuSO₄ solution, the electrode potential is the tendency of Cu²⁺ to deposit as Cu or Cu to dissolve as Cu²⁺.
Standard electrode potential (E⁰)
The electrode potential when the concentrations of all species involved in the half-cell are unity (1 M) and gases are at 1 bar pressure.
e.g. E⁰ for Cu²⁺|Cu is +0.34 V, meaning Cu²⁺ is reduced more easily than H⁺.
Standard hydrogen electrode (SHE)
A half-cell represented by Pt(s)|H₂(g)|H⁺(aq), assigned a zero potential at all temperatures, used as a reference for measuring other electrode potentials.
e.g. It consists of a platinum electrode coated with platinum black, dipped in 1 M H⁺ solution, with H₂ gas at 1 bar pressure bubbled through it.
Cell potential (emf)
The potential difference between the two electrodes of a galvanic cell when no current is drawn through the cell.
e.g. For the Daniell cell, E_cell = E⁰_Cu²⁺|Cu – E⁰_Zn²⁺|Zn = 0.34 V – (–0.76 V) = 1.10 V.
Nernst equation
An equation that relates the electrode potential or cell potential to the concentration of the species involved.
e.g. For the cell reaction aA + bB → cC + dD, E_cell = E⁰_cell – (RT/nF) ln Q.
Conductivity (κ)
The inverse of resistivity; the conductance of a material when it is 1 m long and has an area of cross-section of 1 m². SI unit: S m⁻¹.
e.g. 0.1 M KCl solution has a conductivity of 1.29 S m⁻¹ at 298 K.
Molar conductivity (Λₘ)
The conductivity of a volume of solution containing one mole of electrolyte kept between two electrodes with unit distance and sufficient area. Λₘ = κ / c.
e.g. For 0.1 M KCl, Λₘ = 129.0 S cm² mol⁻¹.
Limiting molar conductivity (Λ⁰ₘ)
The molar conductivity of an electrolyte when the concentration approaches zero (infinite dilution).
e.g. Λ⁰ₘ for KCl is 150.0 S cm² mol⁻¹ at 298 K.
Kohlrausch law of independent migration of ions
The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the cation and anion of the electrolyte.
e.g. Λ⁰ₘ(NaCl) = λ⁰_Na⁺ + λ⁰_Cl⁻ = 50.1 + 76.3 = 126.4 S cm² mol⁻¹.
Degree of dissociation (α)
The fraction of the total number of moles of an electrolyte that dissociates into ions in solution.
e.g. For 0.001028 M acetic acid, α = Λₘ / Λ⁰ₘ = 48.15 / 390.5 = 0.1233.
Faraday constant (F)
The charge carried by one mole of electrons, equal to 96487 C mol⁻¹.
e.g. Used in the Nernst equation and in calculating Gibbs energy: Δ_rG = –nFE_cell.
Salt bridge
A device that connects the two half-cells internally, allowing the flow of ions to maintain electrical neutrality.
e.g. In a Daniell cell, a salt bridge containing KCl or KNO₃ connects ZnSO₄ and CuSO₄ solutions.
Cell constant (G*)
The ratio of the distance between the electrodes (l) to the area of cross-section (A) of the solution column. G* = l/A.
e.g. For a conductivity cell with l = 1 cm and A = 1 cm², G* = 1 cm⁻¹.
Resistivity (ρ)
The resistance of a material when it is 1 m long and has an area of cross-section of 1 m². SI unit: Ω m.
e.g. For a 0.05 M NaOH solution, ρ = 87.135 Ω cm.
Conductance (G)
The inverse of resistance. G = 1/R. SI unit: siemens (S).
e.g. If R = 100 Ω, then G = 0.01 S.
Formulas & Equations (17)
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Reads as: Zinc solid plus copper two plus aqueous yields zinc two plus aqueous plus copper solid
Variables: Zn(s): solid zinc; Cu²⁺(aq): aqueous copper ions; Zn²⁺(aq): aqueous zinc ions; Cu(s): solid copper
Use when: Writing the overall cell reaction for the Daniell cell.
⚠ Forgetting to balance the charges — the number of electrons lost must equal the number gained.
E_cell = E_right – E_left
Reads as: E cell equals E right minus E left
Variables: E_cell: cell potential (V); E_right: reduction potential of the right-hand electrode (V); E_left: reduction potential of the left-hand electrode (V)
Use when: Calculating the emf of a galvanic cell from standard reduction potentials.
⚠ Using oxidation potentials instead of reduction potentials, or subtracting in the wrong order.
E(Mⁿ⁺/M) = E⁰(Mⁿ⁺/M) – (RT/nF) ln (1/[Mⁿ⁺])
Reads as: E of M n plus over M equals E standard of M n plus over M minus RT over nF times natural log of one over concentration of M n plus
Variables: E: electrode potential (V); E⁰: standard electrode potential (V); R: gas constant (8.314 J K⁻¹ mol⁻¹); T: temperature (K); n: number of electrons transferred; F: Faraday constant (96487 C mol⁻¹); [Mⁿ⁺]: concentration of metal ion (M)
Use when: Calculating the potential of a single electrode under non-standard conditions.
⚠ Forgetting to take the concentration of solid metal as unity, or using the wrong sign for the log term.
E_cell = E⁰_cell – (RT/nF) ln Q
Reads as: E cell equals E standard cell minus RT over nF times natural log of Q
Variables: E_cell: cell potential (V); E⁰_cell: standard cell potential (V); R: gas constant (8.314 J K⁻¹ mol⁻¹); T: temperature (K); n: number of electrons transferred; F: Faraday constant (96487 C mol⁻¹); Q: reaction quotient
Use when: Calculating the cell potential for any concentration of reactants and products.
⚠ Writing Q incorrectly — products over reactants, each raised to its stoichiometric coefficient.
E_cell = E⁰_cell – (0.059/n) log Q (at 298 K)
Reads as: E cell equals E standard cell minus 0.059 over n times log Q at 298 Kelvin
Variables: E_cell: cell potential (V); E⁰_cell: standard cell potential (V); n: number of electrons transferred; Q: reaction quotient
Use when: Quick calculation of cell potential at 298 K using log base 10.
⚠ Using natural log instead of log base 10, or forgetting to convert temperature to Kelvin.
E⁰_cell = (2.303RT/nF) log K_c
Reads as: E standard cell equals 2.303 RT over nF times log K c
Variables: E⁰_cell: standard cell potential (V); R: gas constant (8.314 J K⁻¹ mol⁻¹); T: temperature (K); n: number of electrons transferred; F: Faraday constant (96487 C mol⁻¹); K_c: equilibrium constant
Use when: Calculating the equilibrium constant from the standard cell potential.
⚠ Forgetting that K_c is the equilibrium constant for the cell reaction, not the reverse.
Δ_rG = –nFE_cell
Reads as: Delta r G equals minus n F E cell
Variables: Δ_rG: Gibbs energy change for the reaction (J mol⁻¹); n: number of electrons transferred; F: Faraday constant (96487 C mol⁻¹); E_cell: cell potential (V)
Use when: Relating the Gibbs energy of a cell reaction to its cell potential.
⚠ Forgetting the negative sign — a positive E_cell gives a negative Δ_rG (spontaneous).
Δ_rG⁰ = –nFE⁰_cell
Reads as: Delta r G standard equals minus n F E standard cell
Variables: Δ_rG⁰: standard Gibbs energy change (J mol⁻¹); n: number of electrons transferred; F: Faraday constant (96487 C mol⁻¹); E⁰_cell: standard cell potential (V)
Use when: Calculating standard Gibbs energy from standard cell potential.
⚠ Using E_cell instead of E⁰_cell for standard conditions.
R = ρ (l/A)
Reads as: R equals rho times l over A
Variables: R: resistance (Ω); ρ: resistivity (Ω m); l: length (m); A: area of cross-section (m²)
Use when: Calculating resistance from the dimensions of a conductor.
⚠ Using inconsistent units — ensure l and A are in SI units.
G = 1/R = κ (A/l)
Reads as: G equals one over R equals kappa times A over l
Variables: G: conductance (S); R: resistance (Ω); κ: conductivity (S m⁻¹); A: area of cross-section (m²); l: length (m)
Use when: Relating conductance to conductivity and cell dimensions.
⚠ Confusing conductance (G) with conductivity (κ).
Λₘ = κ / c
Reads as: Lambda m equals kappa over c
Variables: Λₘ: molar conductivity (S m² mol⁻¹); κ: conductivity (S m⁻¹); c: concentration (mol m⁻³)
Use when: Calculating molar conductivity from measured conductivity and concentration.
⚠ Using concentration in mol L⁻¹ without converting to mol m⁻³, or mixing units of κ (S cm⁻¹ vs S m⁻¹).
Λₘ (S cm² mol⁻¹) = (κ (S cm⁻¹) × 1000 (cm³ L⁻¹)) / molarity (mol L⁻¹)
Reads as: Lambda m in S cm squared per mole equals kappa in S per cm times 1000 cm cubed per liter divided by molarity in moles per liter
Variables: Λₘ: molar conductivity (S cm² mol⁻¹); κ: conductivity (S cm⁻¹); molarity: concentration (mol L⁻¹)
Use when: Calculating molar conductivity in common units used in NCERT problems.
⚠ Forgetting the factor of 1000 to convert L to cm³.
Λₘ = Λ⁰ₘ – A√c
Reads as: Lambda m equals lambda zero m minus A times square root of c
Variables: Λₘ: molar conductivity at concentration c; Λ⁰ₘ: limiting molar conductivity; A: constant depending on the type of electrolyte; c: concentration
Use when: Describing the variation of molar conductivity with concentration for strong electrolytes.
⚠ Applying this equation to weak electrolytes — it is only valid for strong electrolytes.
Λ⁰ₘ = n⁺ λ⁰₊ + n⁻ λ⁰₋
Reads as: Lambda zero m equals n plus times lambda zero plus plus n minus times lambda zero minus
Variables: Λ⁰ₘ: limiting molar conductivity; n⁺: number of cations per formula unit; λ⁰₊: limiting molar conductivity of cation; n⁻: number of anions per formula unit; λ⁰₋: limiting molar conductivity of anion
Use when: Calculating the limiting molar conductivity of any electrolyte using Kohlrausch law.
⚠ Forgetting to multiply by the number of ions (e.g., for CaCl₂, use 2 × λ⁰_Cl⁻).
α = Λₘ / Λ⁰ₘ
Reads as: Alpha equals lambda m over lambda zero m
Variables: α: degree of dissociation; Λₘ: molar conductivity at concentration c; Λ⁰ₘ: limiting molar conductivity
Use when: Calculating the degree of dissociation of a weak electrolyte.
⚠ Using this for strong electrolytes where α is nearly 1 at all concentrations.
K_a = (c α²) / (1 – α)
Reads as: K a equals c alpha squared over one minus alpha
Variables: K_a: dissociation constant; c: concentration (mol L⁻¹); α: degree of dissociation
Use when: Calculating the dissociation constant of a weak electrolyte from its molar conductivity.
⚠ Forgetting that this formula applies only to weak electrolytes and assumes α is small.
Cell constant G* = l/A = κ × R
Reads as: Cell constant G star equals l over A equals kappa times R
Variables: G*: cell constant (m⁻¹); l: distance between electrodes (m); A: area of cross-section (m²); κ: conductivity (S m⁻¹); R: resistance (Ω)
Use when: Determining the cell constant from a solution of known conductivity.
⚠ Using the wrong units — ensure κ and R are in consistent units.
How It Works
Working of a Daniell cell
  1. Step 1: The zinc electrode (anode) undergoes oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻.
  2. Step 2: The electrons flow through the external circuit from the zinc electrode to the copper electrode.
  3. Step 3: At the copper electrode (cathode), reduction occurs: Cu²⁺(aq) + 2e⁻ → Cu(s).
  4. Step 4: The salt bridge allows the flow of ions (e.g., K⁺ and Cl⁻) to maintain electrical neutrality in both half-cells.
  5. Step 5: The overall cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), producing a potential of 1.1 V under standard conditions.
💡 The anode is negative (oxidation) and the cathode is positive (reduction). Electrons flow from anode to cathode in the external circuit.
Measurement of electrode potential using SHE
  1. Step 1: Construct a cell with the standard hydrogen electrode (SHE) as one half-cell and the electrode of interest as the other half-cell.
  2. Step 2: Connect the two half-cells with a salt bridge and a voltmeter.
  3. Step 3: Measure the emf of the cell. If the SHE is the anode (left), the measured emf is the reduction potential of the other half-cell.
  4. Step 4: For standard conditions (1 M concentration, 1 bar pressure), the measured emf equals the standard electrode potential (E⁰) of the half-cell.
💡 The SHE is assigned a potential of exactly 0 V. The sign of the measured emf gives the sign of the electrode potential relative to SHE.
Applying the Nernst equation to calculate cell potential
  1. Step 1: Write the balanced overall cell reaction.
  2. Step 2: Determine the number of electrons transferred (n).
  3. Step 3: Write the reaction quotient Q (products over reactants, each raised to its stoichiometric coefficient).
  4. Step 4: Substitute into the Nernst equation: E_cell = E⁰_cell – (RT/nF) ln Q.
  5. Step 5: At 298 K, use the simplified form: E_cell = E⁰_cell – (0.059/n) log Q.
💡 For a spontaneous reaction, E_cell is positive. The Nernst equation shows that E_cell decreases as the reaction proceeds toward equilibrium.
Measurement of conductivity of an ionic solution
  1. Step 1: Fill a conductivity cell with the solution and connect it to a Wheatstone bridge.
  2. Step 2: Use an alternating current (AC) source to avoid changes in solution composition.
  3. Step 3: Balance the bridge by adjusting a variable resistance until no current flows through the detector.
  4. Step 4: Calculate the unknown resistance using the bridge balance condition: R₂ = (R₁ × R₄) / R₃.
  5. Step 5: Determine the cell constant (G*) using a standard KCl solution of known conductivity.
  6. Step 6: Calculate the conductivity of the unknown solution: κ = G* / R.
  7. Step 7: Calculate molar conductivity: Λₘ = κ / c (with appropriate unit conversions).
💡 Always use AC to prevent electrolysis. The cell constant must be determined before measuring unknown solutions.
Determination of dissociation constant of a weak electrolyte using conductivity
  1. Step 1: Measure the conductivity (κ) of the weak electrolyte solution at a known concentration (c).
  2. Step 2: Calculate its molar conductivity: Λₘ = κ / c (with unit conversion).
  3. Step 3: Obtain the limiting molar conductivity (Λ⁰ₘ) using Kohlrausch law (sum of ionic contributions).
  4. Step 4: Calculate the degree of dissociation: α = Λₘ / Λ⁰ₘ.
  5. Step 5: Calculate the dissociation constant: K_a = (c α²) / (1 – α).
💡 This method works only for weak electrolytes. For strong electrolytes, α is nearly 1 and this approach is not valid.
Exam Traps — Don't Make These Mistakes
⚠ Confusing anode and cathode in a galvanic cell
✗ In a galvanic cell, the anode is positive and the cathode is negative.
→ In a galvanic cell, the anode is where oxidation occurs and it has a negative potential relative to the solution; the cathode is where reduction occurs and it has a positive potential.
✓ In a galvanic cell, the anode is negative (oxidation) and the cathode is positive (reduction). Electrons flow from the anode to the cathode.
⚠ Incorrect sign in Nernst equation
✗ E_cell = E⁰_cell + (RT/nF) ln Q
→ The Nernst equation has a minus sign: E_cell = E⁰_cell – (RT/nF) ln Q. Adding the term gives an incorrect potential that increases with Q, which is not physically correct.
✓ E_cell = E⁰_cell – (RT/nF) ln Q. As the reaction proceeds, Q increases, so E_cell decreases.
⚠ Using concentration instead of activity in Nernst equation
✗ Using molar concentration directly without considering that for solids and pure liquids, activity is 1.
→ In the Nernst equation, the concentration of solids (like metals) and pure liquids is taken as unity. Students often include them in the reaction quotient.
✓ For the half-cell reaction Mⁿ⁺(aq) + ne⁻ → M(s), the Nernst equation is E = E⁰ – (RT/nF) ln (1/[Mⁿ⁺]). The solid M has activity = 1.
⚠ Mixing units of conductivity and concentration
✗ Λₘ = κ / c where κ is in S cm⁻¹ and c is in mol L⁻¹, without the factor of 1000.
→ When κ is in S cm⁻¹ and c is in mol L⁻¹, the volume conversion factor of 1000 cm³ L⁻¹ must be included: Λₘ = (κ × 1000) / c.
✓ Λₘ (S cm² mol⁻¹) = (κ (S cm⁻¹) × 1000 (cm³ L⁻¹)) / molarity (mol L⁻¹).
⚠ Applying Kohlrausch law to weak electrolytes at finite concentration
✗ Λₘ for a weak electrolyte at any concentration is the sum of the ionic conductivities at that concentration.
→ Kohlrausch law applies only to limiting molar conductivity (at infinite dilution). At finite concentrations, ions interact and the law does not hold.
✓ Kohlrausch law states that the limiting molar conductivity (Λ⁰ₘ) of an electrolyte is the sum of the limiting molar conductivities of its ions. It is used to find Λ⁰ₘ for weak electrolytes.
⚠ Forgetting the negative sign in Δ_rG = –nFE_cell
✗ Δ_rG = nFE_cell
→ The correct relation is Δ_rG = –nFE_cell. The negative sign is crucial because a spontaneous reaction (positive E_cell) must have a negative Δ_rG.
✓ Δ_rG = –nFE_cell. For a spontaneous cell reaction, E_cell is positive, so Δ_rG is negative.
Compare & Contrast
Type of chemical reaction
Galvanic cell
Electrolytic cell
A galvanic cell converts chemical energy from a spontaneous redox reaction into electrical energy, while an electrolytic cell uses electrical energy to drive a non-spontaneous chemical reaction.
Definition and units
Conductivity (κ)
Molar conductivity (Λₘ)
Conductivity (κ) is the conductance of a unit volume of solution (units: S m⁻¹), while molar conductivity (Λₘ) is the conductance of a volume containing one mole of electrolyte (units: S m² mol⁻¹). Λₘ = κ / c.
Variation of molar conductivity with concentration
Strong electrolyte
Weak electrolyte
For strong electrolytes, Λₘ decreases slowly with increasing concentration and follows Λₘ = Λ⁰ₘ – A√c. For weak electrolytes, Λₘ increases steeply at low concentrations due to increased dissociation, and Λ⁰ₘ cannot be obtained by extrapolation.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1 short answer question (2-3 marks) on Nernst equation or Kohlrausch law, and 1 long answer question (5 marks) on numerical problems involving cell potential, Gibbs energy, or conductivity calculations. Total: approximately 8-10 marks.
Most asked: Nernst equation and its application to calculate cell potential · Relation between E⁰_cell, Δ_rG⁰, and equilibrium constant · Kohlrausch law and its applications (calculating Λ⁰ₘ for weak electrolytes, degree of dissociation) · Conductivity and molar conductivity variation with concentration · Construction and working of Daniell cell · Standard electrode potential and SHE
Don't skip: Numerical problems combining Nernst equation, Gibbs energy, and equilibrium constant; Numerical problems on molar conductivity, degree of dissociation, and dissociation constant using Kohlrausch law; Derivation of Nernst equation and its application to a given cell
One-Line Revision (read aloud before the exam)
  1. Electrochemistry is the study of the interconversion of chemical and electrical energy.
  2. A galvanic cell converts spontaneous chemical energy into electrical energy; an electrolytic cell uses electrical energy for non-spontaneous reactions.
  3. In a Daniell cell, the anode is Zn (oxidation) and the cathode is Cu (reduction), with a standard emf of 1.1 V.
  4. Standard hydrogen electrode (SHE) is assigned a potential of 0 V and is used as a reference for measuring electrode potentials.
  5. The Nernst equation for a cell is E_cell = E⁰_cell – (RT/nF) ln Q; at 298 K, it becomes E_cell = E⁰_cell – (0.059/n) log Q.
  6. The standard Gibbs energy change is related to cell potential by Δ_rG⁰ = –nFE⁰_cell.
  7. The equilibrium constant K_c is related to standard cell potential by E⁰_cell = (2.303RT/nF) log K_c.
  8. Conductivity (κ) decreases with dilution, but molar conductivity (Λₘ) increases with dilution.
  9. For strong electrolytes, Λₘ = Λ⁰ₘ – A√c; for weak electrolytes, Λₘ increases sharply at low concentrations.
  10. Kohlrausch law: Λ⁰ₘ of an electrolyte is the sum of the limiting molar conductivities of its ions.
  11. The degree of dissociation of a weak electrolyte is α = Λₘ / Λ⁰ₘ.
  12. The dissociation constant of a weak electrolyte is K_a = (c α²) / (1 – α).
  13. The cell constant G* = l/A is determined using a standard KCl solution of known conductivity.
  14. In a conductivity measurement, AC is used to avoid changes in solution composition due to electrolysis.
  15. A salt bridge maintains electrical neutrality in a galvanic cell by allowing ion flow between half-cells.

MCQs — 10 Questions with Answers

Q1. In a Daniell cell, the standard cell potential E° = 1.1 V at 25°C. What is ΔG° for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)? (F = 96500 C/mol)

  • A. −212.3 kJ/mol ✓
  • B. −106.15 kJ/mol
  • C. +212.3 kJ/mol
  • D. +106.15 kJ/mol

Answer: A — ΔG° = −nFE° = −2 × 96500 × 1.1 = −212,300 J/mol = −212.3 kJ/mol; negative value indicates spontaneity.

Q2. A 0.1 M NaCl solution has conductivity κ = 1.2 × 10⁻² S/cm. What is its molar conductivity?

  • A. 120 S·cm²/mol ✓
  • B. 12 S·cm²/mol
  • C. 1.2 S·cm²/mol
  • D. 0.12 S·cm²/mol

Answer: A — Molar conductivity Λm = (κ/c) × 1000 = (1.2 × 10⁻²/0.1) × 1000 = 120 S·cm²/mol (factor 1000 converts concentration from mol/cm³ to mol/L).

Q3. According to Kohlrausch's law, the molar conductivity at infinite dilution Λ°m for an electrolyte depends on:

  • A. only the cation present
  • B. only the anion present
  • C. the sum of limiting molar conductivities of cations and anions ✓
  • D. the concentration of the solution

Answer: C — Kohlrausch's law states Λ°m = Σ(λ° cation) + Σ(λ° anion); it is independent of concentration and depends on ionic mobilities.

Q4. In the electrolysis of molten NaCl, 0.5 mol of electrons is passed through the cell. How many moles of Cl₂ gas are produced at the anode?

  • A. 0.25 mol ✓
  • B. 0.5 mol
  • C. 1.0 mol
  • D. 2.0 mol

Answer: A — At the anode: 2Cl⁻ → Cl₂ + 2e⁻; 0.5 mol electrons produces 0.5/2 = 0.25 mol Cl₂.

Q5. Which of the following statements about the Nernst equation is NOT correct? (I) Q represents the reaction quotient with all species at their instantaneous concentrations. (II) As Q increases, Ecell decreases. (III) At equilibrium, Ecell = E° because Q = K.

  • A. Only (I) is incorrect
  • B. Only (II) is incorrect
  • C. Only (III) is incorrect ✓
  • D. All three statements are correct

Answer: C — At equilibrium, Ecell = 0 (not E°) because E° = (0.0592/n)log(K), so Ecell = E° − (0.0592/n)log(K) = 0; statements (I) and (II) are correct.

Q6. A galvanic cell has E° = +0.34 V. Which statement is correct? (A) The cell reaction is non-spontaneous and ΔG° > 0. (B) The cell reaction is spontaneous and ΔG° < 0. (C) The cell reaction is at equilibrium with ΔG° = 0. (D) The cell potential will increase with temperature.

  • A. The cell reaction is non-spontaneous and ΔG° > 0
  • B. The cell reaction is spontaneous and ΔG° < 0 ✓
  • C. The cell reaction is at equilibrium with ΔG° = 0
  • D. The cell potential will increase with temperature

Answer: B — Since E° is positive, ΔG° = −nFE° is negative, indicating the reaction is spontaneous; options (A) and (C) contradict this.

Q7. Calculate the mass of copper deposited at the cathode when 2 A current passes through a copper sulfate solution for 30 minutes. (Atomic mass Cu = 64 g/mol, F = 96500 C/mol)

  • A. 1.19 g ✓
  • B. 2.38 g
  • C. 3.57 g
  • D. 4.76 g

Answer: A — Charge Q = I × t = 2 × (30 × 60) = 3600 C; moles of Cu = Q/(nF) = 3600/(2 × 96500) = 0.01865 mol; mass = 0.01865 × 64 = 1.19 g.

Q8. In a galvanic cell, why does molar conductivity increase as the solution becomes more dilute?

  • A. Because ions collide more frequently with solvent molecules and move slower
  • B. Because ions are farther apart and experience less interionic attraction, moving faster ✓
  • C. Because concentration increases and more ions carry charge
  • D. Because the solution becomes more resistive at lower ionic strength

Answer: B — Upon dilution, ions are more dispersed, interionic attractions decrease, and ions move faster with higher mobility, increasing molar conductivity despite lower total ion concentration.

Q9. HOTS: A galvanic cell is constructed using Zn and Cu electrodes. Initially, [Zn²⁺] = 0.1 M and [Cu²⁺] = 1.0 M. Given E°Zn²⁺|Zn = −0.76 V and E°Cu²⁺|Cu = +0.34 V, calculate Ecell using the Nernst equation at 25°C. (R = 8.314 J/(mol·K), F = 96500 C/mol)

  • A. 1.03 V ✓
  • B. 1.10 V
  • C. 1.17 V
  • D. 0.95 V

Answer: A — E°cell = 0.34 − (−0.76) = 1.10 V; Q = [Zn²⁺]/[Cu²⁺] = 0.1/1.0 = 0.1; Ecell = 1.10 − (0.0592/2)log(0.1) = 1.10 − (0.0296 × −1) = 1.10 + 0.0296 ≈ 1.03 V.

Q10. During the electrolysis of acidified water using inert electrodes, oxygen is produced at the anode and hydrogen at the cathode. If 500 cm³ of O₂ (at STP) is collected, what volume of H₂ (at STP) is produced? (Assume 2 mol H₂O → 2 mol H₂ + 1 mol O₂)

  • A. 250 cm³
  • B. 500 cm³
  • C. 1000 cm³ ✓
  • D. 1500 cm³

Answer: C — Stoichiometry: 1 mol O₂ : 2 mol H₂; moles O₂ = 500/22400 mol; moles H₂ = 2 × (500/22400) = 1000/22400; volume H₂ = 1000 cm³ at STP.

Flashcards

What is the definition of a galvanic cell?

A galvanic cell is an electrochemical device that converts the chemical energy of a spontaneous redox reaction into electrical energy.

In a galvanic cell, which electrode is negative and where does oxidation occur?

The anode is negative and oxidation occurs there; electrons flow from anode to cathode through the external circuit.

State the Nernst equation and define each symbol.

Ecell = E° − (RT/nF)ln(Q), where R = gas constant, T = temperature, n = moles of electrons, F = Faraday constant (96500 C/mol), Q = reaction quotient.

What is molar conductivity and how does it vary with dilution?

Molar conductivity Λm = κ/c (conductivity ÷ concentration) and it increases with dilution because ions move faster when dispersed.

State Kohlrausch's law of independent ion migration.

At infinite dilution, the molar conductivity of an electrolyte is the sum of limiting molar conductivities of its constituent cations and anions: Λ°m = Σ(λ° cation) + Σ(λ° anion).

What is Faraday's first law of electrolysis?

The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of charge passed: moles = Q/(nF) where Q is charge and n is electrons transferred.

How are ΔG° and E° cell related?

ΔG° = −nFE°cell; if E° is positive, ΔG° is negative and the reaction is spontaneous.

In an electrolytic cell, which electrode is the anode and what happens there?

The anode is positive and oxidation occurs there; species lose electrons and are oxidized.

What is the relationship between conductivity κ and resistivity ρ?

Conductivity κ is the reciprocal of resistivity: κ = 1/ρ; conductivity measures how well a solution conducts electricity.

How does applying an external voltage greater than the cell potential change the direction of a galvanic cell reaction?

When external voltage exceeds cell potential, the reaction reverses and the cell functions as an electrolytic cell, converting electrical energy into chemical energy.

Important Board Questions

Define electrode potential and standard electrode potential. Give one example. [2 marks]

Electrode potential is the potential difference at electrode-electrolyte interface; standard electrode potential is when all species are at unit concentration (1 M) or unit activity. Example: E°Cu²⁺|Cu = +0.34 V means Cu²⁺ + 2e⁻ → Cu at standard conditions.

Derive the relationship between ΔG°, E° and equilibrium constant K. Show all steps and explain the significance when E° > 0, E° = 0, and E° < 0. [5 marks]

Start with ΔG° = −nFE° and relate to ΔG° = −RT ln(K); set up equilibrium condition Ecell = 0 = E° − (RT/nF)ln(K); analyse sign of E° against ΔG° and K to interpret spontaneity and direction.

A silver coulometer (containing AgNO₃ solution with silver electrodes) is connected in series with a copper sulfate cell during electrolysis. When 1.08 g of silver is deposited at the cathode of the coulometer, calculate: (i) the total charge passed through the circuit in coulombs, (ii) the mass of copper deposited at the cathode of the copper sulfate cell. (Atomic masses: Ag = 108 g/mol, Cu = 64 g/mol; F = 96500 C/mol) [6 marks]

For (i): Ag⁺ + e⁻ → Ag; moles Ag = 1.08/108 = 0.01 mol; charge = 0.01 × F = 965 C. For (ii): In CuSO₄ cell, Cu²⁺ + 2e⁻ → Cu; same charge flows so moles Cu = 965/(2 × 96500) = 0.005 mol; mass Cu = 0.005 × 64 = 0.32 g (show stoichiometric ratio 1 mol e⁻ : Ag but 2 mol e⁻ : Cu).

Assertion-Reason Questions for Electrochemistry

Q1. Assertion (A):

In a galvanic cell, the anode is the electrode where oxidation occurs and has a negative charge with respect to the solution.

Reason (R):

Oxidation is the loss of electrons, and the loss of electrons from the electrode causes it to become negatively charged as electrons accumulate.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The anode is indeed where oxidation occurs (metal atoms lose electrons and enter solution as ions), and this loss of electrons makes the electrode negative relative to the solution. The reason correctly explains why the anode becomes negatively charged.

Q2. Assertion (A):

A salt bridge is essential in all galvanic cells to complete the electrical circuit.

Reason (R):

The salt bridge allows the internal migration of ions between the two half-cells, maintaining electrical neutrality and providing a path for charge flow within the cell.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — While the assertion is true that a salt bridge is essential in many galvanic cells to complete the circuit and maintain neutrality, the reason is incomplete. The salt bridge is not essential in cells where both electrodes dip in the same electrolyte solution (no salt bridge needed). Thus, both statements are true, but the reason does not fully explain why the assertion is true in all cases.

Q3. Assertion (A):

The standard electrode potential of a half-cell is independent of the concentration of ions in the solution.

Reason (R):

By definition, standard electrode potential is measured when all species have unit concentration (1 mol/L), regardless of the actual concentration used in experiments.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true and the reason correctly explains the assertion. Standard electrode potential (E°) is a fixed value defined at 1 mol/L concentration and 25°C; any deviation from these conditions gives a non-standard potential. The reason precisely defines why the standard potential remains constant.

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Case-Based Questions (CBQ) for Electrochemistry

Case Study:

A metallurgical company in India is planning to extract copper electrochemically from copper sulfate solution using a graphite anode and copper cathode. In the initial setup, they notice that at the cathode, copper ions are being deposited as metallic copper, while at the anode, nothing seems to be happening. The plant engineer applies an external potential of 2.0 V to drive the process. However, when the same external potential is applied but with a copper anode instead of graphite, the anode gradually dissolves while copper continues to deposit at the cathode. The company realizes this is actually a refining process rather than pure extraction.

Q1. In the electrolytic process with the graphite anode, what is happening at the anode when 2.0 V is applied?

  • A. Water molecules are being oxidized to produce O₂ gas and H⁺ ions ✓
  • B. Sulfate ions are being oxidized to sulfuric acid
  • C. Copper ions are being reduced to metallic copper
  • D. No reaction occurs at the anode

Answer: A

Q2. When a copper anode replaces the graphite anode, why does the anode dissolve instead of undergoing water oxidation?

  • A. Copper is more easily oxidized than water molecules, so Cu → Cu²⁺ + 2e⁻ occurs preferentially at the anode ✓
  • B. Copper cannot conduct electricity, so water is oxidized instead
  • C. The applied potential is too low to oxidize water but sufficient for copper
  • D. Copper anode causes a chemical reaction with sulfate ions

Answer: A

Q3. In the copper refining process (with copper anode), how does this differ from simple electrochemical extraction?

  • A. Refining produces copper from copper sulfate while extraction removes all impurities
  • B. Refining purifies existing copper by dissolving impure copper at the anode and redepositing pure copper at the cathode ✓
  • C. Refining uses less electrical energy than extraction
  • D. Refining occurs at lower temperatures than extraction

Answer: B

Q4. If the applied potential were increased to 3.5 V with the copper anode, what additional reaction might occur at the anode alongside copper oxidation?

  • A. Copper would be completely oxidized to Cu³⁺ instead of Cu²⁺
  • B. Sulfate ions would begin to oxidize, potentially producing sulfuric acid or peroxodisulfate ions ✓
  • C. Water would be reduced instead of oxidized
  • D. No additional reaction would occur; only Cu oxidation continues

Answer: B

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Previous Year Questions (PYQ) for Electrochemistry

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

The standard electrode potentials of three metals A, B and C are, respectively, + 0·34 V, − 1·18 V and − 0·25 V. The correct order of reducing power of these metals is :

  • B > C > A
  • A > B > C
  • C > B > A
  • A > C > B

CBSE 2026 · Set 56(B) · 3 marks

Define conductivity and molar conductivity for a solution of an electrolyte. Why does the conductivity of a solution decrease with decrease in concentration ?

CBSE 2026 · Set 56(B) · 5 marks

Write the Nernst equation and calculate the emf of the following cell at 298 K : Mg (s)Mg2+ (0·001 M) || Cu2+ (0·0001 M)Cu (s) Given : E°Mg2+/Mg = − 2·36 V, E°Cu2+/Cu = + 0·34 V [log 10 = 1]

CBSE 2026 · Set 56/1/1 · 1 mark

What will happen during the electrolysis of aqueous solution of CuCl2 by using platinum electrodes?

  • Cu will deposit at Anode
  • H2 gas will be released at cathode
  • O2 gas will be released at anode
  • Cl2 gas will be released at anode

CBSE 2026 · Set 56/1/1 · 3 marks

Answer the following: (i) Why is the Equilibrium Constant (Kc) related to E°cell and not to Ecell? (ii) Two metals 'A' and 'B' have standard electrode potential values of –0.24 V and +0.80 V respectively. Which of these will liberate hydrogen gas from dil. H2SO4? (iii) Write the cell reaction which occurs in lead storage battery when it is in charging.

CBSE 2026 · Set 56/1/1 · 5 marks

From the given data of E values, answer the following questions: E°(M2+/M) V: Cr –1.18, Mn –0.91, Fe –1.18, Co –0.44, Ni –0.28, Cu –0.25, Cu +0.34. (I) Why E°(M2+/M) show irregular trend in the above values? (II) Why is E°(Cu2+/Cu) value exceptionally positive? (III) Why E°(Mn2+/Mn) value is highly negative?

CBSE 2026 · Set 56/1/1 · 5 marks

Calculate emf and ΔG for the following cell at 298 K: Mg(s) / Mg2+(0.01 M) // Ag+(0.001 M) / Ag(s). Given: E°(Mg2+/Mg) = –2.37 V, E°(Ag+/Ag) = +0.80 V. [1 F = 96500 C mol–1, log 10 = 1]

CBSE 2026 · Set 56/1/2 · 1 mark

How many Faradays are required to reduce 1 mol of Cr2O7²⁻ to Cr³⁺ in acidic medium?

  • 2
  • 3
  • 6
  • 4

CBSE 2026 · Set 56/1/2 · 5 marks

Calculate emf and ΔG for the following cell at 298 K: Mg(s) / Mg²⁺(0.01 M) // Ag⁺(0.001 M) / Ag(s). Given: E°(Mg²⁺/Mg) = –2.37 V, E°(Ag⁺/Ag) = +0.80 V [Note: 1 F = 96500 C mol⁻¹, log 10 = 1]

CBSE 2026 · Set 56/1/3 · 1 mark

How the conductivity varies on decreasing concentration for both weak and strong electrolytes ?

  • It increases for weak electrolyte and decreases for strong electrolyte.
  • It decreases for weak electrolyte and increases for strong electrolyte.
  • It increases for both weak and strong electrolytes.
  • It decreases for both weak and strong electrolytes.

CBSE 2026 · Set 56/1/3 · 3 marks

(a) Answer the following : (i) Why is the Equilibrium Constant (Kc) related to Ecell and not to E°cell ? (ii) Two metals 'A' and 'B' have standard electrode potential values of –0.24 V and +0.80 V respectively. Which of these will liberate hydrogen gas from dil. H2SO4 ? (iii) Write the cell reaction which occurs in lead storage battery when it is in charging.

CBSE 2026 · Set 56/1/3 · 5 marks

(a) Calculate emf and ΔG for the following cell at 298 K : Mg(s) / Mg2+(0.01 M) // Ag+(0.001 M) / Ag(s). Given : E Mg2+ / Mg = –2.37 V; E Ag+ / Ag = +0.80 V. [1 F = 96500 C mol–1, log 10 = 1]

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Class 12 Chemistry Formula Sheet

Every formula and named reaction from all 10 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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