Class 12 Chemistry | Chapter 2
Electrochemistry
Galvanic Cells • Nernst Equation • Kohlrausch Law • Faraday's Laws • Batteries
1. Electrochemical Cells
Electrochemistry: The study of production of electricity from energy released during
spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical
transformations.
1.1 Galvanic Cells (Voltaic Cells)
Converts chemical energy of a spontaneous redox reaction into electrical energy. Example: Daniell cell.
- Anode: Oxidation occurs here. It has a negative potential. (LOAN: Left, Oxidation, Anode, Negative).
- Cathode: Reduction occurs here. It has a positive potential.
- Salt Bridge: A U-tube containing an inert electrolyte (like KCl, KNO3) in agar-agar. It completes the electrical circuit and maintains electrical neutrality in the two half-cells.
1.2 Cell Potential (EMF)
The difference between the electrode potentials of the cathode and anode. It is called EMF (Electromotive Force) when no current is drawn through the cell.
E°cell = E°cathode - E°anode
E°cell = E°Right - E°Left
(Always use standard REDUCTION potentials).
E°cell = E°Right - E°Left
(Always use standard REDUCTION potentials).
1.3 Standard Hydrogen Electrode (SHE)
Reference electrode assigned a standard electrode potential of exactly zero volts at all temperatures.
- Reaction: H+ (aq, 1M) + e− ⇌ 1/2 H2 (g, 1 bar)
- If E° of an electrode > 0, its reduced form is more stable than hydrogen gas (Stronger oxidizing agent).
- If E° of an electrode < 0, hydrogen gas is more stable than the reduced form (Stronger reducing agent).
2. Nernst Equation
Relates electrode potential to the concentration of ions.
For a half-cell: Mn+(aq) + ne− → M(s)
E = E° - (RT/nF) ln(1 / [Mn+])
At 298 K (25°C):
E = E° - (0.0591 / n) log(1 / [Mn+])
E = E° - (RT/nF) ln(1 / [Mn+])
At 298 K (25°C):
E = E° - (0.0591 / n) log(1 / [Mn+])
2.1 Nernst Equation for Complete Cell
For a general reaction: a A + b B → c C + d D
Ecell = E°cell - (0.0591 / n) log Q
Ecell = E°cell - (0.0591 / n) log ([C]c[D]d / [A]a[B]b)
Ecell = E°cell - (0.0591 / n) log ([C]c[D]d / [A]a[B]b)
(Concentration of pure solids and pure liquids is taken as 1).
2.2 Equilibrium Constant (Kc) from Nernst Equation
At equilibrium, cell potential (Ecell) becomes zero. The reaction quotient Q becomes Kc.
0 = E°cell - (0.0591 / n) log Kc
log Kc = (n · E°cell) / 0.0591 (at 298 K)
log Kc = (n · E°cell) / 0.0591 (at 298 K)
2.3 Electrochemical Cell and Gibbs Energy
Electrical work done in one second is equal to electrical potential multiplied by total charge passed.
ΔG = -nFEcell
ΔG° = -nFE°cell (Standard conditions)
ΔG° = -RT ln Kc
ΔG° = -nFE°cell (Standard conditions)
ΔG° = -RT ln Kc
⚠️ NEET TIP: For a spontaneous reaction, ΔG must be
negative, which means Ecell must be positive. If you multiply a
half-reaction by 2, its E° does NOT change (it is intensive), but its ΔG° is doubled (it is
extensive).
3. Conductance of Electrolytic Solutions
- Resistance (R): R = ρ(l/A). Unit: ohm (Ω).
- Resistivity (ρ): Specific resistance. Unit: Ω m or Ω cm.
- Conductance (G): Reciprocal of resistance. G = 1/R. Unit: ohm−1, mho, or Siemens (S).
- Conductivity (κ - kappa): Specific conductance. κ = 1/ρ = G(l/A). Unit: S m−1 or S cm−1.
- Cell Constant (G*): l/A. Measured using a conductivity cell containing KCl solution of known conductivity. G* = R · κ.
3.1 Molar Conductivity (Λm)
Conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.
Λm = κ / c
If κ is in S cm−1 and c is in mol L−1 (Molarity):
Λm = (κ × 1000) / M (Unit: S cm² mol⁻¹)
If κ is in S cm−1 and c is in mol L−1 (Molarity):
Λm = (κ × 1000) / M (Unit: S cm² mol⁻¹)
3.2 Variation of Conductivity and Molar Conductivity with Concentration
- Conductivity (κ): Always decreases with decrease in concentration (dilution) for both strong and weak electrolytes, because the number of ions per unit volume decreases.
- Molar Conductivity (Λm): Always increases with decrease in concentration (dilution).
- Strong Electrolytes: Increases slowly with dilution due to decrease in interionic attraction. (Debye-Huckel-Onsager equation: Λm = Λ°m - A√c).
- Weak Electrolytes: Increases steeply at lower concentrations due to significant increase in degree of dissociation (α).
4. Kohlrausch Law of Independent Migration of Ions
Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions
of the anion and cation of the electrolyte.
Λ°m = ν+λ°+ +
ν−λ°−
Example: Λ°m(MgCl2) = λ°(Mg2+) + 2λ°(Cl−)
4.1 Applications of Kohlrausch Law
1. Calculation of Λ°m for weak electrolytes (since it cannot be obtained by
extrapolating the Λm vs √c graph).
Example: Λ°m(CH3COOH) =
Λ°m(CH3COONa) + Λ°m(HCl) -
Λ°m(NaCl)
2. Calculation of Degree of Dissociation (α):
α = Λm / Λ°m
3. Calculation of Dissociation Constant (Ka):
Ka = (cα²) / (1 - α) = (cΛm²) /
(Λ°m (Λ°m - Λm))
5. Electrolytic Cells and Electrolysis
Converts electrical energy into chemical energy. Non-spontaneous reaction is driven by external voltage.
5.1 Faraday's Laws of Electrolysis
First Law: Mass of substance deposited/liberated is directly proportional to the quantity
of electricity passed.
m = Z · Q = Z · I · t (Z = electrochemical equivalent)
Second Law: If same current is passed through different electrolytes in series, masses deposited are proportional to their equivalent weights (E).
m1 / m2 = E1 / E2
m = Z · Q = Z · I · t (Z = electrochemical equivalent)
Second Law: If same current is passed through different electrolytes in series, masses deposited are proportional to their equivalent weights (E).
m1 / m2 = E1 / E2
- Faraday (F): Charge on 1 mole of electrons. F = NA × e = 96487 C mol−1 ≈ 96500 C mol−1.
- Charge required to reduce 1 mole of Mn+ is nF. (e.g., Al3+ requires 3F).
5.2 Products of Electrolysis
Depends on the nature of material being electrolysed and the type of electrodes. The species with higher reduction potential gets reduced at cathode (preferentially). The species with lower reduction potential gets oxidized at anode.
- Electrolysis of molten NaCl: Na at cathode, Cl2 at anode.
- Electrolysis of aqueous NaCl: H2 at cathode (since H2O has higher reduction potential than Na+), Cl2 at anode (due to overvoltage of oxygen, Cl− is oxidized instead of water). NaOH is left in solution.
6. Batteries and Fuel Cells
6.1 Primary Batteries
Reaction occurs only once. Cannot be recharged (dead).
Dry Cell: Anode = Zinc container. Cathode = Carbon (graphite) rod surrounded by
MnO2+C. Electrolyte = NH4Cl + ZnCl2 paste. Cell potential ≈ 1.5
V.
Mercury Cell: Used in hearing aids/watches. Anode = Zn(Hg). Cathode = Paste of HgO + C.
Electrolyte = Paste of KOH + ZnO. Potential ≈ 1.35 V (remains constant because no ions in solution
change concentration).
6.2 Secondary Batteries
Can be recharged by passing current in the opposite direction.
Lead Storage Battery: Anode = Pb. Cathode = PbO2. Electrolyte = 38%
H2SO4.
- Discharge reaction: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
6.3 Fuel Cells
Galvanic cells designed to convert the energy of combustion of fuels (like H2, CH4,
CH3OH) directly into electrical energy. High efficiency (~70%) and pollution-free.
H2-O2 Fuel Cell: Used in Apollo space programme. Water vapor produced
was condensed and added to drinking water supply. Electrodes: Porous carbon with Pt/Pd catalyst.
Electrolyte: Aqueous KOH/NaOH.
7. Corrosion
Destruction of metals by chemical or electrochemical reaction with the environment. Example: Rusting of iron
(hydrated ferric oxide, Fe2O3·xH2O).
Mechanism (Rusting):
At Anode (spot on iron): 2Fe(s) → 2Fe2+ + 4e−
At Cathode: O2(g) + 4H+(aq) + 4e− → 2H2O(l)
Fe2+ ions are further oxidized by atmospheric oxygen to Fe3+ to form rust.
Prevention: Barrier protection (painting, oiling), Galvanization (coating with zinc), Cathodic protection (connecting iron to a more active metal like Mg or Zn, which acts as a sacrificial anode).
🎓 NEET Previous Year Questions
Q1. [NEET 2022] Find the emf of the cell in which the following reaction takes place
at 298 K: Ni(s) + 2Ag⁺(0.001 M) → Ni²⁺(0.001 M) + 2Ag(s). (Given: E°cell =
10.5 V, 2.303 RT/F = 0.059 at 298 K). Note: The E° value given in actual NEET paper had a typo, but
solving format remains same. Let's assume standard values or given 1.05V.
Answer E = E° - (0.0591/n) log( [Ni²⁺]
/ [Ag⁺]² ). n = 2.
E = 1.05 - (0.0591/2) log( 10⁻³ / (10⁻³)² ) = 1.05 - 0.0295 log(10³) = 1.05 - 0.0295(3) = 1.05 - 0.0885 = 0.9615 V.
E = 1.05 - (0.0591/2) log( 10⁻³ / (10⁻³)² ) = 1.05 - 0.0295 log(10³) = 1.05 - 0.0295(3) = 1.05 - 0.0885 = 0.9615 V.
Q2. [NEET 2020] The number of Faradays (F) required to produce 20.0 g of calcium from
molten CaCl₂ (Atomic mass of Ca = 40 g mol⁻¹) is:
Answer Ca²⁺ + 2e⁻ → Ca.
1 mol (40g) Ca requires 2F charge.
20g Ca (which is 0.5 mol) will require: 0.5 × 2F = 1 F.
1 mol (40g) Ca requires 2F charge.
20g Ca (which is 0.5 mol) will require: 0.5 × 2F = 1 F.
Q3. [NEET 2018] For given cell: Zn | Zn²⁺ (C₁) || Cu²⁺
(C₂) | Cu. Change in free energy is ΔG. What happens if C₁ = C₂?
Answer E = E° - (0.059/2) log(C₁/C₂).
If C₁ = C₂, log(1) = 0. Hence E = E°. Then ΔG = -nFE = -nFE° =
ΔG°.
Q4. [NEET 2017] Limiting molar conductivity of NH₄OH is equal to:
Answer According to Kohlrausch's Law for weak
electrolytes:
Λ°m(NH₄OH) = Λ°m(NH₄Cl) + Λ°m(NaOH) - Λ°m(NaCl).
Λ°m(NH₄OH) = Λ°m(NH₄Cl) + Λ°m(NaOH) - Λ°m(NaCl).
Q5. [NEET 2016] Molar conductivity of a 1.5 M solution of an electrolyte is found to
be 138.9 S cm² mol⁻¹. Calculate the conductivity of this solution.
Answer Λm = (κ × 1000) / M
⇒ κ = (Λm × M) / 1000
κ = (138.9 × 1.5) / 1000 = 208.35 / 1000 = 0.208 S cm⁻¹.
κ = (138.9 × 1.5) / 1000 = 208.35 / 1000 = 0.208 S cm⁻¹.
💡 Rapid Revision
- Galvanic Cell: Anode is Negative (oxidation). Cathode is Positive (reduction).
- Strong oxidizing agent = High positive E°. Strong reducing agent = Highly negative E°.
- In Nernst Eq, always put product ions in numerator, reactant ions in denominator (raised to power of stoichiometry).
- κ decreases on dilution. Λm increases on dilution.
- Faraday's laws: 1 Faraday (96500 C) deposits 1 Equivalent weight of substance.
- Rusting of iron acts as an electrochemical cell where pure iron acts as anode and oxygen acts at cathode.
CLASS 12 CHEMISTRY | NCERT SOLUTIONS
Chapter 2 — Electrochemistry
22 Solved Numericals — Step-by-Step Breakdown
Key Formulas: E = E° - (0.0591/n)log(Q) | ΔG° = -nFE° |
Λm = (κ×1000)/M | m = Z·I·t | α =
Λm/Λ°m | F = 96500 C
📝 Nernst Equation & Gibbs Free Energy (Q1 – Q8)
3 MarksQ1. Represent the cell in which the following
reaction takes place: Mg(s) + 2Ag⁺(0.0001M) → Mg²⁺(0.130M) + 2Ag(s). Calculate
its E(cell) if E°(cell) = 3.17 V.
✓ Solution
Cell representation: Mg(s) | Mg²⁺(0.130 M) || Ag⁺(0.0001 M) | Ag(s)
Number of electrons transferred (n) = 2.
Nernst equation: Ecell = E°cell - (0.0591/2) log( [Mg²⁺] / [Ag⁺]² )
Ecell = 3.17 - (0.0295) log( 0.130 / (10⁻⁴)² )
Ecell = 3.17 - 0.0295 log( 0.130 × 10⁸ ) = 3.17 - 0.0295 log( 1.3 × 10⁷ )
log(1.3 × 10⁷) = log(1.3) + 7 = 0.1139 + 7 = 7.1139
Cell representation: Mg(s) | Mg²⁺(0.130 M) || Ag⁺(0.0001 M) | Ag(s)
Number of electrons transferred (n) = 2.
Nernst equation: Ecell = E°cell - (0.0591/2) log( [Mg²⁺] / [Ag⁺]² )
Ecell = 3.17 - (0.0295) log( 0.130 / (10⁻⁴)² )
Ecell = 3.17 - 0.0295 log( 0.130 × 10⁸ ) = 3.17 - 0.0295 log( 1.3 × 10⁷ )
log(1.3 × 10⁷) = log(1.3) + 7 = 0.1139 + 7 = 7.1139
Ecell = 3.17 - (0.0295 × 7.1139) = 3.17 - 0.21 = 2.96
V
3 MarksQ2. Calculate the standard cell potentials of
galvanic cell in which the following reaction takes place: 2Cr(s) + 3Cd²⁺(aq) →
2Cr³⁺(aq) + 3Cd(s). (Given E°(Cr³⁺/Cr) = -0.74V, E°(Cd²⁺/Cd)
= -0.40V). Calculate ΔrG° and equilibrium constant of the reaction.
✓ Solution
1. Standard Cell Potential:
Anode (oxidation) = Cr; Cathode (reduction) = Cd.
E°cell = E°cathode - E°anode = (-0.40) - (-0.74) = +0.34 V.
2. Gibbs Free Energy (ΔG°):
Reaction involves transfer of 6 electrons (n = 6). F = 96500 C mol⁻¹.
ΔG° = -nFE°cell = -6 × 96500 × 0.34 = -196860 J mol⁻¹
ΔG° = -2.303 RT log Kc ⇒ -196860 = -2.303 × 8.314 × 298 × log Kc
log Kc = 196860 / 5705.8 = 34.5
1. Standard Cell Potential:
Anode (oxidation) = Cr; Cathode (reduction) = Cd.
E°cell = E°cathode - E°anode = (-0.40) - (-0.74) = +0.34 V.
2. Gibbs Free Energy (ΔG°):
Reaction involves transfer of 6 electrons (n = 6). F = 96500 C mol⁻¹.
ΔG° = -nFE°cell = -6 × 96500 × 0.34 = -196860 J mol⁻¹
ΔG° = -196.86 kJ mol⁻¹
3. Equilibrium Constant (Kc):ΔG° = -2.303 RT log Kc ⇒ -196860 = -2.303 × 8.314 × 298 × log Kc
log Kc = 196860 / 5705.8 = 34.5
Kc = Antilog(34.5) = 3.16 × 10³⁴
3 MarksQ3. Calculate the potential of hydrogen
electrode in contact with a solution whose pH is 10.
✓ Solution
Reduction half-reaction: H⁺ + e⁻ → 1/2 H₂(g)
If pH = 10, then [H⁺] = 10⁻¹⁰ M.
Nernst equation for single electrode:
E = E° - (0.0591/n) log( 1 / [H⁺] )
E° for standard hydrogen electrode is 0.00 V. n = 1.
E = 0 - 0.0591 log( 1 / 10⁻¹⁰ ) = -0.0591 log(10¹⁰)
Reduction half-reaction: H⁺ + e⁻ → 1/2 H₂(g)
If pH = 10, then [H⁺] = 10⁻¹⁰ M.
Nernst equation for single electrode:
E = E° - (0.0591/n) log( 1 / [H⁺] )
E° for standard hydrogen electrode is 0.00 V. n = 1.
E = 0 - 0.0591 log( 1 / 10⁻¹⁰ ) = -0.0591 log(10¹⁰)
E = -0.0591 × 10 = -0.591 V
3 MarksQ4. Calculate the emf of the cell: Cu(s) |
Cu²⁺(0.130M) || Ag⁺(1.0 × 10⁻⁴ M) | Ag(s). Given
E°(Cu²⁺/Cu) = 0.34 V and E°(Ag⁺/Ag) = 0.80 V.
✓ Solution
E°cell = E°cathode - E°anode = 0.80 - 0.34 = 0.46 V.
Cell reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s). Here n = 2.
Nernst Eq: E = E° - (0.0591/2) log( [Cu²⁺] / [Ag⁺]² )
E = 0.46 - 0.02955 log( 0.130 / (10⁻⁴)² )
E = 0.46 - 0.02955 log( 0.130 × 10⁸ ) = 0.46 - 0.02955 log(1.3 × 10⁷)
E = 0.46 - 0.02955(7.1139) = 0.46 - 0.21
E°cell = E°cathode - E°anode = 0.80 - 0.34 = 0.46 V.
Cell reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s). Here n = 2.
Nernst Eq: E = E° - (0.0591/2) log( [Cu²⁺] / [Ag⁺]² )
E = 0.46 - 0.02955 log( 0.130 / (10⁻⁴)² )
E = 0.46 - 0.02955 log( 0.130 × 10⁸ ) = 0.46 - 0.02955 log(1.3 × 10⁷)
E = 0.46 - 0.02955(7.1139) = 0.46 - 0.21
E = 0.25 V
3 MarksQ5. The standard electrode potential for
Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu²⁺(aq)
→ Zn²⁺(aq) + Cu(s)
✓ Solution
E°cell = 1.1 V. Change in electrons (n) = 2.
Faraday constant (F) = 96487 C mol⁻¹.
ΔG° = -nFE°cell
ΔG° = -2 × 96487 × 1.1 = -212271.4 J mol⁻¹
E°cell = 1.1 V. Change in electrons (n) = 2.
Faraday constant (F) = 96487 C mol⁻¹.
ΔG° = -nFE°cell
ΔG° = -2 × 96487 × 1.1 = -212271.4 J mol⁻¹
ΔG° = -212.27 kJ mol⁻¹
(Note: Using F=96500 gives -212.3 kJ, which is also accepted.)3 MarksQ6. Calculate the equilibrium constant of the
reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s); E°(cell) = 0.46 V.
✓ Solution
Here n = 2.
log Kc = (n × E°cell) / 0.0591
log Kc = (2 × 0.46) / 0.0591 = 0.92 / 0.0591 = 15.567
Kc = Antilog(15.567) = Antilog(0.567) × 10¹⁵
Here n = 2.
log Kc = (n × E°cell) / 0.0591
log Kc = (2 × 0.46) / 0.0591 = 0.92 / 0.0591 = 15.567
Kc = Antilog(15.567) = Antilog(0.567) × 10¹⁵
Kc = 3.69 × 10¹⁵
3 MarksQ7. A cell is prepared by dipping copper rod
in 1M CuSO₄ solution and zinc rod in 1M ZnSO₄ solution. The standard reduction potentials of
Cu and Zn are 0.34 V and -0.76 V. What will be the cell reaction? Can copper displace zinc from its
solution?
✓ Solution
Zn has a more negative reduction potential, meaning it acts as the anode (oxidation). Cu acts as the cathode (reduction).
Cell Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Zn has a more negative reduction potential, meaning it acts as the anode (oxidation). Cu acts as the cathode (reduction).
Cell Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Can Cu displace Zn? No. Copper is below Zinc in the
reactivity series (higher reduction potential). Copper cannot displace a more active metal like Zinc
from its solution.
3 MarksQ8. Calculate the cell potential at 298 K:
Fe(s) | Fe²⁺(0.001M) || H⁺(1M) | H₂(g)(1 bar) | Pt(s). (Given
E°(Fe²⁺/Fe) = -0.44V).
✓ Solution
Reaction: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g). (n = 2).
E°cell = E°H⁺/H2 - E°Fe²⁺/Fe = 0.00 - (-0.44) = 0.44 V.
Nernst Eq: E = E° - (0.0591/n) log( [Fe²⁺] P(H₂) / [H⁺]² )
E = 0.44 - (0.0591/2) log( (0.001 × 1) / (1)² )
E = 0.44 - 0.02955 log(10⁻³)
E = 0.44 - 0.02955(-3) = 0.44 + 0.08865
Reaction: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g). (n = 2).
E°cell = E°H⁺/H2 - E°Fe²⁺/Fe = 0.00 - (-0.44) = 0.44 V.
Nernst Eq: E = E° - (0.0591/n) log( [Fe²⁺] P(H₂) / [H⁺]² )
E = 0.44 - (0.0591/2) log( (0.001 × 1) / (1)² )
E = 0.44 - 0.02955 log(10⁻³)
E = 0.44 - 0.02955(-3) = 0.44 + 0.08865
E = 0.528 V
💡 Conductance & Kohlrausch Law (Q9 – Q16)
3 MarksQ9. Resistance of a conductivity cell filled
with 0.1 mol L⁻¹ KCl solution is 100 Ω. If the resistance of the same cell when filled
with 0.02 mol L⁻¹ KCl solution is 520 Ω, calculate the conductivity and molar
conductivity of 0.02 mol L⁻¹ KCl solution. The conductivity of 0.1 mol L⁻¹ KCl
solution is 1.29 S/m.
✓ Solution
1. Find Cell Constant (G*):
G* = R × κ (for 0.1M solution)
R = 100 Ω; κ = 1.29 S/m = 1.29 × 10⁻² S cm⁻¹.
G* = 100 × 1.29 × 10⁻² = 1.29 cm⁻¹.
2. Conductivity of 0.02M solution:
κ = G* / R = 1.29 / 520 = 0.248 × 10⁻² S cm⁻¹.
3. Molar Conductivity:
Λm = (κ × 1000) / M = (0.00248 × 1000) / 0.02
1. Find Cell Constant (G*):
G* = R × κ (for 0.1M solution)
R = 100 Ω; κ = 1.29 S/m = 1.29 × 10⁻² S cm⁻¹.
G* = 100 × 1.29 × 10⁻² = 1.29 cm⁻¹.
2. Conductivity of 0.02M solution:
κ = G* / R = 1.29 / 520 = 0.248 × 10⁻² S cm⁻¹.
3. Molar Conductivity:
Λm = (κ × 1000) / M = (0.00248 × 1000) / 0.02
Λm = 2.48 / 0.02 = 124 S cm²
mol⁻¹
3 MarksQ10. The molar conductivity of 0.025 mol
L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation
and dissociation constant. Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and
λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.
✓ Solution
Limiting molar conductivity:
Λ°m(HCOOH) = λ°(H⁺) + λ°(HCOO⁻) = 349.6 + 54.6 = 404.2 S cm² mol⁻¹.
Degree of dissociation (α):
α = Λm / Λ°m = 46.1 / 404.2 = 0.114.
Dissociation constant (Ka):
Ka = (cα²) / (1 - α) = (0.025 × (0.114)²) / (1 - 0.114)
Ka = (0.025 × 0.012996) / 0.886 = 0.0003249 / 0.886
Limiting molar conductivity:
Λ°m(HCOOH) = λ°(H⁺) + λ°(HCOO⁻) = 349.6 + 54.6 = 404.2 S cm² mol⁻¹.
Degree of dissociation (α):
α = Λm / Λ°m = 46.1 / 404.2 = 0.114.
Dissociation constant (Ka):
Ka = (cα²) / (1 - α) = (0.025 × (0.114)²) / (1 - 0.114)
Ka = (0.025 × 0.012996) / 0.886 = 0.0003249 / 0.886
Ka = 3.67 × 10⁻⁴
3 MarksQ11. Calculate Λ°m for
acetic acid. Given that Λ°m for HCl, NaCl, and CH₃COONa are 425.9, 126.4, and
91.0 S cm² mol⁻¹ respectively.
✓ Solution
According to Kohlrausch's Law:
Λ°m(CH₃COOH) = λ°(CH₃COO⁻) + λ°(H⁺)
This can be obtained by:
Λ°m(CH₃COONa) + Λ°m(HCl) - Λ°m(NaCl)
= [λ°(CH₃COO⁻) + λ°(Na⁺)] + [λ°(H⁺) + λ°(Cl⁻)] - [λ°(Na⁺) + λ°(Cl⁻)]
According to Kohlrausch's Law:
Λ°m(CH₃COOH) = λ°(CH₃COO⁻) + λ°(H⁺)
This can be obtained by:
Λ°m(CH₃COONa) + Λ°m(HCl) - Λ°m(NaCl)
= [λ°(CH₃COO⁻) + λ°(Na⁺)] + [λ°(H⁺) + λ°(Cl⁻)] - [λ°(Na⁺) + λ°(Cl⁻)]
Λ°m(CH₃COOH) = 91.0 + 425.9 - 126.4 = 516.9 - 126.4
= 390.5 S cm² mol⁻¹
3 MarksQ12. The electrical resistance of a column of
0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 10³ Ω. Calculate its
resistivity, conductivity, and molar conductivity.
✓ Solution
Radius = 0.5 cm. Area (A) = πr² = 3.14 × (0.5)² = 0.785 cm². Length (l) = 50 cm.
1. Resistivity (ρ): R = ρ(l/A) ⇒ ρ = R·A / l
ρ = (5.55 × 10³ × 0.785) / 50 = 4356.75 / 50 = 87.135 Ω cm.
2. Conductivity (κ): κ = 1 / ρ
κ = 1 / 87.135 = 0.01148 S cm⁻¹.
3. Molar conductivity:
Radius = 0.5 cm. Area (A) = πr² = 3.14 × (0.5)² = 0.785 cm². Length (l) = 50 cm.
1. Resistivity (ρ): R = ρ(l/A) ⇒ ρ = R·A / l
ρ = (5.55 × 10³ × 0.785) / 50 = 4356.75 / 50 = 87.135 Ω cm.
2. Conductivity (κ): κ = 1 / ρ
κ = 1 / 87.135 = 0.01148 S cm⁻¹.
3. Molar conductivity:
Λm = (κ × 1000) / M = (0.01148 × 1000) / 0.05 =
11.48 / 0.05 = 229.6 S cm² mol⁻¹
3 MarksQ13. Predict the products of electrolysis of:
(a) An aqueous solution of AgNO₃ with silver electrodes. (b) An aqueous solution of AgNO₃
with platinum electrodes.
✓ Solution
(a) With Silver electrodes (Active):
Cathode: Ag⁺ has higher reduction potential than water, so Ag is deposited. (Ag⁺ + e⁻ → Ag)
Anode: Silver anode dissolves since Ag is easily oxidized compared to water. (Ag → Ag⁺ + e⁻)
(b) With Platinum electrodes (Inert):
Cathode: Same as above. Ag is deposited.
(a) With Silver electrodes (Active):
Cathode: Ag⁺ has higher reduction potential than water, so Ag is deposited. (Ag⁺ + e⁻ → Ag)
Anode: Silver anode dissolves since Ag is easily oxidized compared to water. (Ag → Ag⁺ + e⁻)
(b) With Platinum electrodes (Inert):
Cathode: Same as above. Ag is deposited.
Anode: Water is oxidized to release Oxygen gas because Pt is unreactive.
(2H₂O → O₂ + 4H⁺ + 4e⁻)
3 MarksQ14. The conductivity of 0.20M solution of KCl
at 298K is 0.0248 S cm⁻¹. Calculate its molar conductivity.
✓ Solution
κ = 0.0248 S cm⁻¹; M = 0.20 mol L⁻¹.
Λm = (κ × 1000) / M
Λm = (0.0248 × 1000) / 0.20
Λm = 24.8 / 0.20
κ = 0.0248 S cm⁻¹; M = 0.20 mol L⁻¹.
Λm = (κ × 1000) / M
Λm = (0.0248 × 1000) / 0.20
Λm = 24.8 / 0.20
Λm = 124.0 S cm² mol⁻¹
3 MarksQ15. Conductivity of 0.00241 M acetic acid is
7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity. If Λ°m for
acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?
✓ Solution
1. Molar conductivity:
Λm = (7.896 × 10⁻⁵ × 1000) / 0.00241 = 0.07896 / 0.00241 = 32.76 S cm² mol⁻¹.
2. Degree of dissociation (α):
α = Λm / Λ°m = 32.76 / 390.5 = 0.084.
3. Dissociation constant (Ka):
Ka = (Cα²) / (1 - α) = (0.00241 × (0.084)²) / (1 - 0.084)
1. Molar conductivity:
Λm = (7.896 × 10⁻⁵ × 1000) / 0.00241 = 0.07896 / 0.00241 = 32.76 S cm² mol⁻¹.
2. Degree of dissociation (α):
α = Λm / Λ°m = 32.76 / 390.5 = 0.084.
3. Dissociation constant (Ka):
Ka = (Cα²) / (1 - α) = (0.00241 × (0.084)²) / (1 - 0.084)
Ka = (0.00241 × 0.007056) / 0.916 = 1.85 ×
10⁻⁵
3 MarksQ16. Why does the conductivity of a solution
decrease with dilution?
✓ Solution
Conductivity (κ) is the conductance of one unit volume (1 cm³ or 1 m³) of the solution.
Upon dilution, the concentration of the electrolyte decreases. As more solvent is added, the number of current-carrying ions present in one unit volume of the solution decreases.
Conductivity (κ) is the conductance of one unit volume (1 cm³ or 1 m³) of the solution.
Upon dilution, the concentration of the electrolyte decreases. As more solvent is added, the number of current-carrying ions present in one unit volume of the solution decreases.
Therefore, the conductivity of the solution consistently decreases with
dilution for both strong and weak electrolytes.
📈 Faraday's Laws of Electrolysis (Q17 – Q22)
3 MarksQ17. A solution of CuSO₄ is electrolysed
for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
(Atomic mass Cu = 63.5 g)
✓ Solution
Current (I) = 1.5 A; Time (t) = 10 × 60 = 600 s.
Total charge passed (Q) = I × t = 1.5 × 600 = 900 C.
Cathode reaction: Cu²⁺ + 2e⁻ → Cu(s). (n = 2).
Charge required to deposit 1 mol (63.5 g) of Cu = 2F = 2 × 96487 C. (Let's use 96487 as per NCERT).
Mass deposited (m) = (Q × Molar Mass) / (n × F)
m = (900 × 63.5) / (2 × 96487) = 57150 / 192974
Current (I) = 1.5 A; Time (t) = 10 × 60 = 600 s.
Total charge passed (Q) = I × t = 1.5 × 600 = 900 C.
Cathode reaction: Cu²⁺ + 2e⁻ → Cu(s). (n = 2).
Charge required to deposit 1 mol (63.5 g) of Cu = 2F = 2 × 96487 C. (Let's use 96487 as per NCERT).
Mass deposited (m) = (Q × Molar Mass) / (n × F)
m = (900 × 63.5) / (2 × 96487) = 57150 / 192974
m = 0.296 g (Using F=96500 gives 0.2961 g)
3 MarksQ18. How much charge is required for the
following reductions: (i) 1 mol of Al³⁺ to Al (ii) 1 mol of MnO₄⁻ to
Mn²⁺?
✓ Solution
(i) Al³⁺ + 3e⁻ → Al
1 mol of Al³⁺ requires 3 moles of electrons = 3F.
Charge = 3 × 96500 = 2.895 × 10⁵ C.
(ii) MnO₄⁻ → Mn²⁺
Oxidation state of Mn in MnO₄⁻ is +7. It reduces to +2. Change = 5 electrons.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
(i) Al³⁺ + 3e⁻ → Al
1 mol of Al³⁺ requires 3 moles of electrons = 3F.
Charge = 3 × 96500 = 2.895 × 10⁵ C.
(ii) MnO₄⁻ → Mn²⁺
Oxidation state of Mn in MnO₄⁻ is +7. It reduces to +2. Change = 5 electrons.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
1 mol requires 5F. Charge = 5 × 96500 = 4.825 × 10⁵
C.
3 MarksQ19. How much electricity in terms of Faraday
is required to produce (i) 20.0 g of Ca from molten CaCl₂ (ii) 40.0 g of Al from molten
Al₂O₃? (Mass: Ca=40, Al=27).
✓ Solution
(i) Ca²⁺ + 2e⁻ → Ca. (2F gives 1 mol = 40g Ca).
20.0 g Ca = 0.5 mol. Electricity required = 0.5 × 2F = 1 F.
(ii) Al³⁺ + 3e⁻ → Al. (3F gives 1 mol = 27g Al).
40.0 g Al = 40/27 mol.
Electricity required = (40/27) × 3F = 120/27 F = 4.44 F.
(i) Ca²⁺ + 2e⁻ → Ca. (2F gives 1 mol = 40g Ca).
20.0 g Ca = 0.5 mol. Electricity required = 0.5 × 2F = 1 F.
(ii) Al³⁺ + 3e⁻ → Al. (3F gives 1 mol = 27g Al).
40.0 g Al = 40/27 mol.
Electricity required = (40/27) × 3F = 120/27 F = 4.44 F.
3 MarksQ20. If a current of 0.5 amperes flows through
a metallic wire for 2 hours, then how many electrons would flow through the wire?
✓ Solution
I = 0.5 A. t = 2 × 60 × 60 = 7200 s.
Charge (Q) = I × t = 0.5 × 7200 = 3600 C.
Charge on one electron (e⁻) = 1.6 × 10⁻¹⁹ C.
Number of electrons = Total charge / charge on one electron
Number = 3600 / (1.6 × 10⁻¹⁹)
I = 0.5 A. t = 2 × 60 × 60 = 7200 s.
Charge (Q) = I × t = 0.5 × 7200 = 3600 C.
Charge on one electron (e⁻) = 1.6 × 10⁻¹⁹ C.
Number of electrons = Total charge / charge on one electron
Number = 3600 / (1.6 × 10⁻¹⁹)
Number of e⁻ = 2.25 × 10²² electrons
4 MarksQ21. Three electrolytic cells A,B,C containing
solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady
current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell
B. How long did the current flow? What mass of copper and zinc were deposited?
✓ Solution
1. Time for Ag deposition: Ag⁺ + e⁻ → Ag (n=1). M=108. m=1.45g. I=1.5A.
m = (M × I × t) / (n × F) ⇒ 1.45 = (108 × 1.5 × t) / (1 × 96500)
t = (1.45 × 96500) / 162 = 139925 / 162 = 863.7 s (14.4 min).
2. Second Law for Cu: m(Cu)/m(Ag) = E(Cu)/E(Ag) ⇒ m(Cu)/1.45 = (63.5/2) / (108/1)
m(Cu) = 1.45 × 31.75 / 108 = 0.426 g.
3. Second Law for Zn: m(Zn)/1.45 = (65.3/2) / 108
1. Time for Ag deposition: Ag⁺ + e⁻ → Ag (n=1). M=108. m=1.45g. I=1.5A.
m = (M × I × t) / (n × F) ⇒ 1.45 = (108 × 1.5 × t) / (1 × 96500)
t = (1.45 × 96500) / 162 = 139925 / 162 = 863.7 s (14.4 min).
2. Second Law for Cu: m(Cu)/m(Ag) = E(Cu)/E(Ag) ⇒ m(Cu)/1.45 = (63.5/2) / (108/1)
m(Cu) = 1.45 × 31.75 / 108 = 0.426 g.
3. Second Law for Zn: m(Zn)/1.45 = (65.3/2) / 108
m(Zn) = 1.45 × 32.65 / 108 = 0.438 g
3 MarksQ22. Consider the button cell used in hearing
aids: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq).
Determine ΔG° and E° for the cell. (Given E°(Ag⁺/Ag) = +0.80V,
E°(Zn²⁺/Zn) = -0.76 V).
✓ Solution
Anode (oxidation) = Zn; Cathode (reduction) = Ag₂O/Ag.
E°cell = E°cathode - E°anode = 0.80 - (-0.76) = 1.56 V.
Number of electrons transferred, n = 2.
ΔG° = -nFE°cell
ΔG° = -2 × 96487 × 1.56 = -301,039 J mol⁻¹
Anode (oxidation) = Zn; Cathode (reduction) = Ag₂O/Ag.
E°cell = E°cathode - E°anode = 0.80 - (-0.76) = 1.56 V.
Number of electrons transferred, n = 2.
ΔG° = -nFE°cell
ΔG° = -2 × 96487 × 1.56 = -301,039 J mol⁻¹
ΔG° = -301.04 kJ mol⁻¹
✍ Score Guide — 22 Questions
All questions curated from NCERT Exercise & Exemplar with clear explanations matching CBSE marking scheme constraints.
All questions curated from NCERT Exercise & Exemplar with clear explanations matching CBSE marking scheme constraints.
High-Yield Facts & Formulas: Electrochemistry
Galvanic Cell
Converts chemical energy into electrical energy using spontaneous redox reactions.
Converts chemical energy into electrical energy using spontaneous redox reactions.
Electrolytic Cell
Converts electrical energy into chemical energy for non-spontaneous reactions.
Converts electrical energy into chemical energy for non-spontaneous reactions.
Standard Electrode Potential (E0)
Measured at 298 K, 1 atm pressure, and 1 M concentration.
Measured at 298 K, 1 atm pressure, and 1 M concentration.
Cell Potential Formula
E0cell = E0cathode - E0anode
E0cell = E0cathode - E0anode
Nernst Equation (Electrode)
E = E0 - (RT/nF) ln(1 / [Mn+])
E = E0 - (RT/nF) ln(1 / [Mn+])
Nernst Equation at 298 K
E = E0 - (0.0591 / n) log(1 / [Mn+])
E = E0 - (0.0591 / n) log(1 / [Mn+])
Equilibrium Constant Relationship
log Kc = (n E0cell) / 0.0591
log Kc = (n E0cell) / 0.0591
Gibbs Free Energy
ΔG0 = -n F E0cell
ΔG0 = -n F E0cell
Resistance (R)
R = ρ (l/A). Unit: Ohm (Ω).
R = ρ (l/A). Unit: Ohm (Ω).
Conductance (G)
G = 1 / R. Unit: Siemens (S) or Ohm-1.
G = 1 / R. Unit: Siemens (S) or Ohm-1.
Conductivity (κ)
κ = (1/R) × (l/A). Inverse of resistivity.
κ = (1/R) × (l/A). Inverse of resistivity.
Cell Constant (G*)
G* = l/A. Often calculated as G* = R × κ.
G* = l/A. Often calculated as G* = R × κ.
Molar Conductivity (Λm)
Λm = (κ × 1000) / M. Unit: S cm2 mol-1.
Λm = (κ × 1000) / M. Unit: S cm2 mol-1.
Kohlrausch's Law
Λm0 = ν+λ+0 + ν-λ-0
Λm0 = ν+λ+0 + ν-λ-0
Degree of Dissociation (α)
α = Λm / Λm0
α = Λm / Λm0
Dissociation Constant (Ka)
Ka = (C α2) / (1 - α)
Ka = (C α2) / (1 - α)
Faraday's First Law
W = Z Q = Z I t. (Z is electrochemical equivalent).
W = Z Q = Z I t. (Z is electrochemical equivalent).
Faraday's Constant (F)
Charge on 1 mole of electrons ≈ 96500 C.
Charge on 1 mole of electrons ≈ 96500 C.
Faraday's Second Law
W1 / W2 = E1 / E2 (for same Q).
W1 / W2 = E1 / E2 (for same Q).
Dry Cell (Leclanche Cell)
Anode: Zn; Cathode: Graphite rod surrounded by MnO2 and carbon.
Anode: Zn; Cathode: Graphite rod surrounded by MnO2 and carbon.
Mercury Cell
Constant potential (1.35 V) throughout life; used in hearing aids.
Constant potential (1.35 V) throughout life; used in hearing aids.
Lead Storage Battery
Rechargeable. Anode: Lead; Cathode: Lead dioxide; Electrolyte: 38% H2SO4.
Rechargeable. Anode: Lead; Cathode: Lead dioxide; Electrolyte: 38% H2SO4.
H2-O2 Fuel Cell
Efficiency ≈ 70% compared to thermal plants (40%).
Efficiency ≈ 70% compared to thermal plants (40%).
Corrosion: Anodic Reaction
Fe(s) → Fe2+ + 2e-
Fe(s) → Fe2+ + 2e-
Corrosion: Cathodic Reaction
O2(g) + 4H+(aq) + 4e- → 2H2O(l)
O2(g) + 4H+(aq) + 4e- → 2H2O(l)
Prevention of Corrosion
Barrier protection, sacrificial protection (Galvanization), or using anti-corrosive solutions.
Barrier protection, sacrificial protection (Galvanization), or using anti-corrosive solutions.
Strong Electrolytes
Λm increases slightly with dilution (HCl, NaCl, KNO3).
Λm increases slightly with dilution (HCl, NaCl, KNO3).
Weak Electrolytes
Λm increases steeply with dilution (CH3COOH, NH4OH).
Λm increases steeply with dilution (CH3COOH, NH4OH).
Standard Hydrogen Electrode (SHE)
Reference electrode with E0 = 0.00 V.
Reference electrode with E0 = 0.00 V.
EMF (Electromotive Force)
Potential difference when no current is drawn from the cell.
Potential difference when no current is drawn from the cell.
Electrochemical Series
Arrangement of electrodes in increasing order of standard reduction potentials.
Arrangement of electrodes in increasing order of standard reduction potentials.
Strongest Reducing Agent
Lithium (lowest reduction potential, most negative E0).
Lithium (lowest reduction potential, most negative E0).
Strongest Oxidizing Agent
Fluorine (highest reduction potential, most positive E0).
Fluorine (highest reduction potential, most positive E0).
Overpotential
Extra voltage required for reactions like O2 evolution during electrolysis of water.
Extra voltage required for reactions like O2 evolution during electrolysis of water.
Rust Composition
Hydrated ferric oxide, Fe2O3.xH2O.
Hydrated ferric oxide, Fe2O3.xH2O.
Electrochemical Equivalent (Z)
Mass deposited by 1 Coulomb of charge. Z = Equivalent Weight / 96500.
Mass deposited by 1 Coulomb of charge. Z = Equivalent Weight / 96500.
Current Efficiency
(Actual mass deposited / Theoretical mass deposited) × 100.
(Actual mass deposited / Theoretical mass deposited) × 100.
Salt Bridge Composition
Inert electrolyte like KCl or KNO3 in agar-agar gel.
Inert electrolyte like KCl or KNO3 in agar-agar gel.
Primary cell vs Secondary cell
Primary cell reactions are irreversible; Secondary cell reactions are reversible.
Primary cell reactions are irreversible; Secondary cell reactions are reversible.
Conductivity Cell
Device used to measure electrolytic resistance/conductance.
Device used to measure electrolytic resistance/conductance.
Wheatstone Bridge
Used for measuring unknown resistance of electrolytic solution.
Used for measuring unknown resistance of electrolytic solution.
Conductance of Ions
Smaller ions like H+ and OH- have exceptionally high molar conductivities.
Smaller ions like H+ and OH- have exceptionally high molar conductivities.
Effect of Purity
Corrosion occurs more rapidly in the presence of impurities in the metal.
Corrosion occurs more rapidly in the presence of impurities in the metal.
Nicad Cell Potential
Produces a constant potential of about 1.4 V.
Produces a constant potential of about 1.4 V.
Fuel Cell byproduct
Pure water, which is often used for drinking purposes in space programs.
Pure water, which is often used for drinking purposes in space programs.
Relationship: R and κ
κ = (1/R) (l/A).
κ = (1/R) (l/A).
Hydrogen Evolution
Standard reduction potential of Hydrogen is 0.00 V.
Standard reduction potential of Hydrogen is 0.00 V.
Debye-Huckel Equation
Λm = Λm0 - A √C (for strong electrolytes).
Λm = Λm0 - A √C (for strong electrolytes).
Electrolysis of Water
Requires small amount of H2SO4 to become conducting.
Requires small amount of H2SO4 to become conducting.
Batteries in Series
Potentials are added up.
Potentials are added up.
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