NEET UG Chemistry · Chapter 9

Electrochemistry

Galvanic cells, electrode potential, Nernst equation, conductance, Kohlrausch’s law and electrolysis.

NCERT Class 12NEET: 1–2 Qs per paper 12 min read

1. Chapter at a Glance

Electrochemistry studies the link between chemical reactions and electricity. In a galvanic cell a spontaneous redox reaction produces electricity; in an electrolytic cell electricity drives a non-spontaneous reaction. This chapter covers electrode potentials, the Nernst equation, the relation between cell EMF, Gibbs energy and K, conductance of solutions and Kohlrausch's law, Faraday's laws of electrolysis, and batteries. NEET asks one or two questions every year.

What NEET asks most
  • E°cell from standard reduction potentials; which metal displaces which.
  • Nernst equation for cell EMF at non-standard concentrations.
  • ΔG° = −nFE° and log K = nE°/0.0591.
  • Conductivity, molar conductivity and Kohlrausch's law.
  • Faraday's laws; products of electrolysis; batteries.

2. Galvanic Cells and Electrode Potential

  • In the Daniell cell, Zn + Cu²⁺ → Zn²⁺ + Cu. Zinc is the anode (oxidation, negative terminal); copper is the cathode (reduction, positive terminal). Electrons flow from anode to cathode through the external wire.
  • The salt bridge (KCl or KNO₃ in agar) completes the circuit and keeps both solutions electrically neutral.
  • Cell notation: anode | anode solution || cathode solution | cathode, e.g. Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
  • The standard hydrogen electrode (SHE), Pt | H₂(1 bar) | H⁺(1 M), is assigned E° = 0.
  • E°cell = E°cathode − E°anode (both as reduction potentials). A positive E°cell means the reaction is spontaneous.
  • In the electrochemical series, a more negative E° (Li, K, Zn) means a stronger reducing agent; a more positive E° (F₂, Au, Ag) means a stronger oxidising agent.
salt bridge e⁻ flow → Zn anode (−)Cu cathode (+) ZnSO₄CuSO₄
Daniell cell. Zinc is oxidised at the anode and releases electrons, which travel through the wire to the copper cathode, where Cu²⁺ is reduced. E°cell = 0.34 − (−0.76) = 1.10 V.

3. Nernst Equation and Thermodynamics

Ecell = E°cell − (0.0591/n) log Q    (at 298 K)
ΔG° = −nFE°cell    log K = nE°cell/0.0591    F = 96 487 C mol⁻¹
  • For Mn+ + ne⁻ → M: E = E° − (0.0591/n) log(1/[Mn+]).
  • At equilibrium Ecell = 0 and Q = K.
  • A concentration cell has identical electrodes in solutions of different concentration; E° = 0, and E = (0.0591/n) log(Chigher/Clower).
  • For a hydrogen electrode in contact with H⁺: E = −0.0591 pH.

4. Conductance of Electrolytic Solutions

QuantityDefinitionUnitNEET Note
Resistance RR = ρl/AΩ—
Conductance G1/RS (siemens)—
Cell constant G*l/Acm⁻¹Found using KCl of known κ
Conductivity κG × G* = 1/ρS cm⁻¹Decreases on dilution
Molar conductivity Λm1000κ/CS cm² mol⁻¹Increases on dilution
  • Strong electrolytes (KCl, HCl): Λm rises slowly on dilution and follows Λm = Λ°m − A√C. Λ°m is found by extrapolating to zero concentration.
  • Weak electrolytes (CH₃COOH): Λm rises sharply at high dilution because ionisation increases. Λ°m cannot be found by extrapolation.
  • Kohlrausch's law: at infinite dilution, each ion contributes independently: Λ°m = ν₊λ°₊ + ν₋λ°₋. It gives Λ°m of weak electrolytes indirectly and the degree of dissociation α = Λm/Λ°m.

5. Electrolysis and Faraday's Laws

  • First law: the mass deposited is proportional to the charge passed: w = ZIt, where Z = M/(nF) is the electrochemical equivalent.
  • Second law: when the same charge passes through different electrolytes, the masses deposited are in the ratio of their equivalent masses (M/n).
  • 1 faraday (96 487 C) is the charge of 1 mole of electrons.
  • Products depend on the ease of discharge. For aqueous NaCl: H₂ at the cathode (water is reduced more easily than Na⁺) and Cl₂ at the anode (because of overpotential for O₂). For molten NaCl: Na at the cathode and Cl₂ at the anode. For aqueous CuSO₄ with inert electrodes: Cu at the cathode and O₂ at the anode.

6. Batteries and Fuel Cells

CellTypeKey reaction / factNEET Note
Dry (Leclanché) cellPrimaryZn anode, graphite cathode with MnO₂; about 1.5 VNot rechargeable
Mercury cellPrimaryZn–Hg amalgam, HgO; constant 1.35 VHearing aids, watches
Lead storage batterySecondaryPb anode, PbO₂ cathode, 38% H₂SO₄; forms PbSO₄ on dischargeRechargeable; car batteries
Nickel–cadmium cellSecondaryCd anode, NiO(OH) cathodeLonger life than lead cells
H₂–O₂ fuel cellFuel cell2H₂ + O₂ → 2H₂OUsed in Apollo missions; about 70% efficient

7. Formula Master Table

QuantityFormula
Standard cell EMFE°cell = E°cathode − E°anode
Nernst equationE = E° − (0.0591/n) log Q
Gibbs energyΔG° = −nFE°
Equilibrium constantlog K = nE°/0.0591
Hydrogen electrodeE = −0.0591 pH
Conductivityκ = G × (l/A)
Molar conductivityΛm = 1000κ/C
Debye–Hückel–OnsagerΛm = Λ°m − A√C
Kohlrausch's lawΛ°m = ν₊λ°₊ + ν₋λ°₋
Degree of dissociationα = Λm/Λ°m; Ka = Cα²/(1 − α)
Faraday's first laww = (M/nF) × I × t

8. Master Summary Table

ConceptGalvanic cellElectrolytic cell
Energy changeChemical → electricalElectrical → chemical
ReactionSpontaneous (ΔG < 0)Non-spontaneous (ΔG > 0)
AnodeNegative, oxidationPositive, oxidation
CathodePositive, reductionNegative, reduction
NEET NoteOxidation always at the anodeOnly the signs of the electrodes swap

9. Mind Map

Electrochemistry
Cells
  • Daniell
  • SHE, E°
  • Series
Thermodynamics
  • Nernst
  • ΔG° = −nFE°
  • log K
Conductance
  • κ, Λm
  • Dilution
  • Kohlrausch
Electrolysis
  • Faraday's laws
  • Products
  • Batteries

10. Tips and Tricks

Shortcuts
  • "An Ox, Red Cat": oxidation at the anode, reduction at the cathode.
  • E° is an intensive property: do not multiply it when you multiply a half-reaction by a number.
  • A metal with lower E° displaces a metal with higher E° from its salt solution (Zn displaces Cu).
  • Charge (C) = I (A) × t (s); moles of electrons = charge/96 487.
Common NEET traps
  • Multiplying E° by the stoichiometric coefficient.
  • Saying conductivity increases on dilution (molar conductivity does; conductivity falls).
  • Expecting Na at the cathode in electrolysis of aqueous NaCl.
  • Getting the sign of the anode wrong in electrolytic cells.

11. Important One-liners

  • The E° of the standard hydrogen electrode is taken as zero at all temperatures.
  • Lithium has the most negative standard reduction potential, making it the strongest reducing agent in water.
  • Fluorine has the most positive standard reduction potential.
  • A salt bridge maintains electrical neutrality in the two half-cells.
  • Molar conductivity increases on dilution for all electrolytes.
  • The lead storage battery is a secondary (rechargeable) cell.
  • 1 faraday = 96 487 C = charge of 1 mol of electrons.

12. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take F = 96 487 C mol⁻¹ and 2.303RT/F = 0.0591 V at 298 K. Click to reveal each answer and working.

Average level

N1Average

E°(Mg²⁺/Mg) = −2.37 V and E°(Ag⁺/Ag) = +0.80 V. Write the cell for the reaction between Mg and Ag⁺ and find its E°cell.

Show answer & solution

Concept: the more negative electrode is the anode.

  1. Anode: Mg → Mg²⁺ + 2e⁻; cathode: 2Ag⁺ + 2e⁻ → 2Ag.
  2. Cell: Mg(s) | Mg²⁺ || Ag⁺ | Ag(s).
  3. E°cell = 0.80 − (−2.37) = 3.17 V (E° is not doubled for 2Ag⁺).

Answer: 3.17 V

N2Average

For the Mg–Ag cell above (n = 2), find ΔG°.

Show answer & solution

Concept: ΔG° = −nFE°.

  1. ΔG° = −2 × 96 487 × 3.17.
  2. ΔG° ≈ −6.12 × 10⁵ J ≈ −612 kJ mol⁻¹ (highly spontaneous).

Answer: about −612 kJ mol⁻¹

N3Average

The conductivity of 0.05 M KCl solution is 6.67 × 10⁻³ S cm⁻¹. Find its molar conductivity.

Show answer & solution

Concept: Λm = 1000κ/C (κ in S cm⁻¹, C in mol L⁻¹).

  1. Λm = 1000 × 6.67 × 10⁻³/0.05.
  2. Λm = 133.4 S cm² mol⁻¹.

Answer: 133.4 S cm² mol⁻¹

N4Average

A current of 2 A is passed through CuSO₄ solution for 965 s. Find the mass of copper deposited (Cu = 63.5).

Show answer & solution

Concept: Cu²⁺ + 2e⁻ → Cu; moles of e⁻ = Q/F.

  1. Q = 2 × 965 = 1930 C; moles of e⁻ = 1930/96 487 ≈ 0.02.
  2. Moles of Cu = 0.02/2 = 0.01.
  3. Mass = 0.01 × 63.5 ≈ 0.635 g.

Answer: about 0.635 g

N5Average

A conductivity cell filled with 0.1 M KCl (κ = 0.0129 S cm⁻¹) has a resistance of 100 Ω. Find the cell constant.

Show answer & solution

Concept: G* = κ × R.

  1. G* = 0.0129 × 100.
  2. G* = 1.29 cm⁻¹.

Answer: 1.29 cm⁻¹

Advanced level

N6Advanced

Find the EMF of the cell Ni(s) | Ni²⁺(0.1 M) || Ag⁺(0.01 M) | Ag(s) at 298 K. E°(Ni²⁺/Ni) = −0.25 V, E°(Ag⁺/Ag) = +0.80 V.

Show answer & solution

Concept: Ni + 2Ag⁺ → Ni²⁺ + 2Ag; n = 2; Q = [Ni²⁺]/[Ag⁺]².

  1. E°cell = 0.80 − (−0.25) = 1.05 V.
  2. Q = 0.1/(0.01)² = 1000; log Q = 3.
  3. E = 1.05 − (0.0591/2) × 3 = 1.05 − 0.089 ≈ 0.961 V.

Answer: about 0.96 V

N7Advanced

Λ°m values (S cm² mol⁻¹): HCl 426.2, NaCl 126.4, CH₃COONa 91.0. The molar conductivity of 0.01 M acetic acid is 16.3 S cm² mol⁻¹. Find Λ°m(CH₃COOH), its degree of dissociation and Ka.

Show answer & solution

Concept: Kohlrausch: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl).

  1. Λ° = 426.2 + 91.0 − 126.4 = 390.8 S cm² mol⁻¹.
  2. α = 16.3/390.8 ≈ 0.0417.
  3. Ka = Cα²/(1 − α) = 0.01 × (0.0417)²/0.958 ≈ 1.8 × 10⁻⁵.

Answer: 390.8 S cm² mol⁻¹; α ≈ 0.042; Ka ≈ 1.8 × 10⁻⁵

N8Advanced

A two-electron cell reaction has E°cell = 0.59 V at 298 K. Find its equilibrium constant.

Show answer & solution

Concept: log K = nE°/0.0591.

  1. log K = 2 × 0.59/0.0591 ≈ 19.97.
  2. K ≈ 10²⁰ (about 9.3 × 10¹⁹) — the reaction goes practically to completion.

Answer: about 1 × 10²⁰

N9Advanced

A current of 3 A is passed through acidified water for 32 min 10 s. Find the mass of hydrogen produced and its volume at STP (22.7 L mol⁻¹).

Show answer & solution

Concept: 2H⁺ + 2e⁻ → H₂.

  1. t = 1930 s; Q = 3 × 1930 = 5790 C; moles of e⁻ = 5790/96 487 ≈ 0.06.
  2. Moles of H₂ = 0.03 ⇒ mass = 0.06 g.
  3. Volume = 0.03 × 22.7 ≈ 0.68 L (O₂ at the anode would be half this, 0.015 mol).

Answer: 0.06 g; about 0.68 L

N10Advanced

A hydrogen electrode dipped in a solution gives a potential of −0.295 V relative to the SHE at 298 K. Find the pH of the solution.

Show answer & solution

Concept: for 2H⁺ + 2e⁻ → H₂ (1 bar), E = −0.0591 pH.

  1. −0.295 = −0.0591 × pH.
  2. pH = 0.295/0.0591 ≈ 4.99 ≈ 5.

Answer: pH about 5

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

In a galvanic cell, oxidation takes place at the:

  1. Cathode, which is positive
  2. Anode, which is negative
  3. Anode, which is positive
  4. Salt bridge
Show solution

Oxidation always happens at the anode. In a galvanic cell the anode releases electrons, so it is the negative terminal. The anode is positive only in an electrolytic cell.

Answer: (B)

Q2Easy

On dilution of a solution of a strong electrolyte:

  1. Both κ and Λm increase
  2. κ decreases and Λm increases
  3. κ increases and Λm decreases
  4. Both decrease
Show solution

On dilution, there are fewer ions per cm³, so κ falls. But Λm (per mole) rises, because the ions interact less with each other.

Answer: (B)

Q3Medium

Given E°: Zn²⁺/Zn = −0.76 V, Fe²⁺/Fe = −0.44 V, Cu²⁺/Cu = +0.34 V. Which statement is correct?

  1. Cu can reduce Zn²⁺
  2. Zn can reduce Fe²⁺ and Cu²⁺
  3. Fe can reduce Zn²⁺
  4. Cu is the strongest reducing agent
Show solution

Zn has the most negative E°, so it is the strongest reducing agent and can reduce both Fe²⁺ and Cu²⁺. A metal can reduce only the ions of metals that lie above it (more positive E°). So Cu cannot reduce Zn²⁺, and Fe cannot reduce Zn²⁺ either.

Answer: (B)

Q4Medium

During electrolysis of aqueous NaCl with inert electrodes, the product at the cathode is:

  1. Na
  2. H₂
  3. Cl₂
  4. O₂
Show solution

Water is reduced (E° −0.83 V) far more easily than Na⁺ (−2.71 V), so H₂ forms at the cathode and NaOH is left in solution. Na forms only when molten NaCl is electrolysed.

Answer: (B)

Q5Medium

The unit of molar conductivity is:

  1. S cm⁻¹
  2. S cm² mol⁻¹
  3. Ω cm
  4. S mol⁻¹
Show solution

Λm = κ/C gives (S cm⁻¹)/(mol cm⁻³) = S cm² mol⁻¹. The unit S cm⁻¹ belongs to conductivity κ, and Ω cm belongs to resistivity.

Answer: (B)

Q6Medium

Which of the following is a secondary (rechargeable) cell?

  1. Leclanché cell
  2. Mercury cell
  3. Lead storage battery
  4. Dry cell
Show solution

The lead storage battery can be recharged by passing current through it in the reverse direction. Leclanché (dry) cells and mercury cells are primary cells and cannot be recharged.

Answer: (C)

Q7Hard

The same charge is passed through molten AlCl₃ and molten NaCl. If 4.5 g of Al is deposited, the mass of Na deposited is (Al = 27, Na = 23):

  1. 3.83 g
  2. 11.5 g
  3. 23 g
  4. 6.9 g
Show solution

Faraday's second law says the masses are proportional to the equivalent masses. Moles of electrons = 3 × (4.5/27) = 0.5, so the moles of Na = 0.5 and its mass = 11.5 g. (A) forgets that Al is trivalent.

Answer: (B)

Q8Hard

For a cell reaction with E°cell > 0, which set is correct?

  1. ΔG° > 0, K < 1
  2. ΔG° < 0, K > 1
  3. ΔG° = 0, K = 1
  4. ΔG° < 0, K < 1
Show solution

ΔG° = −nFE° is negative when E° is positive. Then log K = nE°/0.0591 > 0, so K > 1. The three quantities always agree with each other in this way.

Answer: (B)

14. Quick Revision Summary

Remember before the exam
  • Anode = oxidation (negative in galvanic cell); E°cell = E°cathode − E°anode.
  • Nernst: E = E° − (0.0591/n) log Q; ΔG° = −nFE°; log K = nE°/0.0591.
  • κ falls on dilution; Λm = 1000κ/C rises; Kohlrausch for weak electrolytes.
  • w = MIt/(nF); 1 F = 96 487 C.
  • Aqueous NaCl gives H₂ and Cl₂; lead battery is secondary; H₂–O₂ fuel cell.
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