Is it correct to say: Writing Nernst equation as E = E° + (0.0591/n) log Q (wrong sign).?

It is MINUS: E = E° − (0.0591/n) log Q. Physical meaning — increase in [products] lowers driving force.

Is it correct to say: Confusing electrode potentials — using oxidation potentials in E°(cell) = E°(cathode) − E°(anode).?

Standard tables give REDUCTION potentials. Both terms in the formula are reduction potentials (SRP).

Is it correct to say: Mixing up κ (specific conductivity, S/cm) and Λm (molar conductivity, S·cm²/mol).?

Λm = 1000 κ / c. κ decreases on dilution, Λm INCREASES on dilution.

Is it correct to say: Applying Kohlrausch's law directly on weak electrolyte data without the salt-cycle.?

For weak electrolytes use combinations of strong-electrolyte Λ°m: Λ°(HA) = Λ°(NaA) + Λ°(HCl) − Λ°(NaCl).

Is it correct to say: Using n = 1 for every reaction in Nernst / ΔG° = −nFE°.?

n is the total electrons transferred in the balanced reaction. Zn + Cu²⁺ → n = 2, Al³⁺ + Fe → carefully balance.

Is it correct to say: In electrolysis numericals, ignoring the valency of the ion.?

Equivalent mass E = molar mass / n where n = charge on the ion. For Cu²⁺ → Cu, E = 63.5/2 = 31.75 g/eq.

Electrochemistry

Galvanic cells, Nernst equation, conductance, Kohlrausch's law, electrolysis and batteries — NCERT Class 12 Chemistry Ch 2

📐 10 formulas✏️ 3 examples🎯 6 practice⚖️ 5-7 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Nernst equation numerical is a near-certain 3-mark or 5-mark question. Learn the 298 K form E = E° − (0.0591/n) log Q.
→Sign of ΔG° = −nFE° — if E°(cell) is positive, ΔG° is negative and the reaction is spontaneous.
→Kohlrausch's law lets you find Λ°m of a weak electrolyte (like CH₃COOH) — very common HOTS.
→Distinguish molar conductivity (Λm, S·cm²/mol) from specific conductivity (κ, S/cm). Unit mix-ups are the biggest marks-lost error.
→In electrolysis: use Faraday's laws, always convert time to seconds and mass to grams.

📐 Formulas(10)

Standard EMF of Cell★ Board fav

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}

Symbol	Meaning
$E_{ce l l}^{\circ}$	Standard cell potential (V)
$E_{c a t h o d e}^{\circ}$	SRP of the cathode (reduction site)
$E_{an o d e}^{\circ}$	SRP of the anode (oxidation site)

Nernst Equation (general)★ Board fav

E_{ce l l} = E_{ce l l}^{\circ} - \frac{R T}{n F} ln Q

Nernst Equation at 298 K★ Board fav

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.0591}{n} lo g Q

Gibbs Free Energy and EMF★ Board fav

Δ G^{\circ} = - n F E_{ce l l}^{\circ} = - R T ln K_{c}

Symbol	Meaning
$n$	Moles of electrons transferred
$F$	Faraday's constant = 96500 C/mol
$K_{c}$	Equilibrium constant of cell reaction

Conductance & Specific Conductivity

κ = \frac{1}{ρ} = G \times \frac{l}{A}

Symbol	Meaning
$κ$	Specific conductivity (S·cm⁻¹)
$G$	Conductance = 1/R (siemens, S)
$l / A$	Cell constant (cm⁻¹)

Molar Conductivity★ Board fav

Λ_{m} = \frac{κ \times 1000}{c}

Symbol	Meaning
$Λ_{m}$	Molar conductivity (S·cm²·mol⁻¹)
$c$	Molar concentration (mol/L)

Kohlrausch's Law of Independent Migration★ Board fav

Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}

Degree of Dissociation (weak electrolyte)

α = \frac{Λ _{m}^{c}}{Λ _{m}^{\circ}}, K_{a} = \frac{c α ^{2}}{1 - α}

Faraday's First Law of Electrolysis★ Board fav

w = Z \cdot Q = Z \cdot I \cdot t = \frac{E \cdot I \cdot t}{F}

Symbol	Meaning
$w$	Mass deposited (g)
$E$	Equivalent mass = molar mass / n
$I$	Current (A)
$t$	Time (s)
$F$	Faraday = 96500 C/mol

Faraday's Second Law

\frac{w _{1}}{w _{2}} = \frac{E _{1}}{E _{2}}

✏️ Solved Examples

1Solved Exampleeasy3 steps

Calculate the standard EMF of the Daniell cell: Zn | Zn²⁺ || Cu²⁺ | Cu. Given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

Identify cathode (higher SRP = Cu) and anode (lower SRP = Zn)

Cathode: Cu^{2 +} / Cu, Anode: Zn^{2 +} / Zn

2Solved Exampleboard4 steps

Calculate the EMF of the cell Zn | Zn²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu at 298 K. E°(cell) = 1.10 V.

Overall reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2

Q = \frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]} = \frac{0.001}{0.1} = 1 0^{- 2}

3Solved ExampleHOTS4 steps

The molar conductivity of 0.025 mol/L acetic acid is 45.0 S·cm²/mol. Calculate its degree of dissociation and dissociation constant. Given λ°(H⁺) = 349.6 and λ°(CH₃COO⁻) = 40.9 S·cm²/mol.

Find Λ°m of acetic acid using Kohlrausch's law

Λ_{m}^{\circ} (CH_{3} COOH) = λ^{\circ} (H^{+}) + λ^{\circ} (CH_{3} COO^{-}) = 349.6 + 40.9 = 390.5

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing Nernst equation as E = E° + (0.0591/n) log Q (wrong sign).
✓It is MINUS: E = E° − (0.0591/n) log Q. Physical meaning — increase in [products] lowers driving force.
2
Confusing electrode potentials — using oxidation potentials in E°(cell) = E°(cathode) − E°(anode).
✓Standard tables give REDUCTION potentials. Both terms in the formula are reduction potentials (SRP).
3
Mixing up κ (specific conductivity, S/cm) and Λm (molar conductivity, S·cm²/mol).
✓Λm = 1000 κ / c. κ decreases on dilution, Λm INCREASES on dilution.
4
Applying Kohlrausch's law directly on weak electrolyte data without the salt-cycle.
✓For weak electrolytes use combinations of strong-electrolyte Λ°m: Λ°(HA) = Λ°(NaA) + Λ°(HCl) − Λ°(NaCl).
5
Using n = 1 for every reaction in Nernst / ΔG° = −nFE°.
✓n is the total electrons transferred in the balanced reaction. Zn + Cu²⁺ → n = 2, Al³⁺ + Fe → carefully balance.
6
In electrolysis numericals, ignoring the valency of the ion.
✓Equivalent mass E = molar mass / n where n = charge on the ion. For Cu²⁺ → Cu, E = 63.5/2 = 31.75 g/eq.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
The EMF of a cell Zn|Zn²⁺(1 M)||Ag⁺(1 M)|Ag at 298 K is 1.56 V. If E°(Zn²⁺/Zn) = −0.76 V, find E°(Ag⁺/Ag).
Q2
Calculate ΔG° for a cell with E°(cell) = 1.10 V and n = 2. (F = 96500 C/mol)
$Δ G^{\circ} = - n F E^{\circ}$
Q3
The resistance of a conductivity cell filled with 0.1 M KCl solution is 100 Ω. If the cell constant is 1.29 cm⁻¹, calculate the specific conductivity.
Q4
How many grams of copper will be deposited when a current of 2 A is passed for 30 minutes through CuSO₄ solution? (Cu = 63.5)
Q5
Λ°m(HCl) = 425.9, Λ°m(NaCl) = 126.4, Λ°m(CH₃COONa) = 91.0 S·cm²/mol. Find Λ°m(CH₃COOH).
Q6
For a cell reaction with E°(cell) = +0.59 V at 298 K, n = 2. Find the equilibrium constant K_c.

📝 Notes

Electrochemistry

Study of the interconversion of chemical and electrical energy, and of the conduction of electricity by ionic solutions.

Two systems in one chapter

Galvanic (voltaic) cell: spontaneous redox reaction produces electricity. ΔG < 0, E(cell) > 0.
Electrolytic cell: external electricity forces a non-spontaneous reaction. ΔG > 0.

Cell notation and sign convention

Standard convention: anode on the left, cathode on the right, with double bar (||) representing the salt bridge:

Zn | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu

The anode is where oxidation happens (loses electrons); the cathode is where reduction happens (gains electrons). Electrons flow through external wire from anode to cathode; current flows opposite.

Nernst equation — what it tells you

At non-standard concentrations the cell potential differs from E°. The Nernst equation:

Predicts direction of change if concentrations shift.
Lets you compute E at any composition.
At equilibrium, E = 0 and Q = K, giving log K = nE°/0.0591.

Conductance behaviour

For an electrolyte solution:

Strong electrolytes (KCl, NaOH): Λ_m rises slowly with dilution and Λ°m is obtained by extrapolation (Debye-Hückel-Onsager plot: Λm vs √c is linear).
Weak electrolytes (CH₃COOH, NH₄OH): Λ_m rises sharply on dilution because α increases; Λ°m obtained via Kohlrausch's law.

Batteries and corrosion

Primary cell (dry cell, mercury cell): non-rechargeable, use up reagents.
Secondary cell (lead storage, Ni–Cd, Li-ion): rechargeable, reactions reversible.
Fuel cell (H₂–O₂): continuous supply of reactants, pollution-free, high efficiency.
Corrosion of iron is an electrochemical process: Fe → Fe²⁺ at anodic patches, O₂ reduced at cathodic patches; rust = hydrated Fe₂O₃.