Write the equations relating E° cell with Δ r G° and K c . For the reaction 2Fe 3+ (aq) + 2I − (aq) → 2Fe 2+ (aq) + I 2 (s) , E° cell = 0.236 V at 298 K. Calculate Δ r G° and K c .

Relations: Δ r G° = −nFE° cell and Δ r G° = −2.303 RT log K c (or E° cell = (0.059/n) log K c at 298 K). Here n = 2, F = 96487 C mol −1 : Δ r G° = −2 × 96487 × 0.236 = −45,542 J mol −1 = −45.54 kJ mol −1 . log K c = 45542 / (2.303 × 8.314 × 298) = 7.983 ⇒ K c = 9.62 × 10 7 .

Resistance of a conductivity cell filled with 0.1 mol L −1 KCl solution is 100 Ω. If the resistance of the same cell with 0.02 mol L −1 KCl is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L −1 KCl. (κ of 0.1 M KCl = 1.29 S m −1 )

1. Cell constant G* = κ × R = 1.29 S m −1 × 100 Ω = 129 m −1 (= 1.29 cm −1 ) . 2. Conductivity of 0.02 M KCl: κ = G* / R = 129 / 520 = 0.248 S m −1 (= 0.248 × 10 −2 S cm −1 ) . 3. Molar conductivity: c = 1000 × 0.02 = 20 mol m −3 Λ m = κ / c = 0.248 / 20 = 124 × 10 −4 S m 2 mol −1 = 124 S cm 2 mol −1 .

NCERT Plus Two (Class 12) Chemistry • Chapter 2

Electrochemistry

Q: What is a salt bridge? Mention two important functions of a salt bridge, and depict the Daniell cell using IUPAC convention.

A salt bridge is an inverted U-tube filled with a concentrated solution of an inert electrolyte (such as KCl, KNO 3 , or NH 4 NO 3 ) mixed with agar-agar jelly. Two Functions: (1) It completes the internal electrical circuit by allowing migration of ions without mixing the two half-cell solutions. (2) It maintains electrical neutrality in both half-cells. IUPAC Cell Representation: Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s)

Q: Calculate the EMF of the cell in which the following reaction takes place at 298 K: Ni(s) + 2Ag + (0.002 M) → Ni 2+ (0.160 M) + 2Ag(s) [Given E° cell = 1.05 V]

Here n = 2 . By the Nernst equation at 298 K: E cell = E° cell − (0.059 / 2) log([Ni 2+ ] / [Ag + ] 2 ) E cell = 1.05 − 0.0295 log(0.160 / (0.002) 2 ) = 1.05 − 0.0295 log(4 × 10 4 ) = 1.05 − 0.0295 × 4.6021 = 1.05 − 0.1358 = 0.914 V (≈ 0.91 V).

Q: Define conductivity (κ) and molar conductivity (Λ m ). How do both vary with dilution?

Conductivity (κ): Conductance of one unit volume of solution kept between two electrodes of unit cross-sectional area separated by unit distance ( κ = G × l/A , SI unit: S m −1 ). Molar Conductivity (Λ m ): Conductance of the volume V of solution containing 1 mole of electrolyte placed between electrodes at unit distance ( Λ m = κ / c , unit: S cm 2 mol −1 or S m 2 mol −1 ). Variation with Dilution: Conductivity (κ) decreases on dilution (fewer ions per unit volume), whereas molar conductivity (Λ m ) increases on dilution (volume V increases more than κ falls; steeply for weak electrolytes due to higher α).

Q: Why is AC used instead of DC in measuring the resistance of an electrolytic solution? Define cell constant (G*).

Why AC? Direct current (DC) causes electrolysis of the solution over time, altering its ionic concentration and causing polarization at the electrodes. Using alternating current (AC) avoids net chemical change. Cell Constant (G*): Ratio of distance between the two electrodes ( l ) to their cross-sectional area ( A ): G* = l / A = R × κ (unit: m −1 or cm −1 ).

Q: State Kohlrausch's Law of Independent Migration of Ions. Given Λ° m values of NaCl, HCl and CH 3 COONa as 126.4, 425.9 and 91.0 S cm 2 mol −1 respectively, calculate Λ° m for acetic acid (CH 3 COOH).

Statement: At infinite dilution, the limiting molar conductivity (Λ° m ) of an electrolyte equals the sum of the individual contributions of its cation and anion: Λ° m = ν + λ° + + ν − λ° − . Calculation: Λ° m (CH 3 COOH) = Λ° m (HCl) + Λ° m (CH 3 COONa) − Λ° m (NaCl) = 425.9 + 91.0 − 126.4 = 390.5 S cm 2 mol −1 .

Q: State Faraday's First and Second Laws of electrolysis. A solution of CuSO 4 is electrolysed for 10 minutes with a current of 1.5 A. What mass of copper is deposited at the cathode? [Atomic mass of Cu = 63 g mol −1 ]

First Law: Amount of chemical reaction at any electrode is proportional to the quantity of electricity passed ( w = ZIt ). Second Law: Masses of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights. Calculation: t = 10 × 60 = 600 s ; Q = 1.5 × 600 = 900 C . Cu 2+ + 2e − → Cu (2F = 2 × 96487 C deposits 63 g Cu). Mass deposited = (63 × 900) / (2 × 96487) = 0.2938 g .

Galvanic Cells, Electrode Potentials, Nernst Equation, Conductivity & Kohlrausch Law, Faraday Electrolysis, Fuel Cells & Corrosion

📄 Official NCERT Textbook PDF

Topic 2.1 – 2.2: Electrochemical Cells: Daniell Cell, Galvanic vs Electrolytic Cells, Salt Bridge & SHE

Every time you unlock your smartphone, start a scooter, or even feel a nerve impulse flash from your fingertips to your brain, tiny chemical reactions are pushing electrons across wires or membranes! Put a zinc rod into blue copper sulphate solution directly and heat simply escapes into the beaker. Separate the two halves into two beakers connected by a wire and a glowing U-tube called a Salt Bridge, and suddenly that same chemistry lights a bulb at 1.10 V—welcome to the Daniell Cell!

Electrochemistry studies both the production of electricity from spontaneous chemical reactions (Galvanic / Voltaic cells) and the use of electricity to drive non-spontaneous chemical changes (Electrolytic cells).

1. The Daniell Cell & External Opposing Potential (E_ext)

Overall spontaneous redox reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (E°_cell = 1.10 V at 1 M concentration).

Applied Opposing Voltage (E_ext)	Cell Behaviour	Electron & Current Direction	Electrode Changes
E_ext < 1.10 V	Acts as Galvanic Cell	Electrons: Zn → Cu; Current: Cu → Zn	Zn dissolves at anode; Cu deposits at cathode
E_ext = 1.10 V	Equilibrium	No flow of e⁻ or current (I = 0)	No net chemical reaction
E_ext > 1.10 V	Acts as Electrolytic Cell	Electrons: Cu → Zn; Current: Zn → Cu	Zn deposits at Zn(−); Cu dissolves at Cu(+)

2. IUPAC Cell Representation, Salt Bridge & LOAN Rule

Anode (Left): Oxidation takes place (Zn(s) → Zn²⁺(aq) + 2e⁻) and carries a negative charge in a galvanic cell. Memory aid: LOAN (Left, Oxidation, Anode, Negative).
Cathode (Right): Reduction takes place (Cu²⁺(aq) + 2e⁻ → Cu(s)) and carries a positive potential.
IUPAC Notation: Single vertical line | separates phase boundary (metal | ion); double vertical line || denotes the salt bridge:
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) and Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)
Salt Bridge Functions: Filled with inert electrolyte (KCl, KNO₃, or NH₄NO₃ in agar-agar), it (i) completes the internal electrical circuit by allowing ion migration without mixing the two solutions, and (ii) maintains electrical neutrality in both half-cells. (Not needed if both electrodes dip in the same electrolyte!)

E_cell = E_right − E_left = E_cathode − E_anode (both as Standard Reduction Potentials!)
Daniell Cell: E°_cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = (+0.34 V) − (−0.76 V) = +1.10 V

3. Standard Hydrogen Electrode (SHE) & Electrochemical Series

A single half-cell potential cannot be measured alone. By convention, the Standard Hydrogen Electrode (SHE)—Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M) using a platinised platinum foil—is assigned E° = 0.00 V at all temperatures (H⁺(aq) + e⁻ → ½H₂(g)).

Coupling SHE as anode with Cu²⁺/Cu gives +0.34 V (Cu²⁺ reduces more easily than H⁺; Cu does not dissolve in HCl, only oxidised by NO₃⁻ in HNO₃).
Coupling SHE with Zn²⁺/Zn gives −0.76 V (Zn is a stronger reducing agent than H₂ and liberates H₂ from dilute acids).
Electrochemical Series (Table 2.1): Higher E° = stronger oxidising agent (F₂ has highest E° = +2.87 V, strongest oxidant). Lower/more negative E° = stronger reducing agent (Li has lowest E° = −3.05 V, strongest reductant in aqueous solution!).

Watch out for two classic traps: (1) Under IUPAC rules, ALL standard electrode potentials are reduction potentials—if a question gives oxidation potential, flip its sign before using E°_cell = E°_cathode − E°_anode! (2) You CANNOT store CuSO₄ in a zinc pot (NCERT Intext 2.2) because Zn has lower E° (−0.76 V < +0.34 V) and displaces Cu²⁺ spontaneously!

Topic 2.3: Nernst Equation, Equilibrium Constant (Kc) & Gibbs Energy (ΔrG°)

Why does a brand-new battery run your toy car at full speed, yet slow to a crawl hours later? Because cell voltage is NOT frozen at E°_cell! As reactants get consumed and products pile up, the chemical "push" weakens—and the exact mathematical speedometer tracking voltage at ANY concentration is the famous Nernst Equation!

Walther Nernst showed how electrode and cell potentials depend on ion concentration and temperature. For a single electrode reaction Mⁿ⁺(aq) + ne⁻ → M(s) (pure solid concentration = 1):

E_(Mⁿ⁺/M) = E°_(Mⁿ⁺/M) − (RT / nF) ln (1 / [Mⁿ⁺])
At 298 K (using R = 8.314 J K⁻¹ mol⁻¹, F = 96487 C mol⁻¹, ln x = 2.303 log₁₀ x):
E_cell = E°_cell − (0.059 / n) log Q where Q = [Products] / [Reactants] (pure solids/liquids = 1)

For Daniell cell (n = 2): E_cell = E°_cell − (0.059 / 2) log([Zn²⁺] / [Cu²⁺]). Increasing reactant ion [Cu²⁺] or decreasing product ion [Zn²⁺] raises E_cell!

NCERT Example 2.1: Represent the cell and calculate E_cell for:
Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s), given E°_cell = 3.17 V.
• Cell notation: Mg(s) | Mg²⁺(0.130 M) || Ag⁺(0.0001 M) | Ag(s) (here n = 2; note [Ag⁺]²!)
• E_cell = 3.17 − (0.059 / 2) log(0.130 / (0.0001)²) = 3.17 − 0.0295 log(1.3 × 10⁷) = 3.17 − 0.21 = 2.96 V.

1. Equilibrium Constant (K_c) from Nernst Equation

As a galvanic cell runs, [reactants] fall and [products] rise until the voltmeter reads E_cell = 0 and Q = K_c:

0 = E°_cell − (2.303 RT / nF) log K_c ⇒ E°_cell = (0.059 / n) log K_c (at 298 K)

NCERT Example 2.2: For Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°_cell = 0.46 V.
log K_c = (0.46 × 2) / 0.059 = 15.6 ⇒ K_c = 3.92 × 10¹⁵. (For Daniell cell at 1.1 V, K_c = 2 × 10³⁷!)

2. Gibbs Energy & Maximum Electrical Work (Δ_rG°)

Maximum reversible electrical work done by a galvanic cell equals the decrease in Gibbs energy:

Δ_rG = −nFE_cell | Δ_rG° = −nFE°_cell = −2.303 RT log K_c

NCERT Example 2.3: For Daniell cell (E°_cell = 1.1 V, n = 2, F = 96487 C mol⁻¹):
Δ_rG° = −2 × 1.1 × 96487 = −212,271 J mol⁻¹ = −212.27 kJ mol⁻¹.

Intensive vs Extensive Trap! E_cell is an intensive property (independent of equation coefficients), whereas Δ_rG is extensive and depends on n! If you multiply the Daniell equation by 2 (2Zn + 2Cu²⁺ → 2Zn²⁺ + 2Cu), E° stays 1.10 V, while Δ_rG° doubles to −4FE°! Also, NEVER forget the stoichiometric exponent in Q (e.g., [Ag⁺]²) and for a hydrogen electrode at pH 10 (Intext 2.4): E = −0.059 × pH = −0.59 V!

Topic 2.4 – 2.4.1: Conductance of Electrolytic Solutions: Resistivity, Conductivity (κ), Cell Constant & Molar Conductivity (Λm)

In copper wires, tiny electrons race through a stationary metal lattice. Inside salt water, there are NO free electrons—bulky hydrated ions like Na⁺ and Cl⁻ physically swim through viscous water! Try measuring that solution's resistance with a regular DC battery and disaster strikes: electrolysis immediately bubbles gases and alters the liquid! That is why chemists measure ionic resistance using an AC Wheatstone Bridge and a Conductivity Cell.

1. Key Definitions & Units

Quantity	Formula	SI Unit	Common Lab Unit & Conversion
Resistance (R)	R = ρ (l / A)	ohm (Ω)	Ω = (kg m²)/(s³ A²)
Resistivity (ρ)	ρ = R (A / l)	Ω m	Ω cm (1 Ω m = 100 Ω cm)
Conductance (G)	G = 1 / R = κ (A / l)	siemens (S) = Ω⁻¹	mho or Ω⁻¹
Conductivity (κ)	κ = 1 / ρ = G × (l / A) = G* / R	S m⁻¹	S cm⁻¹ (1 S cm⁻¹ = 100 S m⁻¹)
*Cell Constant (G)**	G* = l / A = R × κ	m⁻¹	cm⁻¹

2. Electronic (Metallic) vs Electrolytic (Ionic) Conductance

Metallic Conductance: Flow of electrons; depends on metal structure, valence electrons, and temperature (decreases as temperature rises due to kernel vibrations). Chemical composition stays unchanged.
Electrolytic Conductance: Movement of ions; depends on electrolyte nature, ion size & solvation, solvent viscosity, concentration, and temperature (increases as temperature rises!). Pure water has tiny conductivity (3.5 × 10⁻⁵ S m⁻¹).
Special Note: Superconductors have zero resistivity (ceramic oxides up to 150 K). Conducting organic polymers (polyacetylene, polyaniline) won MacDiarmid, Heeger & Shirakawa the 2000 Nobel Prize!

3. Measuring κ (AC Wheatstone Bridge) & Molar Conductivity (Λ_m)

Two problems arise when measuring ionic resistance: (1) DC changes solution composition → solved by using an AC oscillator (550–5000 Hz); (2) a liquid cannot clip into wire terminals → solved using a Conductivity Cell with platinised Pt electrodes. Because measuring physical l and A directly is unreliable, we first calibrate the cell constant G* = R × κ using a standard KCl solution of known κ.

Molar Conductivity: Λ_m = κ / c
• Lab Units: Λ_m (S cm² mol⁻¹) = [κ (S cm⁻¹) × 1000] / Molarity (mol L⁻¹)
• SI Units: Λ_m (S m² mol⁻¹) = κ (S m⁻¹) / [1000 × Molarity]
• Conversion: 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹

NCERT Example 2.4: A cell filled with 0.1 M KCl (κ = 1.29 S m⁻¹) has R = 100 Ω. When filled with 0.02 M KCl, R = 520 Ω.
• Step 1: Cell constant G* = κ × R = 1.29 × 100 = 129 m⁻¹ = 1.29 cm⁻¹.
• Step 2: κ(0.02 M) = G* / R = 129 / 520 = 0.248 S m⁻¹ (= 0.248 × 10⁻² S cm⁻¹).
• Step 3: c = 1000 × 0.02 = 20 mol m⁻³ → Λ_m = 0.248 / 20 = 124 × 10⁻⁴ S m² mol⁻¹ = 124 S cm² mol⁻¹!

Check your units in every Λ_m problem! If κ is in S cm⁻¹, multiply by 1000 in the numerator: Λ_m = (κ × 1000) / M. If κ is in S m⁻¹, 1000 goes in the denominator because c = 1000 × M mol m⁻³! Also remember: temperature ↑ decreases metallic conductivity, but increases electrolytic conductivity!

Topic 2.4.2: Variation of Conductivity & Molar Conductivity with Dilution + Kohlrausch's Law

Pour a glass of water into saltwater—does the solution conduct MORE or LESS? Trick question! Conductivity (κ) measures ions inside a fixed 1 cm³ cube, so dilution spreads ions apart and κ drops. Meanwhile, Molar Conductivity (Λ_m) stretches the electrodes wide enough to hold an entire mole of ions—so they zoom more freely without traffic jams, and Λ_m rises!

1. Effect of Dilution on κ and Λ_m

Conductivity (κ) ALWAYS decreases with dilution (for both strong & weak electrolytes) because the number of ions per unit volume decreases.
Molar Conductivity (Λ_m = κV) ALWAYS increases with dilution because volume V containing 1 mole of electrolyte increases and more than compensates for the fall in κ.

2. Strong vs Weak Electrolytes (Fig. 2.6)

Strong Electrolytes (KCl, NaCl, HCl): Completely ionised even at moderate concentration. Λ_m rises slowly upon dilution obeying the Debye-Hückel-Onsager equation:
Λ_m = Λ°_m − A c^1/2 (straight line of slope −A and intercept = Λ°_m, limiting molar conductivity at zero concentration; Example 2.6 gives Λ°_m(KCl) = 150.0 S cm² mol⁻¹). Constant A depends on electrolyte charge type (1-1, 2-1, 2-2).
Weak Electrolytes (CH₃COOH): Low degree of dissociation (α) at higher concentration. Dilution increases α sharply, causing Λ_m to shoot up steeply near c → 0! Thus Λ°_m CANNOT be found by extrapolation!

3. Kohlrausch's Law of Independent Migration of Ions

Statement: 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°₋
E.g., Example 2.7: Λ°_m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻) = 119.0 + 2(76.3) = 271.6 S cm² mol⁻¹

4. Three Applications of Kohlrausch's Law

Finding Λ°_m of Weak Electrolytes: Combine strong electrolytes! (Example 2.8):
Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(CH₃COONa) − Λ°_m(NaCl) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹.
Similarly, Λ°_m(H₂O) = Λ°_m(HCl) + Λ°_m(NaOH) − Λ°_m(NaCl).
Degree of Dissociation (α): α = Λ_m / Λ°_m
Dissociation Constant (K_a): K_a = cα² / (1 − α) = c Λ_m² / [Λ°_m(Λ°_m − Λ_m)]

NCERT Example 2.9: Conductivity of 0.001028 M acetic acid is 4.95 × 10⁻⁵ S cm⁻¹; Λ°_m = 390.5 S cm² mol⁻¹.
• Λ_m = (4.95 × 10⁻⁵ × 1000) / 0.001028 = 48.15 S cm² mol⁻¹.
• α = 48.15 / 390.5 = 0.1233 ⇒ K_a = 0.001028 × (0.1233)² / (1 − 0.1233) = 1.78 × 10⁻⁵ mol L⁻¹.

Never confuse the two dilution trends in 1-mark questions: Conductivity (κ) DECREASES on dilution (fewer ions per cm³), while Molar Conductivity (Λ_m) INCREASES on dilution! Also, remember why weak electrolyte graphs cannot extrapolate to Λ°_m: at very high dilution Λ_m rises almost vertically AND κ is too tiny to measure accurately!

Topic 2.5 – 2.5.1: Electrolytic Cells, Faraday's Laws of Electrolysis & Products of Electrolysis

How do factories turn dull, blistered copper into 99.99% pure electrical wiring, coat shiny chrome onto car bumpers, and extract reactive sodium or aluminium metal that refuses to yield to carbon furnaces? By forcing DC electricity through molten salts or aqueous baths—and Michael Faraday discovered the exact "exchange rate" between Coulombs of electricity and grams of metal!

In an electrolytic cell, external DC voltage drives a non-spontaneous redox reaction. Impure copper anode dissolves and pure Cu plates at cathode (electrorefining); Na & Mg are extracted from fused chlorides, and Al from Al₂O₃ + cryolite.

1. Faraday's Two Laws of Electrolysis (1833–34)

First Law: The amount of chemical reaction (mass deposited/liberated, w) at any electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte:
w ∝ Q ⇒ w = Z I t = (M × I × t) / (n × F) (where Q = I × t in Coulombs; Z = electrochemical equivalent)
Second Law: When the same quantity of electricity passes through different electrolytic cells connected in series, the masses liberated are proportional to their chemical equivalent weights (Atomic Mass / n): w₁ / E₁ = w₂ / E₂.

1 Faraday (F) = Charge on 1 mole of electrons = N_A × e = 6.022 × 10²³ × 1.6021 × 10⁻¹⁹ C = 96487 C mol⁻¹ (≈ 96500 C mol⁻¹)
• 1 mol Ag⁺ → Ag requires 1 F | 1 mol Cu²⁺ → Cu requires 2 F | 1 mol Al³⁺ → Al requires 3 F | 1 mol MnO₄⁻ → Mn²⁺ requires 5 F

NCERT Example 2.10: CuSO₄ is electrolysed for 10 min with 1.5 A. Calculate mass of Cu deposited.
• Q = I × t = 1.5 A × (10 × 60 s) = 900 C.
• From Cu²⁺ + 2e⁻ → Cu, 2 × 96487 C deposits 63 g Cu ⇒ Mass = (63 × 900) / (2 × 96487) = 0.2938 g.

2. Products of Electrolysis (Inert vs Reactive Electrodes & Overpotential)

Electrode products depend on: (i) nature of electrolyte & concentration, (ii) inert (Pt/Au) vs reactive electrodes (Cu/Ag), (iii) standard E° values (higher E° reduces first at cathode; lower E° oxidises first at anode), and (iv) kinetic overpotential (extra voltage needed for slow gas evolution like O₂).

Electrolyte (Pt Electrodes)	Cathode Product	Anode Product	Key Reason
Molten NaCl	Na(l)	Cl₂(g)	Only Na⁺ and Cl⁻ present
Aqueous NaCl (Brine)	H₂(g) (+ OH⁻)	Cl₂(g)	H⁺/H₂ (0.00 V) beats Na⁺ (−2.71 V); Cl⁻ beats H₂O due to O₂ overpotential!
Dilute H₂SO₄	H₂(g)	O₂(g)	`2H₂O → O₂ + 4H⁺ + 4e⁻` (E° = 1.23 V) preferred
Conc. H₂SO₄	H₂(g)	S₂O₈²⁻	At high [SO₄²⁻], `2SO₄²⁻ → S₂O₈²⁻ + 2e⁻` (1.96 V) dominates
Aq. AgNO₃ (Ag vs Pt)	Ag(s)	Ag⁺ (with Ag) / O₂ (with Pt)	Reactive Ag anode dissolves itself; inert Pt oxidises H₂O

Overpotential Trap (Brine Electrolysis)! Comparing standard potentials alone, water oxidation (E° = +1.23 V) looks easier at the anode than chloride oxidation (E° = +1.36 V). Yet during electrolysis of aqueous NaCl, Cl₂ gas evolves instead of O₂ because oxygen evolution is kinetically slow and requires extra voltage called overpotential!

Topic 2.6 – 2.6.2: Commercial Batteries: Primary (Dry Cell, Mercury Cell) & Secondary (Lead Storage, Ni-Cd)

Your TV remote uses a cheap ₹15 cylinder you throw away once it dies (Primary Battery). Your hearing aid needs a button cell that delivers an exact, rock-solid 1.35 V without wavering for months (Mercury Cell). Meanwhile, your family car or home inverter carries heavy lead plates bathed in acid that you recharge thousands of times (Secondary Battery)!

A practical battery contains one or more galvanic cells connected in series; it should be compact, reasonably light, and maintain a fairly steady voltage during use.

1. Primary Batteries (Reaction Occurs Once — Non-Rechargeable)

Dry Cell (Leclanché Cell, ~1.5 V): Used in clocks and transistors.
- Anode: Zinc container: Zn(s) → Zn²⁺ + 2e⁻
- Cathode: Carbon (graphite) rod surrounded by powdered MnO₂ + C: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃ (Mn drops from +4 to +3).
- Electrolyte: Moist paste of NH₄Cl + ZnCl₂. NH₃ complexes with Zn²⁺ to form [Zn(NH₃)₄]²⁺ (prevents gas pressure buildup!).
Mercury Cell (~1.35 V): Used in hearing aids, watches.
- Anode: Zn–Hg amalgam: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
- Cathode: Paste of HgO + C: HgO + H₂O + 2e⁻ → Hg(l) + 2OH⁻
- Electrolyte: Paste of KOH + ZnO. Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l).

2. Secondary Batteries (Rechargeable by Reversing Current)

Lead Storage Battery (Automobiles & Inverters): Lead (Pb) anode, lead grid packed with PbO₂ cathode, and 38% H₂SO₄ electrolyte.
- Discharging:
  Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
  Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
  Overall Discharge: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
- Recharging: External current reverses reactions, converting solid PbSO₄ back to Pb (anode) and PbO₂ (cathode) while regenerating H₂SO₄!
Nickel–Cadmium (Ni–Cd) Cell: Jelly-roll plates separated by layer soaked in moist NaOH/KOH. Longer life than lead-acid, more expensive. Overall discharge:
Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)

Top 1-Mark Exam Question! Why does a Mercury cell give a constant voltage (1.35 V) throughout its lifetime, while a Dry cell's voltage drops? Answer: Look at the overall mercury cell equation: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)—it involves NO ions in solution whose concentration can change during use!

Topic 2.7: H₂–O₂ Fuel Cell & The Hydrogen Economy

When Apollo astronauts flew to the Moon in 1969, they couldn't haul giant lead-acid batteries—and couldn't pack a coal generator! Instead, NASA hooked up tanks of hydrogen and oxygen inside an H₂–O₂ Fuel Cell: it generated clean electricity silently at ~70% efficiency, and the "exhaust" condensed straight into pure drinking water for the crew!

Traditional thermal power plants burn fossil fuels to boil water into steam that spins turbines—losing huge heat energy with only ~40% efficiency and heavy CO₂ pollution. A Fuel Cell is a galvanic cell engineered to convert the energy of combustion of fuels (H₂, CH₄, CH₃OH) directly into electricity while reactants are fed continuously.

1. Construction & Reactions of the H₂–O₂ Fuel Cell (Fig. 2.12)

Electrodes: Porous carbon electrodes embedded with finely divided Pt or Pd catalysts (to accelerate electrode reactions).
Electrolyte: Concentrated aqueous NaOH (or KOH) solution.

Anode (Oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Cathode (Reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall Cell Reaction: 2H₂(g) + O₂(g) → 2H₂O(l)

2. Advantages & The Hydrogen Economy

High Efficiency: Converts chemical energy directly into electrical work at ~70% efficiency (vs ~40% for thermal plants).
Continuous Operation: Runs indefinitely as long as H₂ and O₂ are supplied.
Pollution-Free: Zero greenhouse gases or toxic smoke—only pure water vapour! (Alternative fuels like methane CH₄ and methanol CH₃OH can also power fuel cells.)
Hydrogen Economy Vision: Producing H₂ by solar-powered water splitting and burning it cleanly in fuel cells combats global warming, melting polar ice, and coastal flooding caused by CO₂ emissions.

Be careful when writing half-reactions for the NCERT H₂–O₂ fuel cell! Because the electrolyte is concentrated NaOH(aq) (alkaline), balance both half-reactions using OH⁻ and H₂O—NOT H⁺! Check: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻ at anode, and O₂ + 2H₂O + 4e⁻ → 4OH⁻ at cathode.

Topic 2.8: Corrosion: Electrochemical Mechanism of Rusting & Prevention

Leave a steel gate outside during Kerala's monsoon rains, and reddish-brown flakes soon eat away at the iron! That familiar enemy—rusting—isn't simple air touching metal. Every tiny drop of rainwater sitting on iron sets up a microscopic short-circuited Galvanic Cell right on the metal's surface, quietly dissolving iron into rust!

Corrosion slowly coats metallic surfaces with oxides or salts—like rusting of iron, tarnishing of silver (Ag₂S), and green patina on copper/bronze—costing crores of rupees every year.

1. Electrochemical Mechanism of Rusting of Iron (Fig. 2.13)

Rusting occurs in the presence of water and air (O₂). Atmospheric CO₂ dissolves in moisture to form carbonic acid (H₂CO₃), supplying H⁺ ions (acidic oxides like SO₂ also supply H⁺).

Anodic Spot (Oxidation): 2Fe(s) → 2Fe²⁺(aq) + 4e⁻   (E°_Fe²⁺/Fe = −0.44 V)
Cathodic Spot (Reduction): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)   (E° = +1.23 V)
Overall Cell Reaction: 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)   (E°_cell = +1.67 V!)

Electrons released at the anode travel straight through the iron metal to the cathodic spot. Next, ferrous ions undergo atmospheric oxidation to form reddish-brown rust (hydrated ferric oxide, Fe₂O₃·xH₂O)—and notice how H⁺ ions are regenerated to keep rusting going:

2Fe²⁺(aq) + 2H₂O(l) + ½O₂(g) → Fe₂O₃(s) + 4H⁺(aq)

2. Three Key Methods to Prevent Corrosion

Barrier Protection: Cover the surface with paint, grease, or special anti-rust chemicals like bisphenol so air and moisture cannot contact the metal.
Metallic Coating: Coat with inert or protective metals like Tin (Sn) or Zinc (Zn, galvanisation).
Electrochemical / Sacrificial Protection: Connect a more reactive metal (more negative E°) like Magnesium (Mg, −2.36 V) or Zinc (Zn, −0.76 V) to the iron structure (underground pipes, ship hulls). That metal acts as the anode and corrodes ("sacrifices itself"), forcing the iron to stay safe as the cathode!

Why does galvanised iron (Zn-coated) stay rust-free even if scratched, whereas tin-plated iron (Sn-coated) rusts rapidly at a scratch? Because E°(Zn²⁺/Zn) = −0.76 V < E°(Fe²⁺/Fe) = −0.44 V < E°(Sn²⁺/Sn) = −0.14 V! At a scratch, Zn acts as anode and protects Fe; with Sn, Fe becomes the anode and rusts faster! Also remember rust is Fe₂O₃·xH₂O, NOT FeO or Fe₃O₄!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Most Repeated) • 2 Marks): What is a salt bridge? Mention two important functions of a salt bridge, and depict the Daniell cell using IUPAC convention.

A salt bridge is an inverted U-tube filled with a concentrated solution of an inert electrolyte (such as KCl, KNO₃, or NH₄NO₃) mixed with agar-agar jelly.

Two Functions: (1) It completes the internal electrical circuit by allowing migration of ions without mixing the two half-cell solutions. (2) It maintains electrical neutrality in both half-cells.

IUPAC Cell Representation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Q2 (Kerala HSE SAY / Model • 3 Marks): Calculate the EMF of the cell in which the following reaction takes place at 298 K:
Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.160 M) + 2Ag(s) [Given E°_cell = 1.05 V]

Here n = 2. By the Nernst equation at 298 K:
E_cell = E°_cell − (0.059 / 2) log([Ni²⁺] / [Ag⁺]²)
E_cell = 1.05 − 0.0295 log(0.160 / (0.002)²) = 1.05 − 0.0295 log(4 × 10⁴)
= 1.05 − 0.0295 × 4.6021 = 1.05 − 0.1358 = 0.914 V (≈ 0.91 V).

Q3 (Kerala HSE March (High Yield) • 3 Marks): Write the equations relating E°_cell with Δ_rG° and K_c. For the reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s), E°_cell = 0.236 V at 298 K. Calculate Δ_rG° and K_c.

Relations: Δ_rG° = −nFE°_cell and Δ_rG° = −2.303 RT log K_c (or E°_cell = (0.059/n) log K_c at 298 K).

Here n = 2, F = 96487 C mol⁻¹:
Δ_rG° = −2 × 96487 × 0.236 = −45,542 J mol⁻¹ = −45.54 kJ mol⁻¹.
log K_c = 45542 / (2.303 × 8.314 × 298) = 7.983 ⇒ K_c = 9.62 × 10⁷.

Q4 (Kerala HSE March (Repeated Favorite) • 2 Marks): Define conductivity (κ) and molar conductivity (Λ_m). How do both vary with dilution?

Conductivity (κ): Conductance of one unit volume of solution kept between two electrodes of unit cross-sectional area separated by unit distance (κ = G × l/A, SI unit: S m⁻¹).

Molar Conductivity (Λ_m): Conductance of the volume V of solution containing 1 mole of electrolyte placed between electrodes at unit distance (Λ_m = κ / c, unit: S cm² mol⁻¹ or S m² mol⁻¹).

Variation with Dilution: Conductivity (κ) decreases on dilution (fewer ions per unit volume), whereas molar conductivity (Λ_m) increases on dilution (volume V increases more than κ falls; steeply for weak electrolytes due to higher α).

Q5 (Kerala HSE SAY • 2 Marks): Why is AC used instead of DC in measuring the resistance of an electrolytic solution? Define cell constant (G*).

Why AC? Direct current (DC) causes electrolysis of the solution over time, altering its ionic concentration and causing polarization at the electrodes. Using alternating current (AC) avoids net chemical change.

Cell Constant (G*): Ratio of distance between the two electrodes (l) to their cross-sectional area (A): G* = l / A = R × κ (unit: m⁻¹ or cm⁻¹).

Q6 (Kerala HSE March (Guaranteed Question) • 3 Marks): State Kohlrausch's Law of Independent Migration of Ions. Given Λ°_m values of NaCl, HCl and CH₃COONa as 126.4, 425.9 and 91.0 S cm² mol⁻¹ respectively, calculate Λ°_m for acetic acid (CH₃COOH).

Statement: At infinite dilution, the limiting molar conductivity (Λ°_m) of an electrolyte equals the sum of the individual contributions of its cation and anion: Λ°_m = ν₊λ°₊ + ν₋λ°₋.

Calculation:
Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(CH₃COONa) − Λ°_m(NaCl)
= 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹.

Q7 (Kerala HSE Model / NCERT • 3 Marks): Resistance of a conductivity cell filled with 0.1 mol L⁻¹ KCl solution is 100 Ω. If the resistance of the same cell with 0.02 mol L⁻¹ KCl is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L⁻¹ KCl. (κ of 0.1 M KCl = 1.29 S m⁻¹)

1. Cell constant G* = κ × R = 1.29 S m⁻¹ × 100 Ω = 129 m⁻¹ (= 1.29 cm⁻¹).

2. Conductivity of 0.02 M KCl: κ = G* / R = 129 / 520 = 0.248 S m⁻¹ (= 0.248 × 10⁻² S cm⁻¹).

3. Molar conductivity: c = 1000 × 0.02 = 20 mol m⁻³
Λ_m = κ / c = 0.248 / 20 = 124 × 10⁻⁴ S m² mol⁻¹ = 124 S cm² mol⁻¹.

Q8 (Kerala HSE March • 3 Marks): State Faraday's First and Second Laws of electrolysis. A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 A. What mass of copper is deposited at the cathode? [Atomic mass of Cu = 63 g mol⁻¹]

First Law: Amount of chemical reaction at any electrode is proportional to the quantity of electricity passed (w = ZIt).
Second Law: Masses of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights.

Calculation: t = 10 × 60 = 600 s; Q = 1.5 × 600 = 900 C.
Cu²⁺ + 2e⁻ → Cu (2F = 2 × 96487 C deposits 63 g Cu).
Mass deposited = (63 × 900) / (2 × 96487) = 0.2938 g.

Q9 (Kerala HSE SAY • 2 Marks): Write the anode and cathode products when (a) molten NaCl and (b) aqueous NaCl are electrolysed using platinum electrodes. Why is Cl₂ obtained instead of O₂ in aqueous NaCl?

(a) Molten NaCl: Cathode: Na(l); Anode: Cl₂(g).

(b) Aqueous NaCl: Cathode: H₂(g) (since E°(H⁺/H₂) = 0.00 V > −2.71 V); Anode: Cl₂(g).

Reason: Although water oxidation (E° = 1.23 V) is thermodynamically more favourable than Cl⁻ oxidation (1.36 V), oxygen evolution is kinetically slow and requires extra overpotential, so Cl₂ forms preferentially.

Q10 (Kerala HSE March • 1 Marks): Why does the cell potential of a mercury cell remain constant (1.35 V) throughout its lifetime?

Because the overall reaction Zn(Hg) + HgO(s) → ZnO(s) + Hg(l) involves no ions in solution whose concentration can change during its working life.

Q11 (Kerala HSE March (Very Important) • 3 Marks): Write the anode, cathode, and overall reactions taking place during the discharging and recharging of a Lead Storage Battery.

Electrolyte: 38% aqueous H₂SO₄; Anode: Pb; Cathode: Pb grid packed with PbO₂.

During Discharging:
Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Overall: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

During Recharging: Reaction reverses under external voltage:
2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

Q12 (Kerala HSE March / SAY • 4 Marks): (a) Draw/describe the H₂–O₂ fuel cell with electrode reactions and list two advantages over thermal power plants.
(b) Explain rusting of iron as an electrochemical phenomenon and name two prevention methods.

(a) H₂–O₂ Fuel Cell: H₂ and O₂ are bubbled through porous carbon electrodes (with Pt/Pd catalysts) into conc. NaOH(aq).
Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall: 2H₂(g) + O₂(g) → 2H₂O(l)
Advantages: High efficiency (~70% vs ~40% thermal plants) and zero pollution (only pure water produced).

(b) Rusting: Anode: 2Fe(s) → 2Fe²⁺ + 4e⁻; Cathode: O₂(g) + 4H⁺ + 4e⁻ → 2H₂O(l) (H⁺ from H₂CO₃); Fe²⁺ oxidises in air to rust Fe₂O₃·xH₂O.
Prevention: (1) Barrier coating with paint or bisphenol; (2) Sacrificial protection using Mg or Zn blocks.