ELECTROCHEMISTRY - DETAILED NEET/JEE NOTES
1. INTRODUCTION TO ELECTROCHEMISTRY
Definition: Electrochemistry is 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.
Applications:
- Production of metals: Na, Mg, Al
- Production of chemicals: NaOH, Cl₂, F₂
- Batteries and fuel cells convert chemical energy into electrical energy
- Electroplating and purification of metals
- Transmission of sensory signals in biological systems
- Environmentally friendly chemical processes
2. ELECTROCHEMICAL CELLS
Electrochemical Cell: A device that converts chemical energy into electrical energy (Galvanic cell) or electrical energy into chemical energy (Electrolytic cell).
2.1 GALVANIC (VOLTAIC) CELLS
Galvanic Cell: An electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy.
Example: Daniell Cell
Components:
- Anode (negative electrode): Zinc rod in ZnSO₄ solution
- Cathode (positive electrode): Copper rod in CuSO₄ solution
- Salt bridge: Connects the two solutions
Half-cell reactions:
Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Cell potential: E°(cell) = 1.1 V (when [Zn²⁺] = [Cu²⁺] = 1 M)
Key Concepts:
- Anode: Electrode where oxidation occurs (negative terminal)
- Cathode: Electrode where reduction occurs (positive terminal)
- Electrons flow: From anode to cathode through external circuit
- Current flow: From cathode to anode (opposite to electron flow)
- Salt bridge: Maintains electrical neutrality and completes the circuit
Cell Representation Convention:
Format: Anode | Anode solution || Cathode solution | Cathode
Example: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Notation:
- Single vertical line (|) = Phase boundary
- Double vertical line (||) = Salt bridge
- Comma (,) = Two species in same phase
2.2 ELECTROLYTIC CELLS
Electrolytic Cell: A device that uses electrical energy to carry out non-spontaneous chemical reactions.
Working Principle:
- External voltage is applied that opposes the natural cell potential
- When E(external) > E(cell), the reaction reverses
- Current flows in opposite direction to galvanic cell
- Used for electroplating, metal extraction, and electrolysis
3. ELECTRODE POTENTIAL
Electrode Potential: The potential difference that develops between the electrode and the electrolyte when an electrode is dipped in an electrolyte solution.
3.1 Standard Hydrogen Electrode (SHE)
Reference Electrode: Assigned zero potential at all temperatures
Conditions:
- Platinum electrode coated with platinum black
- Pure H₂ gas at 1 bar pressure
- H⁺ ion concentration = 1 M (pH = 0)
- Temperature = 298 K
Half reaction: H⁺(aq) + e⁻ → ½H₂(g), E° = 0.00 V
3.2 Standard Electrode Potential (E°)
Standard Electrode Potential: The potential of an electrode when all species are at unit concentration (1 M for solutions, 1 bar for gases) measured against SHE at 298 K.
Measurement:
Cell setup: Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M) || M^n⁺(aq, 1 M) | M(s)
E°(cell) = E°(cathode) - E°(anode)
Since E°(SHE) = 0, therefore E°(cell) = E°(cathode) = E°(M^n⁺/M)
Important Points:
- Positive E°: Species is reduced more easily than H⁺ → Stronger oxidizing agent
- Negative E°: H⁺ is reduced more easily → Species is stronger reducing agent
- Highest E° = F₂/F⁻ (+2.87 V): Strongest oxidizing agent, weakest reducing agent
- Lowest E° = Li⁺/Li (-3.05 V): Weakest oxidizing agent, strongest reducing agent
3.3 Standard Electrode Potentials (Important Values)
| Half Reaction |
E° (V) |
Nature |
| F₂(g) + 2e⁻ → 2F⁻ |
+2.87 |
Strongest oxidizing agent |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O |
+1.51 |
Strong oxidizing agent |
| Cl₂(g) + 2e⁻ → 2Cl⁻ |
+1.36 |
Strong oxidizing agent |
| O₂(g) + 4H⁺ + 4e⁻ → 2H₂O |
+1.23 |
Oxidizing agent |
| Br₂ + 2e⁻ → 2Br⁻ |
+1.09 |
Oxidizing agent |
| Ag⁺ + e⁻ → Ag(s) |
+0.80 |
Moderate oxidizing agent |
| Fe³⁺ + e⁻ → Fe²⁺ |
+0.77 |
Moderate oxidizing agent |
| I₂ + 2e⁻ → 2I⁻ |
+0.54 |
Weak oxidizing agent |
| Cu²⁺ + 2e⁻ → Cu(s) |
+0.34 |
Weak oxidizing agent |
| 2H⁺ + 2e⁻ → H₂(g) |
0.00 |
Reference |
| Pb²⁺ + 2e⁻ → Pb(s) |
-0.13 |
Weak reducing agent |
| Ni²⁺ + 2e⁻ → Ni(s) |
-0.25 |
Reducing agent |
| Fe²⁺ + 2e⁻ → Fe(s) |
-0.44 |
Reducing agent |
| Zn²⁺ + 2e⁻ → Zn(s) |
-0.76 |
Strong reducing agent |
| Al³⁺ + 3e⁻ → Al(s) |
-1.66 |
Strong reducing agent |
| Mg²⁺ + 2e⁻ → Mg(s) |
-2.36 |
Very strong reducing agent |
| Na⁺ + e⁻ → Na(s) |
-2.71 |
Very strong reducing agent |
| Li⁺ + e⁻ → Li(s) |
-3.05 |
Strongest reducing agent |
3.4 Cell EMF Calculation
EMF of Cell:
E°(cell) = E°(cathode) - E°(anode)
E°(cell) = E°(right) - E°(left)
E°(cell) = E°(reduction) - E°(oxidation)
Example: Calculate E°(cell) for the reaction:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
Solution:
Cathode (reduction): 2Ag⁺ + 2e⁻ → 2Ag, E° = +0.80 V
Anode (oxidation): Cu → Cu²⁺ + 2e⁻, E° = +0.34 V
E°(cell) = 0.80 - 0.34 = 0.46 V
Cell representation: Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)
4. NERNST EQUATION
Nernst Equation: Relates electrode potential at any concentration to standard electrode potential.
4.1 For Electrode (Single Half-Cell)
For the reaction: M^n⁺(aq) + ne⁻ → M(s)
General form:
E(M^n⁺/M) = E°(M^n⁺/M) - (RT/nF) ln(1/[M^n⁺])
At 298 K (using log₁₀):
E(M^n⁺/M) = E°(M^n⁺/M) - (0.0591/n) log(1/[M^n⁺])
Or:
E(M^n⁺/M) = E°(M^n⁺/M) + (0.0591/n) log[M^n⁺]
Constants:
- R = 8.314 J K⁻¹ mol⁻¹ (Gas constant)
- F = 96487 C mol⁻¹ (Faraday constant) ≈ 96500 C mol⁻¹
- T = Temperature in Kelvin
- n = Number of electrons transferred
- At 298 K: 2.303RT/F = 0.0591 V
4.2 For Complete Cell
For general reaction: aA + bB → cC + dD
Nernst Equation:
E(cell) = E°(cell) - (RT/nF) ln Q
At 298 K:
E(cell) = E°(cell) - (0.0591/n) log Q
Where Q = [C]^c[D]^d / [A]^a[B]^b (Reaction quotient)
For Daniell Cell:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
E(cell) = E°(cell) - (0.0591/2) log([Zn²⁺]/[Cu²⁺])
Example: Calculate E(cell) for Daniell cell when [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M
Given: E°(cell) = 1.1 V
Solution:
E(cell) = 1.1 - (0.0591/2) log(0.1/0.01)
E(cell) = 1.1 - 0.02955 × log(10)
E(cell) = 1.1 - 0.02955 × 1
E(cell) = 1.1 - 0.03 = 1.07 V
4.3 Applications of Nernst Equation
A. Equilibrium Constant Calculation
At equilibrium: E(cell) = 0 and Q = K(equilibrium constant)
Therefore:
0 = E°(cell) - (0.0591/n) log K
E°(cell) = (0.0591/n) log K
Or:
log K = (n × E°(cell))/0.0591
K = 10^(n × E°(cell)/0.0591)
Example: Calculate K for: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
Given: E°(cell) = 0.46 V
Solution:
log K = (2 × 0.46)/0.0591 = 15.6
K = 10^15.6 = 3.92 × 10¹⁵
B. Gibbs Energy Relationship
Fundamental Equations:
ΔᵣG = -nFE(cell)
ΔᵣG° = -nFE°(cell)
Also:
ΔᵣG° = -RT ln K = -2.303RT log K
Combining:
-nFE°(cell) = -2.303RT log K
E°(cell) = (2.303RT/nF) log K
E°(cell) = (0.0591/n) log K (at 298 K)
Spontaneity Criteria:
- If E°(cell) > 0 → ΔᵣG° < 0 → Reaction is spontaneous
- If E°(cell) < 0 → ΔᵣG° > 0 → Reaction is non-spontaneous
- If E°(cell) = 0 → ΔᵣG° = 0 → Reaction at equilibrium
- If K > 1 → E°(cell) > 0 → Forward reaction favored
- If K < 1 → E°(cell) < 0 → Backward reaction favored
Example: Calculate ΔᵣG° for Daniell cell
Given: E°(cell) = 1.1 V, n = 2
Solution:
ΔᵣG° = -nFE°(cell)
ΔᵣG° = -2 × 96487 × 1.1
ΔᵣG° = -212271 J mol⁻¹
ΔᵣG° = -212.27 kJ mol⁻¹
5. CONDUCTANCE IN ELECTROLYTIC SOLUTIONS
5.1 Basic Definitions and Formulas
A. Resistance (R)
R = ρ(l/A)
Where:
ρ = Resistivity (specific resistance)
l = Length of conductor
A = Area of cross-section
Unit: Ohm (Ω) or kg m²/(s³ A²)
B. Resistivity (ρ)
Resistivity: Resistance of a material of length 1 m and cross-sectional area 1 m²
Unit: Ω m or Ω cm
Conversion: 1 Ω m = 100 Ω cm
C. Conductance (G)
G = 1/R = κ(A/l)
Unit: Siemens (S) or mho (Ω⁻¹)
1 S = 1 Ω⁻¹ = 1 A V⁻¹
D. Conductivity (κ)
κ = 1/ρ = l/(RA) = G × (l/A)
Unit: S m⁻¹ or S cm⁻¹
Conversion: 1 S cm⁻¹ = 100 S m⁻¹
Definition: Conductance of a material of length 1 m and cross-sectional area 1 m²
E. Cell Constant (G*)
G* = l/A = R × κ
Unit: m⁻¹ or cm⁻¹
Determination: Using standard KCl solutions of known conductivity
Measurement of Conductivity:
- Determine cell constant using standard KCl solution
- Measure resistance (R) of unknown solution
- Calculate: κ = G*/R
5.2 Molar Conductivity (Λₘ)
Molar Conductivity: The conductance of volume of solution containing 1 mole of electrolyte kept between two electrodes at unit distance apart.
Λₘ = κ/c
Where:
κ = Conductivity (S m⁻¹ or S cm⁻¹)
c = Concentration (mol m⁻³ or mol L⁻¹)
Units:
If κ in S m⁻¹ and c in mol m⁻³: Λₘ in S m² mol⁻¹
If κ in S cm⁻¹ and c in mol L⁻¹: Λₘ in S cm² mol⁻¹
Conversion:
1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹
Alternative formulas:
Λₘ (S cm² mol⁻¹) = [κ (S cm⁻¹) × 1000 (cm³ L⁻¹)] / Molarity (mol L⁻¹)
Λₘ = κV (where V is volume in cm³ containing 1 mole)
Example: Calculate Λₘ for 0.02 M KCl solution
Given: κ = 0.248 × 10⁻² S cm⁻¹
Solution:
Λₘ = (κ × 1000)/Molarity
Λₘ = (0.248 × 10⁻² × 1000)/0.02
Λₘ = 124 S cm² mol⁻¹
5.3 Variation of Conductivity and Molar Conductivity with Concentration
A. Conductivity (κ)
Effect of Dilution:
- Conductivity decreases with dilution (for both strong and weak electrolytes)
- Reason: Number of ions per unit volume decreases
- κ = G × (l/A), and G decreases as concentration decreases
B. Molar Conductivity (Λₘ)
Effect of Dilution:
- Molar conductivity increases with dilution (for both strong and weak electrolytes)
- Reason: Although κ decreases, volume (V) containing 1 mole increases more
- Since Λₘ = κV, overall Λₘ increases
C. For Strong Electrolytes
D. For Weak Electrolytes
Behavior:
- Steep increase in Λₘ with dilution, especially near lower concentrations
- Non-linear relationship with concentration
- Λ°ₘ cannot be determined by extrapolation
- Increase due to increase in degree of dissociation (α) with dilution
5.4 Limiting Molar Conductivity (Λ°ₘ)
Limiting Molar Conductivity: The molar conductivity at infinite dilution (c → 0), when electrolyte is completely dissociated and interionic interactions are negligible.
5.5 Kohlrausch Law of Independent Migration of Ions
Kohlrausch Law: At infinite dilution, each ion migrates independently and contributes to total molar conductivity regardless of the nature of other ions present.
Mathematical Form:
Λ°ₘ = n₊λ°₊ + n₋λ°₋
Where:
n₊, n₋ = Number of cations and anions per formula unit
λ°₊, λ°₋ = Limiting molar conductivities of individual ions
Examples:
Λ°ₘ(NaCl) = λ°(Na⁺) + λ°(Cl⁻)
Λ°ₘ(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻)
Λ°ₘ(Al₂(SO₄)₃) = 2λ°(Al³⁺) + 3λ°(SO₄²⁻)
Limiting Molar Conductivities of Some Ions at 298 K
| Cation |
λ° (S cm² mol⁻¹) |
Anion |
λ° (S cm² mol⁻¹) |
| H⁺ |
349.6 |
OH⁻ |
199.1 |
| Na⁺ |
50.1 |
Cl⁻ |
76.3 |
| K⁺ |
73.5 |
Br⁻ |
78.1 |
| Ca²⁺ |
119.0 |
CH₃COO⁻ |
40.9 |
| Mg²⁺ |
106.0 |
SO₄²⁻ |
160.0 |
Note: H⁺ and OH⁻ have exceptionally high conductivities due to proton jump mechanism (Grotthuss mechanism).
5.6 Applications of Kohlrausch Law
A. Calculation of Λ°ₘ for Weak Electrolytes
Example: Calculate Λ°ₘ for CH₃COOH (acetic acid)
Method:
Λ°ₘ(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻)
= λ°(H⁺) + λ°(Cl⁻) + λ°(CH₃COO⁻) + λ°(Na⁺) - λ°(Cl⁻) - λ°(Na⁺)
= Λ°ₘ(HCl) + Λ°ₘ(CH₃COONa) - Λ°ₘ(NaCl)
Using values:
= 426.16 + 91.0 - 126.45
= 390.71 S cm² mol⁻¹
B. Calculation of Degree of Dissociation (α)
For Weak Electrolytes:
α = Λₘ/Λ°ₘ
Where:
α = Degree of dissociation
Λₘ = Molar conductivity at concentration c
Λ°ₘ = Limiting molar conductivity
C. Calculation of Dissociation Constant (Ka)
For Weak Acid/Base:
Ka = (cα²)/(1-α)
Since α = Λₘ/Λ°ₘ
Ka = [c(Λₘ)²] / [Λ°ₘ(Λ°ₘ - Λₘ)]
Example: Calculate Ka for 0.001 M CH₃COOH
Given: κ = 4.95 × 10⁻⁵ S cm⁻¹, Λ°ₘ = 390.5 S cm² mol⁻¹
Solution:
Λₘ = (κ × 1000)/c = (4.95 × 10⁻⁵ × 1000)/0.001 = 49.5 S cm² mol⁻¹
α = Λₘ/Λ°ₘ = 49.5/390.5 = 0.127
Ka = (cα²)/(1-α) = (0.001 × 0.127²)/(1-0.127)
Ka = 1.85 × 10⁻⁵ mol L⁻¹
6. ELECTROLYTIC CELLS AND ELECTROLYSIS
6.1 Electrolysis
Electrolysis: The process of using electrical energy to bring about a non-spontaneous chemical reaction.
Example: Electrolysis of Aqueous CuSO₄
With Copper Electrodes:
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
Anode (oxidation): Cu(s) → Cu²⁺(aq) + 2e⁻
Result: Copper dissolves at anode and deposits at cathode (basis for purification)
Example: Electrolysis of Aqueous NaCl
At Cathode (possible reactions):
Na⁺(aq) + e⁻ → Na(s), E° = -2.71 V
H⁺(aq) + e⁻ → ½H₂(g), E° = 0.00 V
Preferred: H₂O(l) + e⁻ → ½H₂(g) + OH⁻(aq) (higher E°)
At Anode (possible reactions):
Cl⁻(aq) → ½Cl₂(g) + e⁻, E° = 1.36 V
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻, E° = 1.23 V
Preferred: Cl⁻ oxidation (due to overpotential of O₂)
Products: H₂ at cathode, Cl₂ at anode, NaOH in solution
6.2 Faraday's Laws of Electrolysis
First Law
The amount of chemical reaction (mass deposited/liberated) at any electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematically: m ∝ Q or m ∝ It
Second Law
When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited/liberated at the electrodes are proportional to their chemical equivalent weights.
Mathematically: m₁/m₂ = E₁/E₂
Where E = Equivalent weight = (Atomic mass)/(Number of electrons)
6.3 Quantitative Aspects
Faraday Constant (F):
F = Charge on 1 mole of electrons
F = NA × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹
F = 96487 C mol⁻¹ ≈ 96500 C mol⁻¹
Charge passed:
Q = I × t
Where: I = Current (A), t = Time (s)
Mass deposited:
m = (Q × M)/(n × F) = (I × t × M)/(n × F)
Where: M = Molar mass, n = Number of electrons
Example: Calculate mass of Cu deposited when 1.5 A current passes through CuSO₄ solution for 10 minutes.
Solution:
Q = I × t = 1.5 × 600 = 900 C
For Cu²⁺ + 2e⁻ → Cu: n = 2, M = 63.5 g/mol
m = (900 × 63.5)/(2 × 96500) = 0.296 g
6.4 Products of Electrolysis
Factors Determining Products:
- Nature of electrode (inert vs reactive)
- Standard electrode potentials of species present
- Overpotential requirements
- Concentration of ions in solution
Preferential Discharge Theory
- At Cathode: Cation with higher (more positive) E° is reduced preferentially
- At Anode: Anion with lower (less positive) E° is oxidized preferentially
- Exception: Overpotential may change the order
7. BATTERIES
Battery: A galvanic cell or series of galvanic cells that converts chemical energy into electrical energy.
7.1 Primary Batteries (Non-rechargeable)
A. Dry Cell (Leclanche Cell)
Components:
- Anode: Zinc container
- Cathode: Graphite rod surrounded by MnO₂ + C
- Electrolyte: Paste of NH₄Cl + ZnCl₂
Reactions:
Anode: Zn(s) → Zn²⁺ + 2e⁻
Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃
Complex formation: Zn²⁺ + 4NH₃ → [Zn(NH₃)₄]²⁺
EMF: ~1.5 V
Uses: Torches, transistors, clocks
B. Mercury Cell
Components:
- Anode: Zinc-mercury amalgam
- Cathode: HgO paste with carbon
- Electrolyte: Paste of KOH + ZnO
Reactions:
Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
Cathode: HgO + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
EMF: ~1.35 V (constant throughout life)
Uses: Hearing aids, watches, calculators
Advantage: Constant voltage (no ions in solution)
7.2 Secondary Batteries (Rechargeable)
A. Lead Storage Battery
Components:
- Anode: Lead (Pb) grid
- Cathode: Lead grid packed with PbO₂
- Electrolyte: 38% H₂SO₄ solution
Discharge Reactions:
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)
Charging Reactions: (Reverse of above)
2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)
EMF: ~2 V per cell (6 cells = 12 V battery)
Uses: Automobiles, inverters
B. Nickel-Cadmium (Ni-Cd) Cell
Components:
- Anode: Cadmium
- Cathode: Nickel hydroxide [Ni(OH)₃]
- Electrolyte: KOH solution
Discharge Reaction:
Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)
Advantages: Longer life than lead storage
Disadvantages: More expensive
Uses: Emergency lighting, portable devices
8. FUEL CELLS
Fuel Cell: A galvanic cell that converts chemical energy of combustion of fuels directly into electrical energy with high efficiency.
8.1 Hydrogen-Oxygen Fuel Cell
Components:
- Electrodes: Porous carbon with Pt or Pd catalyst
- Electrolyte: Concentrated aqueous NaOH or KOH
- Fuel: H₂ gas (anode)
- Oxidant: O₂ gas (cathode)
Reactions:
Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall: 2H₂(g) + O₂(g) → 2H₂O(l)
Efficiency: ~70% (vs ~40% for thermal plants)
Advantages:
- Pollution-free (product is water)
- High efficiency
- Continuous operation as long as fuel is supplied
- No noise
Uses: Apollo space program, experimental vehicles
8.2 Other Fuel Cells
- Methanol fuel cells: Direct oxidation of CH₃OH
- Methane fuel cells: Uses natural gas
- Solid oxide fuel cells: High temperature operation
- Proton exchange membrane (PEM) cells: For vehicles
9. CORROSION
Corrosion: The deterioration of metals by electrochemical oxidation in the presence of air and moisture.
9.1 Rusting of Iron
Mechanism (Electrochemical Process):
At Anode (oxidation):
2Fe(s) → 2Fe²⁺(aq) + 4e⁻, E° = -0.44 V
At Cathode (reduction):
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l), E° = +1.23 V
(H⁺ from H₂CO₃ formed by dissolving CO₂ in water)
Overall Cell Reaction:
2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)
E°(cell) = 1.67 V
Atmospheric Oxidation:
2Fe²⁺(aq) + 2H₂O(l) + ½O₂(g) → Fe₂O₃(s) + 4H⁺(aq)
(Rust: Fe₂O₃·xH₂O)
Conditions Favoring Corrosion
- Presence of moisture (H₂O)
- Presence of oxygen (O₂)
- Presence of electrolytes (salts, acids)
- Presence of CO₂ (forms H₂CO₃)
- Impurities in metal
- Strain in metal structure
9.2 Prevention of Corrosion
A. Barrier Protection
- Painting: Prevents contact with air and moisture
- Greasing/Oiling: Temporary protection
- Coating with chemicals: Bisphenol, polymer coatings
B. Metallic Coating
- Electroplating: Coating with Sn, Zn, Cr, Ni
- Galvanization: Coating with zinc (sacrificial protection)
- Tin plating: For food containers
- Anodizing: For Al (forms protective Al₂O₃ layer)
C. Cathodic Protection
Sacrificial Anode Method:
- Connect a more reactive metal (Mg, Zn) to the iron object
- More reactive metal acts as anode and corrodes
- Iron becomes cathode and is protected
- Used for underground pipes, ship hulls, oil rigs
D. Alloying
- Stainless steel: Fe + Cr + Ni (corrosion resistant)
- Bronze: Cu + Sn (resistant to seawater)
10. IMPORTANT FORMULAS - QUICK REFERENCE
1. Cell EMF:
E°(cell) = E°(cathode) - E°(anode)
2. Nernst Equation (298 K):
E(cell) = E°(cell) - (0.0591/n) log Q
3. Gibbs Energy:
ΔᵣG° = -nFE°(cell)
ΔᵣG° = -RT ln K
4. Equilibrium Constant:
log K = (n × E°(cell))/0.0591
5. Conductivity:
κ = G*/R = l/(RA)
6. Molar Conductivity:
Λₘ = κ/c = (κ × 1000)/M (if κ in S cm⁻¹, c in mol L⁻¹)
7. Kohlrausch Law:
Λ°ₘ = n₊λ°₊ + n₋λ°₋
For strong electrolytes: Λₘ = Λ°ₘ - A√c
8. Degree of Dissociation:
α = Λₘ/Λ°ₘ
9. Dissociation Constant:
Ka = (cα²)/(1-α) = [c(Λₘ)²]/[Λ°ₘ(Λ°ₘ - Λₘ)]
10. Faraday's Law:
m = (I × t × M)/(n × F)
Q = I × t
F = 96500 C mol⁻¹
11. IMPORTANT POINTS FOR NEET/JEE
A. Galvanic vs Electrolytic Cells
| Property |
Galvanic Cell |
Electrolytic Cell |
| Energy conversion |
Chemical → Electrical |
Electrical → Chemical |
| Reaction type |
Spontaneous (ΔG < 0) |
Non-spontaneous (ΔG > 0) |
| E(cell) |
Positive |
Negative |
| Anode charge |
Negative (-) |
Positive (+) |
| Cathode charge |
Positive (+) |
Negative (-) |
| Electron flow |
Anode → Cathode |
Cathode → Anode |
| Salt bridge |
Required |
Not required |
B. Memory Tips
- ANOX REDCAT: ANode = OXidation, REDuction = CAThode
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
- Higher E°: Better oxidizing agent (easily reduced)
- Lower (more negative) E°: Better reducing agent (easily oxidized)
- Conductivity decreases, Molar conductivity increases with dilution
- H⁺ and OH⁻ have highest conductivities (Grotthuss mechanism)
- Strong electrolytes: Linear Λₘ vs √c plot
- Weak electrolytes: Non-linear, use Kohlrausch law for Λ°ₘ
C. Common Mistakes to Avoid
- Confusing anode/cathode charges in galvanic vs electrolytic cells
- Using wrong sign in Nernst equation
- Forgetting to multiply λ° by stoichiometric coefficients
- Wrong conversion between S m⁻¹ and S cm⁻¹
- Forgetting factor of 1000 in Λₘ calculation
- Using E° instead of E in non-standard conditions
- Incorrect number of electrons (n) in calculations
12. SAMPLE PROBLEMS FOR PRACTICE
Problem 1: Calculate E(cell) for Daniell cell when [Zn²⁺] = 0.01 M, [Cu²⁺] = 0.1 M at 298 K. Given: E°(cell) = 1.1 V
Solution:
E(cell) = E°(cell) - (0.0591/n) log([Zn²⁺]/[Cu²⁺])
E(cell) = 1.1 - (0.0591/2) log(0.01/0.1)
E(cell) = 1.1 - 0.02955 × log(0.1)
E(cell) = 1.1 - 0.02955 × (-1)
E(cell) = 1.1 + 0.03 = 1.13 V
Problem 2: Resistance of 0.01 M KCl solution is 200 Ω. If cell constant is 1.0 cm⁻¹, calculate κ and Λₘ.
Solution:
κ = G*/R = 1.0/200 = 0.005 S cm⁻¹
Λₘ = (κ × 1000)/M = (0.005 × 1000)/0.01 = 500 S cm² mol⁻¹
Problem 3: How much charge is required to deposit 1 mole of Al from Al³⁺ solution?
Solution:
Al³⁺ + 3e⁻ → Al
For 1 mole Al: 3 moles of electrons required
Q = 3F = 3 × 96500 = 289500 C
Problem 4: Calculate equilibrium constant for: Zn + Cu²⁺ → Zn²⁺ + Cu
Given: E°(cell) = 1.1 V
Solution:
log K = (n × E°(cell))/0.0591
log K = (2 × 1.1)/0.0591 = 37.23
K = 10³⁷·²³ ≈ 1.7 × 10³⁷
CONCLUSION
Electrochemistry is a fundamental topic combining thermodynamics, kinetics, and chemical reactions. Master the concepts of electrode potentials, Nernst equation, conductance, and Faraday's laws for NEET/JEE success.
*** END OF NOTES ***
For PDF: Use Ctrl+P (Windows) or Cmd+P (Mac) → Save as PDF