1. Introduction to Electrochemistry

Electrochemistry deals with the interconversion of chemical energy and electrical energy.

Types of Cells

Galvanic (Voltaic) Cell	Electrolytic Cell
Converts Chemical → Electrical energy.	Converts Electrical → Chemical energy.
Reaction is Spontaneous (ΔG < 0).	Reaction is Non-spontaneous (ΔG > 0).
Anode is (-)ve, Cathode is (+)ve.	Anode is (+)ve, Cathode is (-)ve.

[Image of Galvanic Cell vs Electrolytic Cell diagram]

2. Electrode Potential & EMF

Potential difference between the electrode and its electrolyte. By convention, Standard Reduction Potential (SRP) is used.

EMF of Cell (E°_cell):

E°_cell = E°_cathode - E°_anode

(Both values must be Reduction Potentials)

Electrochemical Series

Top (Negative E°): Strong Reducing Agents (e.g., Li, K). Easily oxidized.
Bottom (Positive E°): Strong Oxidizing Agents (e.g., F₂, Au). Easily reduced.
Hydrogen Electrode (SHE): E° = 0.00 V (Reference).

3. Nernst Equation

Calculates EMF at non-standard conditions.

E_cell = E°_cell - (0.0591 / n) log Q

At 298 K. Q = Reaction Quotient ([Products]/[Reactants]).
n = Number of electrons transferred.

Relation with Thermodynamics:

ΔG° = -nFE°_cell
E°_cell = (0.0591 / n) log K_eq (at Equilibrium)

4. Conductance of Solutions

Resistance (R)	ρ(l/A)	Unit: Ohm (Ω)
Conductance (G)	1/R	Unit: Siemen (S)
Specific Cond. (κ)	G × (l/A)	Unit: S cm^-1
Molar Cond. (Λ_m)	(κ × 1000) / M	Unit: S cm² mol^-1

Variation with Concentration

Specific Conductance (κ): Decreases with dilution (number of ions per unit volume decreases).
Molar Conductance (Λ_m): Increases with dilution.
• Strong Electrolytes: Linear increase (Debye-Huckel).
• Weak Electrolytes: Steep increase (Ostwald Dilution).

[Image of Conductance vs Concentration Graph]

5. Kohlrausch Law

At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the cation and anion.

Λ°_m = ν₊λ°₊ + ν_-λ°_-

Application: Used to find Λ°_m for weak electrolytes (e.g., CH₃COOH) using strong electrolytes.

6. Faraday's Laws

First Law:

Mass deposited (w) is proportional to charge passed (Q).

w = Z × Q = Z × I × t

(Z = E / 96500)

Second Law:

For same charge passed through different electrolytes:

w₁ / w₂ = E₁ / E₂

7. Batteries & Corrosion

Primary Batteries: Non-rechargeable (e.g., Dry cell, Mercury cell).
Secondary Batteries: Rechargeable (e.g., Lead storage, Ni-Cd).
Lead Storage: Anode (Pb), Cathode (PbO₂), Electrolyte (38% H₂SO₄).
Fuel Cells: Convert combustion energy directly to electricity (e.g., H₂-O₂ cell). High efficiency (~70%).

Corrosion (Rusting of Iron):
Electrochemical process. Fe acts as anode.
Prevention: Galvanization (coating with Zn), Cathodic protection (Mg block).

Numericals & HOTS

Q1. Calculating EMF

Calculate the EMF of the cell at 25°C:
Mg(s) | Mg²⁺(0.001 M) || Cu²⁺(0.0001 M) | Cu(s)
Given: E°(Mg²⁺/Mg) = -2.37 V, E°(Cu²⁺/Cu) = +0.34 V.

Solution:
1. E°_cell = E°_C - E°_A = 0.34 - (-2.37) = 2.71 V.
2. Reaction: Mg + Cu²⁺ → Mg²⁺ + Cu. (n=2)

Nernst Eq: E = E° - (0.059/n) log([Mg²⁺]/[Cu²⁺])
E = 2.71 - (0.059/2) log(10^-3 / 10^-4)
E = 2.71 - 0.0295 log(10)
E = 2.71 - 0.0295 (1)
E = 2.68 V

Q2. ΔG Calculation

The standard electrode potential for a Daniel cell is 1.1 V. Calculate the standard Gibbs energy for the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu. (F = 96500 C/mol).

Solution:
n = 2 electrons transferred.
Formula: ΔG° = -nFE°_cell

ΔG° = -2 × 96500 × 1.1
ΔG° = -212300 J/mol
ΔG° = -212.3 kJ/mol

Q3. Equilibrium Constant (K_c)

Calculate the equilibrium constant for a reaction where E°_cell = 0.46 V and n = 2 at 298 K.

Solution:
Formula: E°_cell = (0.059 / n) log K_c
0.46 = (0.059 / 2) log K_c
0.46 = 0.0295 log K_c
log K_c = 0.46 / 0.0295 = 15.6

K_c = antilog(15.6)
K_c = 3.98 × 10¹⁵

Q4. Mass Deposited

A solution of CuSO₄ is electrolyzed for 10 minutes with a current of 1.5 Amperes. What is the mass of copper deposited at the cathode? (Molar mass Cu = 63.5).

Solution:
Reaction: Cu²⁺ + 2e^- → Cu (n=2).
Q = I × t = 1.5 × (10 × 60) = 900 C.

Formula: w = (M / nF) × Q
w = (63.5 / (2 × 96500)) × 900
w = (63.5 × 900) / 193000
w = 57150 / 193000
w = 0.296 g

Q5. Faraday's Logic (HOTS)

How much charge (in Faradays) is required for the reduction of 1 mole of MnO₄^- to Mn²⁺?

Solution:
Change in Oxidation Number:
MnO₄^- (Mn is +7) → Mn²⁺ (+2).
Difference = 5 electrons.

Reaction: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O
Since 1 mole requires 5 moles of electrons:
Charge = 5 F

Q6. Conductance & Cell Constant

The resistance of a conductivity cell filled with 0.1 M KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 Ω, calculate the conductivity of 0.02 M KCl. (Conductivity of 0.1 M KCl = 1.29 S/m).

Solution:
1. Find Cell Constant (G*) using 0.1 M solution:
κ = G × G* = (1/R) × G*
1.29 = (1/100) × G*
G* = 129 m^-1.

2. Find κ for 0.02 M solution:
κ = (1/R) × G*
κ = (1/520) × 129
κ = 0.248 S/m

Q7. Limiting Molar Conductivity

Λ°_m for NaCl, HCl, and NaAc are 126.4, 425.9, and 91.0 S cm² mol^-1 respectively. Calculate Λ°_m for HAc (Acetic Acid).

Solution:
Target: HAc → H⁺ + Ac^-
We need H⁺ (from HCl) and Ac^- (from NaAc).
We need to remove Na⁺ and Cl^- (subtract NaCl).

Λ°(HAc) = Λ°(HCl) + Λ°(NaAc) - Λ°(NaCl)
Λ°(HAc) = 425.9 + 91.0 - 126.4
Λ°(HAc) = 516.9 - 126.4
Λ°(HAc) = 390.5 S cm² mol^-1

Q8. Calculating Alpha

The molar conductivity of 0.00241 M acetic acid is 32.76 S cm² mol^-1. If its limiting molar conductivity is 390.5 S cm² mol^-1, calculate the degree of dissociation (α).

Solution:
Formula: α = Λ_m / Λ°_m

α = 32.76 / 390.5
α ≈ 0.084

Q9. Competitive Electrolysis (HOTS)

Predict the products of electrolysis of an aqueous solution of AgNO₃ with silver electrodes.

Solution:
Cathode: Competition between Ag⁺ and H₂O. Ag⁺ has higher reduction potential (+0.80V) than H₂O (-0.83V).
Reaction: Ag⁺ + e^- → Ag (Silver deposits).

Anode: Competition between NO₃^-, H₂O, and Ag electrode itself.
The Ag electrode is active (attackable). Oxidation of metal Ag is easier than oxidation of water or nitrate.
Reaction: Ag(s) → Ag⁺ + e^- (Silver dissolves).

Result: Ag dissolves at anode and deposits at cathode (Refining).

Q10. Hydrogen Electrode (HOTS)

Calculate the pH of the following half-cell solution: Pt(s) | H₂(1 atm) | H⁺(aq). The electrode potential is -0.118 V.

Solution:
Reduction half: H⁺ + e^- → ½H₂.
Nernst Eq: E = E° - 0.0591 log(1 / [H⁺])
Since E° = 0:
E = -0.0591 log(1 / [H⁺])
E = -0.0591 × pH (Since pH = log(1/[H⁺]))

-0.118 = -0.0591 × pH
pH = 0.118 / 0.0591
pH = 2

Important Formulae

1. EMF & Nernst Equation

Standard EMF:

E°_cell = E°_cathode - E°_anode

Nernst Equation (at 298 K):

E_cell = E°_cell - (0.0591 / n) log Q

Q = [Products] / [Reactants]

At Equilibrium (E_cell = 0):

E°_cell = (0.0591 / n) log K_c

2. Thermodynamics of Cell

ΔG° = -n F E°_cell

F = 96500 C mol^-1
For Spontaneous reaction: ΔG < 0 (E° > 0)

3. Conductance & Conductivity

Resistance (R)	ρ (l/A)
Conductance (G)	1 / R
Cell Constant (G*)	l / A
Conductivity (κ)	G × G*

4. Molar Conductivity (Λ_m)

Λ_m = (κ × 1000) / M

Unit: S cm² mol^-1
(M = Molarity)

Kohlrausch Law:

Λ°_m = ν₊λ°₊ + ν_-λ°_-

Degree of Dissociation (α):

α = Λ_m / Λ°_m

5. Faraday's Laws of Electrolysis

Mass Deposited (w):

w = Z × I × t

w = (E / 96500) × Q

E = Equivalent Weight (Molar Mass / n-factor)

20 Golden Facts (NEET)

1. Dilution Paradox: Upon dilution, Specific Conductivity (κ) decreases (ions per mL decrease), but Molar Conductivity (Λ_m) increases (mobility increases).
2. Salt Bridge Function: It maintains electrical neutrality in the two half-cells and minimizes liquid-liquid junction potential. Agar-agar with KNO₃ or NH₄NO₃ is used because velocities of K⁺ and NO₃^- are similar.
3. SHE Potential: The electrode potential of the Standard Hydrogen Electrode is arbitrarily assigned as 0.00 V at all temperatures.
4. Spontaneity Condition: For a cell reaction to be spontaneous, E°_cell must be Positive, which implies ΔG° is Negative.
5. Strong vs Weak: For strong electrolytes (KCl), Λ_m vs √C plot is linear (Debye-Huckel). For weak electrolytes (CH₃COOH), it is a steep curve, and limiting value (Λ°_m) cannot be found by extrapolation.
6. 1 Faraday: It is the charge carried by 1 mole of electrons. 1 F ≈ 96500 C. (Exact: 96487 C).
7. Electrochemical Series: Li has the lowest SRP (-3.05 V, Strongest Reductant). F₂ has the highest SRP (+2.87 V, Strongest Oxidant).
8. Fuel Cells: H₂-O₂ fuel cell was used in the Apollo space program. Water produced was used for drinking. Efficiency ≈ 70% (much higher than thermal plants ~40%).
9. Lead Storage Battery:
Discharge: Pb + PbO₂ + H₂SO₄ → PbSO₄ (Anode & Cathode both become PbSO₄).
Recharge: Reaction reverses. H₂SO₄ is regenerated.
10. Overpotential: Oxygen formation at the anode requires a higher potential than theoretically calculated due to kinetic barriers. This is why Cl₂ is preferred over O₂ in NaCl electrolysis.
11. Corrosion Protection: In Galvanization, Zinc coats Iron. Zinc has lower SRP (-0.76 V) than Fe (-0.44 V), so Zn oxidizes first (Sacrificial Anode).
12. Mercury Cell: Provides constant voltage (1.35 V) throughout its life because the overall reaction involves no ions in solution whose concentration can change.
13. Electrolysis of Aq. NaCl:
Cathode: H₂ gas (Reduction of H₂O > Na⁺).
Anode: Cl₂ gas (Oxidation of Cl^- due to overpotential of O₂).
Solution becomes NaOH (Basic).
14. Concentration Cell: A cell with the same electrodes but different concentrations of electrolyte. E°_cell = 0. EMF is generated due to conc. difference.
15. Unit of Cell Constant: G* = l/A. Unit is cm^-1 or m^-1. It depends on the geometry of the cell, not the electrolyte.
16. 96500 C Logic: 1 Mole e^- deposits 1 Gram Equivalent of substance.
Example: To deposit 1 mol Al (3e^- transfer), you need 3 F (3 × 96500 C).
17. Pure Water Conductivity: Pure water has very low conductivity (~ 3.5 × 10^-5 S/m) due to very low ionization.
18. Cu vs Ag: Cu displaces Ag from AgNO₃ solution because Cu (0.34 V) is placed above Ag (0.80 V) in the activity series (Lower reduction potential oxidizes).
19. pH Calculation: For Hydrogen electrode, E = E° - 0.0591 log[1/H⁺]. Since E°=0, E = -0.0591 × pH.
20. H⁺ Mobility: H⁺ ion has exceptionally high conductivity in aqueous solution due to the Grotthuss mechanism (proton jumping via H-bonds).

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