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Electrochemistry
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Class 12ChemistryChapter 3NCERTCBSE
Electrochemistry - Class 12 Chemistry Chapter 3
📌 Quick Overview: Chapter 3 covers Electrochemistry — the study of the relationship between chemical energy and electrical energy. You will learn about electrochemical cells (galvanic and electrolytic), electrode potentials, EMF, Nernst equation, conductance, Kohlrausch's law, electrolysis, Faraday's laws, and corrosion. This chapter has both theory and numericals — Nernst equation, Faraday's law calculations, and conductance numericals are asked in almost every CBSE board exam.
Topics Covered:
- > Electrochemical Cells
- > Galvanic Cell (Daniel Cell)
- > Electrode Potential and EMF
- > Standard Electrode Potential
- > Nernst Equation
- > Gibbs Energy and EMF
- > Equilibrium Constant from EMF
- > Conductance of Electrolytes
- > Molar Conductance
- > Kohlrausch's Law
- > Electrolysis and Faraday's Laws
- > Products of Electrolysis
- > Batteries and Fuel Cells
- > Corrosion
1. Electrochemical Cells
Electrochemical Cell: A device that converts chemical energy into electrical energy (galvanic cell) or electrical energy into chemical energy (electrolytic cell).
| Property | Galvanic (Voltaic) Cell | Electrolytic Cell |
|---|---|---|
| Energy conversion | Chemical → Electrical | Electrical → Chemical |
| Reaction | Spontaneous (Delta G negative) | Non-spontaneous (Delta G positive) |
| Anode | Negative electrode (oxidation) | Positive electrode (oxidation) |
| Cathode | Positive electrode (reduction) | Negative electrode (reduction) |
| Example | Daniel cell, dry cell, lead acid battery | Electroplating, electrolysis of water |
⭐ Memory Aid:
OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons)
An OX and a RED CAT — Oxidation at Anode, Reduction at Cathode
In galvanic cell: Anode is negative, Cathode is positive.
In electrolytic cell: Anode is positive, Cathode is negative.
OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons)
An OX and a RED CAT — Oxidation at Anode, Reduction at Cathode
In galvanic cell: Anode is negative, Cathode is positive.
In electrolytic cell: Anode is positive, Cathode is negative.
2. Daniel Cell (Galvanic Cell) ⭐
The Daniel cell consists of:
> Zinc rod dipped in ZnSO4 solution (anode — oxidation occurs)
> Copper rod dipped in CuSO4 solution (cathode — reduction occurs)
> Salt bridge connects the two half-cells
Cell Reactions:
Anode (oxidation): Zn(s) → Zn2+(aq) + 2e-
Cathode (reduction): Cu2+(aq) + 2e- → Cu(s)
Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cell Notation: Zn | Zn2+(1M) || Cu2+(1M) | Cu
(Anode written on left, Cathode on right, || = salt bridge)
> Zinc rod dipped in ZnSO4 solution (anode — oxidation occurs)
> Copper rod dipped in CuSO4 solution (cathode — reduction occurs)
> Salt bridge connects the two half-cells
Cell Reactions:
Anode (oxidation): Zn(s) → Zn2+(aq) + 2e-
Cathode (reduction): Cu2+(aq) + 2e- → Cu(s)
Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cell Notation: Zn | Zn2+(1M) || Cu2+(1M) | Cu
(Anode written on left, Cathode on right, || = salt bridge)
Salt Bridge: A U-shaped tube filled with a concentrated solution of an inert electrolyte (like KCl, KNO3, NH4NO3 in agar-agar gel).
Functions of Salt Bridge:
> Completes the electrical circuit by allowing ion flow
> Maintains electrical neutrality in both half-cells
> Prevents mixing of two solutions
> Prevents liquid junction potential
Functions of Salt Bridge:
> Completes the electrical circuit by allowing ion flow
> Maintains electrical neutrality in both half-cells
> Prevents mixing of two solutions
> Prevents liquid junction potential
3. Electrode Potential and EMF ⭐
Electrode Potential: The potential difference that develops between the electrode and the electrolyte solution is called electrode potential.
Standard Electrode Potential (E°): Electrode potential measured under standard conditions (1M concentration, 298K, 1 atm pressure).
Standard Hydrogen Electrode (SHE): The reference electrode with E° = 0.00 V by convention.
Reaction: H+(1M) + e- → 1/2 H2(1 atm), E° = 0.00 V
Standard Electrode Potential (E°): Electrode potential measured under standard conditions (1M concentration, 298K, 1 atm pressure).
Standard Hydrogen Electrode (SHE): The reference electrode with E° = 0.00 V by convention.
Reaction: H+(1M) + e- → 1/2 H2(1 atm), E° = 0.00 V
EMF OF CELL:
E°cell = E°cathode - E°anode
= E°reduction(cathode) - E°reduction(anode)
For Daniel Cell:
E°cell = E°(Cu2+/Cu) - E°(Zn2+/Zn)
= +0.34 - (-0.76)
= +1.10 V
SIGN CONVENTION:
E°cell positive → reaction is spontaneous
E°cell negative → reaction is non-spontaneous
⭐ Standard Reduction Potentials (must memorise key values):
Li+/Li = -3.05 V (strongest reducing agent)
K+/K = -2.93 V
Zn2+/Zn = -0.76 V
Fe2+/Fe = -0.44 V
H+/H2 = 0.00 V (reference)
Cu2+/Cu = +0.34 V
Ag+/Ag = +0.80 V
Au3+/Au = +1.50 V (weakest reducing agent / strongest oxidising agent)
F2/F- = +2.87 V
Higher (more positive) E° = better oxidising agent
Lower (more negative) E° = better reducing agent
Li+/Li = -3.05 V (strongest reducing agent)
K+/K = -2.93 V
Zn2+/Zn = -0.76 V
Fe2+/Fe = -0.44 V
H+/H2 = 0.00 V (reference)
Cu2+/Cu = +0.34 V
Ag+/Ag = +0.80 V
Au3+/Au = +1.50 V (weakest reducing agent / strongest oxidising agent)
F2/F- = +2.87 V
Higher (more positive) E° = better oxidising agent
Lower (more negative) E° = better reducing agent
4. Nernst Equation ⭐ (Very Important)
NERNST EQUATION:
E = E° - (RT/nF) ln Q
E = E° - (0.0592/n) log Q [at 298K]
where:
E = cell potential at given conditions
E° = standard cell potential
R = 8.314 J/mol/K
T = temperature in K
n = number of electrons transferred
F = Faraday constant = 96500 C/mol
Q = reaction quotient
For the reaction: aA + bB → cC + dD
Q = [C]^c [D]^d / [A]^a [B]^b
Example 1: Find the EMF of Daniel cell at 298K when [Zn2+] = 0.1 M and [Cu2+] = 0.01 M. (E°cell = 1.10 V, n = 2)
Q = [Zn2+]/[Cu2+] = 0.1/0.01 = 10
E = E° - (0.0592/n) log Q
E = 1.10 - (0.0592/2) log 10
E = 1.10 - (0.0296)(1)
E = 1.10 - 0.0296
E = 1.0704 V
Q = [Zn2+]/[Cu2+] = 0.1/0.01 = 10
E = E° - (0.0592/n) log Q
E = 1.10 - (0.0592/2) log 10
E = 1.10 - (0.0296)(1)
E = 1.10 - 0.0296
E = 1.0704 V
Example 2: Calculate the equilibrium constant for the reaction: Zn + Cu2+ → Zn2+ + Cu. (E°cell = 1.10 V)
At equilibrium, E = 0 and Q = Kc
0 = E° - (0.0592/n) log Kc
log Kc = n x E° / 0.0592 = 2 x 1.10 / 0.0592 = 37.16
Kc = 10^37.16 = 1.45 x 10^37
At equilibrium, E = 0 and Q = Kc
0 = E° - (0.0592/n) log Kc
log Kc = n x E° / 0.0592 = 2 x 1.10 / 0.0592 = 37.16
Kc = 10^37.16 = 1.45 x 10^37
5. Gibbs Energy and EMF ⭐
RELATION BETWEEN GIBBS ENERGY AND EMF:
Delta G = -nFE
Delta G° = -nFE°
where:
Delta G = Gibbs free energy change
n = number of electrons transferred
F = Faraday constant (96500 C/mol)
E = EMF of cell
RELATION WITH EQUILIBRIUM CONSTANT:
Delta G° = -RT ln K = -nFE°
log K = nE° / 0.0592 [at 298K]
SPONTANEITY:
E > 0 → Delta G < 0 → Spontaneous reaction
E < 0 → Delta G > 0 → Non-spontaneous
E = 0 → Delta G = 0 → Equilibrium
Example: Calculate Delta G° for Daniel cell. (E°cell = 1.10 V, n = 2)
Delta G° = -nFE° = -2 x 96500 x 1.10
= -212300 J/mol
= -212.3 kJ/mol
Delta G° = -nFE° = -2 x 96500 x 1.10
= -212300 J/mol
= -212.3 kJ/mol
6. Conductance of Electrolytic Solutions ⭐
RESISTANCE AND CONDUCTANCE:
Resistance R = rho x (l/A)
Conductance G = 1/R = kappa x (A/l)
Cell constant = l/A
kappa (conductivity) = G x (l/A) = G x cell constant
Unit of kappa: S/m or S/cm
MOLAR CONDUCTANCE (Lambda_m):
Lambda_m = (kappa x 1000) / M [if kappa in S/cm, M in mol/L]
Lambda_m = kappa / C [if kappa in S/m, C in mol/m^3]
Unit: S cm^2 mol^-1
EQUIVALENT CONDUCTANCE:
Lambda_eq = kappa x 1000 / N
Effect of Dilution on Conductance:
> Conductivity (kappa) DECREASES with dilution (fewer ions per volume)
> Molar conductance (Lambda_m) INCREASES with dilution (more ions available per mole, more mobility)
> Conductivity (kappa) DECREASES with dilution (fewer ions per volume)
> Molar conductance (Lambda_m) INCREASES with dilution (more ions available per mole, more mobility)
7. Kohlrausch's Law ⭐
Kohlrausch's Law of Independent Migration of Ions: At infinite dilution, each ion makes a definite contribution to the molar conductance of the electrolyte, irrespective of the nature of the other ion with which it is associated.
Lambda°m = v+ lambda°+ + v- lambda°-
where v+ and v- are the number of cations and anions per formula unit, and lambda° values are limiting molar conductances of individual ions.
Lambda°m = v+ lambda°+ + v- lambda°-
where v+ and v- are the number of cations and anions per formula unit, and lambda° values are limiting molar conductances of individual ions.
APPLICATIONS OF KOHLRAUSCH'S LAW:
1. Find Lambda°m of weak electrolytes (cannot be measured directly):
Lambda°m(CH3COOH) = Lambda°m(CH3COONa) + Lambda°m(HCl) - Lambda°m(NaCl)
2. Degree of dissociation of weak electrolyte:
alpha = Lambda_m / Lambda°m
3. Dissociation constant:
Ka = C.alpha^2 / (1-alpha)
Example: Calculate Lambda°m for acetic acid using:
Lambda°m(HCl) = 426 S cm^2/mol
Lambda°m(CH3COONa) = 91 S cm^2/mol
Lambda°m(NaCl) = 126 S cm^2/mol
Lambda°m(CH3COOH) = Lambda°m(CH3COONa) + Lambda°m(HCl) - Lambda°m(NaCl)
= 91 + 426 - 126
= 391 S cm^2/mol
Lambda°m(HCl) = 426 S cm^2/mol
Lambda°m(CH3COONa) = 91 S cm^2/mol
Lambda°m(NaCl) = 126 S cm^2/mol
Lambda°m(CH3COOH) = Lambda°m(CH3COONa) + Lambda°m(HCl) - Lambda°m(NaCl)
= 91 + 426 - 126
= 391 S cm^2/mol
Example 2: The conductivity of 0.001 M acetic acid is 4 x 10^(-5) S/cm. If Lambda°m = 390 S cm^2/mol, find the degree of dissociation.
Lambda_m = kappa x 1000/M = (4 x 10^(-5) x 1000)/0.001 = 40 S cm^2/mol
alpha = Lambda_m / Lambda°m = 40/390 = 0.1026 (about 10.26%)
Lambda_m = kappa x 1000/M = (4 x 10^(-5) x 1000)/0.001 = 40 S cm^2/mol
alpha = Lambda_m / Lambda°m = 40/390 = 0.1026 (about 10.26%)
8. Electrolysis and Faraday's Laws ⭐
Electrolysis: The decomposition of an electrolyte by passing electricity through its molten or aqueous solution is called electrolysis.
8.1 Faraday's First Law
The mass of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of charge (electricity) passed.
m = Z x Q = Z x I x t
where m = mass deposited, Z = electrochemical equivalent, Q = charge in coulombs, I = current in amperes, t = time in seconds.
m = Z x Q = Z x I x t
where m = mass deposited, Z = electrochemical equivalent, Q = charge in coulombs, I = current in amperes, t = time in seconds.
8.2 Faraday's Second Law
When the same quantity of charge is passed through different electrolytes connected in series, the masses of substances deposited at the respective electrodes are proportional to their equivalent masses (molar mass/n factor).
m1/m2 = E1/E2
where E = equivalent mass = Molar mass / n factor
m1/m2 = E1/E2
where E = equivalent mass = Molar mass / n factor
FARADAY'S LAW CALCULATION:
m = (M x I x t) / (n x F)
where:
m = mass deposited (g)
M = molar mass of substance
I = current (amperes)
t = time (seconds)
n = number of electrons in electrode reaction
F = 96500 C/mol (1 Faraday)
ONE FARADAY = charge of 1 mole of electrons = 96500 C
Example 1: How much copper is deposited when 2A current is passed through CuSO4 solution for 30 minutes? (M of Cu = 63.5, n = 2)
t = 30 x 60 = 1800 s
m = (M x I x t) / (n x F)
= (63.5 x 2 x 1800) / (2 x 96500)
= 228600 / 193000
= 1.185 g
t = 30 x 60 = 1800 s
m = (M x I x t) / (n x F)
= (63.5 x 2 x 1800) / (2 x 96500)
= 228600 / 193000
= 1.185 g
Example 2: Calculate the volume of oxygen liberated at STP when 0.5A is passed through dilute H2SO4 for 2 hours.
At anode: 2H2O → O2 + 4H+ + 4e- (n = 4 for O2)
t = 2 x 3600 = 7200 s
m(O2) = (32 x 0.5 x 7200) / (4 x 96500) = 115200/386000 = 0.2984 g
Moles of O2 = 0.2984/32 = 0.00933 mol
Volume at STP = 0.00933 x 22.4 = 0.209 L = 209 mL
At anode: 2H2O → O2 + 4H+ + 4e- (n = 4 for O2)
t = 2 x 3600 = 7200 s
m(O2) = (32 x 0.5 x 7200) / (4 x 96500) = 115200/386000 = 0.2984 g
Moles of O2 = 0.2984/32 = 0.00933 mol
Volume at STP = 0.00933 x 22.4 = 0.209 L = 209 mL
9. Products of Electrolysis ⭐
⭐ Rules for Predicting Products at Cathode (reduction):
The ion with higher reduction potential is preferentially reduced.
> In dilute solution: H+ (from water) reduced → H2 gas
> Ions of metals below H in activity series (Cu2+, Ag+, Au3+): Metal deposited
> Ions of metals above H (Na+, K+, Ca2+, Mg2+, Al3+): Water reduced → H2
Rules for Products at Anode (oxidation):
> Inert electrode (Pt, graphite): anion oxidised
> Active electrode (Cu, Ag): electrode itself dissolves
> In presence of Cl-, Br-, I-: halide ion oxidised preferentially over water
> In dilute solution without halide: water oxidised → O2
The ion with higher reduction potential is preferentially reduced.
> In dilute solution: H+ (from water) reduced → H2 gas
> Ions of metals below H in activity series (Cu2+, Ag+, Au3+): Metal deposited
> Ions of metals above H (Na+, K+, Ca2+, Mg2+, Al3+): Water reduced → H2
Rules for Products at Anode (oxidation):
> Inert electrode (Pt, graphite): anion oxidised
> Active electrode (Cu, Ag): electrode itself dissolves
> In presence of Cl-, Br-, I-: halide ion oxidised preferentially over water
> In dilute solution without halide: water oxidised → O2
| Electrolyte | Electrode | Cathode Product | Anode Product |
|---|---|---|---|
| Molten NaCl | Inert (Pt) | Na metal | Cl2 gas |
| Aqueous NaCl (dil) | Inert (Pt) | H2 gas | O2 gas |
| Aqueous NaCl (conc) | Inert (Pt) | H2 gas | Cl2 gas |
| Aqueous CuSO4 | Inert (Pt) | Cu metal | O2 gas |
| Aqueous CuSO4 | Cu electrode | Cu metal | Cu dissolves |
| Aqueous H2SO4 | Inert (Pt) | H2 gas | O2 gas |
10. Batteries and Fuel Cells
10.1 Primary Batteries (Non-rechargeable)
Dry Cell (Leclanche Cell):
Anode: Zinc container
Cathode: Graphite rod surrounded by MnO2 + NH4Cl paste
EMF = 1.5 V
Use: Torches, clocks, remote controls
Anode: Zinc container
Cathode: Graphite rod surrounded by MnO2 + NH4Cl paste
EMF = 1.5 V
Use: Torches, clocks, remote controls
10.2 Secondary Batteries (Rechargeable)
Lead Storage Battery (Lead Acid Battery):
Anode: Pb
Cathode: PbO2
Electrolyte: 38% H2SO4
EMF per cell = 2V; 6 cells → 12V (car battery)
Discharge:
Anode: Pb + SO4^2- → PbSO4 + 2e-
Cathode: PbO2 + 4H+ + SO4^2- + 2e- → PbSO4 + 2H2O
Charging: Reactions are reversed.
Nickel-Cadmium Cell:
Anode: Cd, Cathode: NiO(OH), longer life than lead battery.
Anode: Pb
Cathode: PbO2
Electrolyte: 38% H2SO4
EMF per cell = 2V; 6 cells → 12V (car battery)
Discharge:
Anode: Pb + SO4^2- → PbSO4 + 2e-
Cathode: PbO2 + 4H+ + SO4^2- + 2e- → PbSO4 + 2H2O
Charging: Reactions are reversed.
Nickel-Cadmium Cell:
Anode: Cd, Cathode: NiO(OH), longer life than lead battery.
10.3 Fuel Cells
Hydrogen-Oxygen Fuel Cell:
> Anode: H2 fed → H2 + 2OH- → 2H2O + 2e-
> Cathode: O2 fed → O2 + 2H2O + 4e- → 4OH-
> Overall: 2H2 + O2 → 2H2O
> Electrolyte: KOH solution
> Electrodes: Porous carbon with Pt catalyst
> Used in NASA space vehicles
> Advantage: Pollution-free, high efficiency (70% vs 40% for thermal)
> Anode: H2 fed → H2 + 2OH- → 2H2O + 2e-
> Cathode: O2 fed → O2 + 2H2O + 4e- → 4OH-
> Overall: 2H2 + O2 → 2H2O
> Electrolyte: KOH solution
> Electrodes: Porous carbon with Pt catalyst
> Used in NASA space vehicles
> Advantage: Pollution-free, high efficiency (70% vs 40% for thermal)
11. Corrosion ⭐
Corrosion: The slow deterioration of metals due to reaction with the environment (air, moisture, chemicals) is called corrosion. Rusting of iron is the most common example.
Electrochemical Theory of Rusting:
Iron acts as a galvanic cell in presence of moisture and CO2/electrolytes.
Anode (Fe, oxidation): Fe → Fe2+ + 2e- (at pits, scratches)
Cathode (O2, reduction): O2 + 4H+ + 4e- → 2H2O (at surface)
Fe2+ + 2OH- → Fe(OH)2 → 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3
Fe(OH)3 → Fe2O3 . xH2O (rust — reddish brown)
Iron acts as a galvanic cell in presence of moisture and CO2/electrolytes.
Anode (Fe, oxidation): Fe → Fe2+ + 2e- (at pits, scratches)
Cathode (O2, reduction): O2 + 4H+ + 4e- → 2H2O (at surface)
Fe2+ + 2OH- → Fe(OH)2 → 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3
Fe(OH)3 → Fe2O3 . xH2O (rust — reddish brown)
⭐ Prevention of Corrosion:
> Barrier protection: painting, oiling, greasing, coating with polymers
> Galvanisation: coating iron with zinc (zinc acts as sacrificial anode)
> Tinning: coating with tin (but if tin scratches, iron corrodes faster)
> Electroplating: coat with Ni, Cr (chromium plating)
> Alloying: making stainless steel (Fe + Cr + Ni)
> Cathodic protection (sacrificial anode): Mg or Zn blocks attached to iron pipes/ships — these oxidise preferentially
> Barrier protection: painting, oiling, greasing, coating with polymers
> Galvanisation: coating iron with zinc (zinc acts as sacrificial anode)
> Tinning: coating with tin (but if tin scratches, iron corrodes faster)
> Electroplating: coat with Ni, Cr (chromium plating)
> Alloying: making stainless steel (Fe + Cr + Ni)
> Cathodic protection (sacrificial anode): Mg or Zn blocks attached to iron pipes/ships — these oxidise preferentially
12. Important Board Exam Questions
Q1. Write the cell notation for Daniel cell. What is the standard EMF? Write the cell reactions at anode and cathode.
Cell notation: Zn(s) | Zn2+(1M) || Cu2+(1M) | Cu(s)
Anode (oxidation): Zn → Zn2+ + 2e-; E° = +0.76 V (as oxidation)
Cathode (reduction): Cu2+ + 2e- → Cu; E° = +0.34 V
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = +1.10 V
Anode (oxidation): Zn → Zn2+ + 2e-; E° = +0.76 V (as oxidation)
Cathode (reduction): Cu2+ + 2e- → Cu; E° = +0.34 V
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = +1.10 V
Q2. Write the Nernst equation for the cell: Zn | Zn2+ || Cu2+ | Cu. Calculate EMF when [Cu2+] = 0.001 M and [Zn2+] = 0.1 M. (E°cell = 1.10 V)
Nernst equation: E = E° - (0.0592/n) log ([Zn2+]/[Cu2+])
E = 1.10 - (0.0592/2) log (0.1/0.001)
E = 1.10 - (0.0296) log (100)
E = 1.10 - (0.0296)(2)
E = 1.10 - 0.0592
E = 1.0408 V
E = 1.10 - (0.0592/2) log (0.1/0.001)
E = 1.10 - (0.0296) log (100)
E = 1.10 - (0.0296)(2)
E = 1.10 - 0.0592
E = 1.0408 V
Q3. The conductivity of 0.02 M KCl solution is 2.768 x 10^(-3) S/cm. Calculate its molar conductance.
Lambda_m = (kappa x 1000) / M
= (2.768 x 10^(-3) x 1000) / 0.02
= 2.768 / 0.02
= 138.4 S cm^2/mol
= (2.768 x 10^(-3) x 1000) / 0.02
= 2.768 / 0.02
= 138.4 S cm^2/mol
Q4. How long must a current of 3A be passed through a solution of silver nitrate to deposit 10.8g of silver? (M of Ag = 108, n = 1, F = 96500 C)
m = (M x I x t) / (n x F)
10.8 = (108 x 3 x t) / (1 x 96500)
t = (10.8 x 96500) / (108 x 3)
t = 1042200 / 324
t = 3216 seconds = 53.6 minutes
10.8 = (108 x 3 x t) / (1 x 96500)
t = (10.8 x 96500) / (108 x 3)
t = 1042200 / 324
t = 3216 seconds = 53.6 minutes
Q5. Calculate the standard Gibbs energy change and equilibrium constant for the reaction: Zn + Cu2+ → Zn2+ + Cu. (E°cell = 1.10 V, n = 2)
Delta G° = -nFE° = -2 x 96500 x 1.10 = -212300 J = -212.3 kJ/mol
log Kc = nE°/0.0592 = (2 x 1.10)/0.0592 = 37.16
Kc = 1.45 x 10^37
log Kc = nE°/0.0592 = (2 x 1.10)/0.0592 = 37.16
Kc = 1.45 x 10^37
Q6. Explain the electrochemical theory of corrosion of iron. How does galvanisation prevent rusting?
In the presence of moisture and CO2, iron acts like a galvanic cell. Pure iron acts as cathode while impurities/strained regions act as anode. At the anode, Fe → Fe2+ + 2e- (oxidation). At the cathode, O2 + 4H+ + 4e- → 2H2O (reduction). Fe2+ combines with OH- to form Fe(OH)2, which is further oxidised to Fe(OH)3 and finally to rust (Fe2O3.xH2O).
Galvanisation: Iron is coated with zinc. Since zinc has a lower reduction potential (-0.76 V) than iron (-0.44 V), zinc acts as a sacrificial anode and corrodes preferentially, protecting the iron beneath even if the coating is scratched.
Galvanisation: Iron is coated with zinc. Since zinc has a lower reduction potential (-0.76 V) than iron (-0.44 V), zinc acts as a sacrificial anode and corrodes preferentially, protecting the iron beneath even if the coating is scratched.
13. Key Formulas at a Glance
CELL EMF:
E°cell = E°cathode - E°anode (reduction potentials)
E°cell positive → spontaneous
NERNST EQUATION (at 298K):
E = E° - (0.0592/n) log Q
GIBBS ENERGY:
Delta G = -nFE
Delta G° = -nFE°
log K = nE°/0.0592 [at 298K]
CONDUCTANCE:
G = kappa x (A/l)
Lambda_m = kappa x 1000 / M [S cm^2/mol]
alpha = Lambda_m / Lambda°m
KOHLRAUSCH'S LAW:
Lambda°m = v+ lambda°+ + v- lambda°-
FARADAY'S LAWS:
m = (M x I x t) / (n x F)
1 Faraday = 96500 C = charge of 1 mole electrons
KEY VALUES:
E°(Cu2+/Cu) = +0.34 V
E°(Zn2+/Zn) = -0.76 V
E°(Ag+/Ag) = +0.80 V
E°(H+/H2) = 0.00 V (reference)
14. MCQ Practice (1 Mark)
1. In a galvanic cell, oxidation occurs at:
(a) Cathode (b) Anode (c) Salt bridge (d) Both electrodes
Answer: (b) Anode
2. The standard EMF of Daniel cell is:
(a) 0.76 V (b) 0.34 V (c) 1.10 V (d) 1.44 V
Answer: (c) 1.10 V = 0.34 - (-0.76)
3. Which quantity increases on dilution of an electrolytic solution?
(a) Conductivity (b) Resistance (c) Molar conductance (d) Both b and c
Answer: (d) Both Resistance and Molar conductance increase on dilution
4. 1 Faraday of electricity deposits how many grams of copper from CuSO4? (M = 63.5, n = 2)
(a) 63.5 g (b) 31.75 g (c) 127 g (d) 6.35 g
Answer: (b) 31.75 g = 63.5/2 (1 Faraday deposits 1 equivalent = M/n)
5. Which of the following is used as anode in dry cell?
(a) Carbon (b) MnO2 (c) Zinc (d) Copper
Answer: (c) Zinc — zinc container acts as anode
6. The Nernst equation at equilibrium gives:
(a) E = E° (b) E = 0 (c) E = 1 (d) E = infinity
Answer: (b) E = 0 — at equilibrium, net cell potential is zero
15. Exam Tips
- Always write E°cell = E°cathode - E°anode (both as reduction potentials) — never mix up reduction and oxidation potentials.
- In Nernst equation: n is the number of electrons transferred in the BALANCED cell reaction — find this from the half-reactions.
- For Faraday's law: always convert time to seconds and use F = 96500 C/mol. Double-check n value for each element (Cu has n=2, Ag has n=1, Al has n=3).
- Molar conductance INCREASES with dilution; Conductivity DECREASES with dilution — very commonly confused in MCQs.
- For Kohlrausch's law application: to find Lambda°m of weak electrolyte, combine strong electrolytes cleverly so that the weak electrolyte's ions appear.
- In galvanic cell: Anode is negative, Cathode is positive. In electrolytic cell: Anode is positive, Cathode is negative.
- Galvanisation uses Zinc because Zn has lower reduction potential than Fe — Zn acts as sacrificial anode.
- Hydrogen-oxygen fuel cell is most likely asked — know both electrode reactions and the electrolyte (KOH).
✅ Summary: Electrochemistry is a high-scoring chapter combining theory and numericals. Focus on: (1) cell notation and EMF calculation, (2) Nernst equation numericals, (3) Gibbs energy-EMF relation, (4) molar conductance and Kohlrausch's law, (5) Faraday's law calculations, and (6) corrosion and prevention. These topics together cover 90% of board exam questions from this chapter. Practice at least 5 Faraday's law and 5 Nernst equation numericals before the exam.
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