1. Exam Strategy
Last-Day Revision
High-Priority Topics (Must Know)
- Nernst Equation numericals — relationship between E°cell, Ecell, Q, and Kc. Expect 1–2 numericals every year.
- Kohlrausch Law calculation — finding Λ°m for weak electrolytes using strong electrolyte data, then calculating α and Ka.
- Faraday's Laws numericals — mass deposited, time, current, charge calculations.
- Galvanic vs Electrolytic cell — differences in anode/cathode polarity, ΔG signs, spontaneity.
- Lead storage battery — complete reactions during charging and discharging.
- Corrosion mechanism — electrochemical explanation of rusting.
Formula Quick-Reference
- Nernst: \(E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q\)
- Equilibrium: \(E^\circ_{cell} = \frac{0.0591}{n}\log K_c\)
- Gibbs Energy: \(\Delta_r G^\circ = -nFE^\circ_{cell}\)
- Molar Conductivity: \(\Lambda_m = \frac{\kappa \times 1000}{M}\)
- Degree of dissociation: \(\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}\)
- Dissociation constant: \(K_a = \frac{c\alpha^2}{1-\alpha}\)
- Faraday's Law: \(m = \frac{M \times I \times t}{n \times F}\)
Exam Technique Tips
- Always write the half-reactions before calculating n (electrons transferred).
- For Nernst: identify what goes in numerator vs denominator of log Q (products over reactants, omit pure solids/liquids).
- In conductivity problems: check units carefully — κ in S cm⁻¹ and M in mol L⁻¹ give Λm in S cm² mol⁻¹.
- For batteries: remember discharging = Galvanic cell; charging = Electrolytic cell.
- Assertion-Reasoning: Always read both statements independently before deciding.
2. Chapter Mindmap
Electrochemistry — Big Picture
The chapter revolves around two central themes:
- Electrochemical Cells → Galvanic (chemical → electrical) and Electrolytic (electrical → chemical)
- Electrolytic Conduction → how ions carry current in solutions
Branch 1: Electrochemical Cells
- Galvanic Cells → Daniell Cell → Standard Electrode Potential (SHE) → Electrochemical Series
- Nernst Equation → Ecell at non-standard conditions → Equilibrium Constant Kc → Gibbs Energy ΔG
- Electrolytic Cells → Faraday's Laws → Products of electrolysis → Commercial cells (Batteries, Fuel Cells)
- Applications → Corrosion (rusting = electrochemical process)
Branch 2: Electrolytic Conduction
- Resistance → Resistivity → Conductance → Conductivity (κ)
- Cell Constant → Molar Conductivity (Λm)
- Variation with concentration: Strong vs Weak electrolytes
- Kohlrausch Law → Λ°m for weak electrolytes → Degree of dissociation (α) → Ka
Key Connections
How Everything Connects
\(E^\circ_{cell}\) connects to \(\Delta G^\circ\) and \(K_c\) through:
\(\Delta_r G^\circ = -nFE^\circ_{cell} = -RT\ln K_c\)
Conductivity connects to dissociation:
\(\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} \rightarrow K_a = \frac{c\alpha^2}{1-\alpha}\)
3. Electrochemical Cells
Concept Overview
Electrochemistry is the study of the production of electricity from the energy released during spontaneous chemical reactions, and the use of electrical energy to bring about non-spontaneous chemical transformations.
An Electrochemical Cell is the device used to perform these conversions. They are broadly classified into two main types based on the direction of energy transfer.
Core Distinction
Galvanic vs. Electrolytic Cells
| Feature | Galvanic (Voltaic) Cell | Electrolytic Cell |
|---|---|---|
| Energy Conversion | Chemical → Electrical | Electrical → Chemical |
| Reaction Type | Spontaneous (ΔG < 0) | Non-Spontaneous (ΔG > 0) |
| Anode Polarity | Negative (−) | Positive (+) |
| Cathode Polarity | Positive (+) | Negative (−) |
| Setup | Two half-cells + salt bridge | One container, external power supply |
| E°cell | Positive | Applied voltage needed |
Universal Rule: Oxidation always at Anode; Reduction always at Cathode — regardless of cell type. (Remember: Anode–Oxidation, Cathode–Reduction — AN OX, RED CAT)
Practice Worksheets
- In an electrolytic cell, the cathode is:
Answer: (B) In an electrolytic cell, the cathode is connected to the negative terminal of the external power source, and reduction always occurs at the cathode.
- Which thermodynamic condition is true for a functioning Galvanic cell?
Answer: (C) Galvanic cells run on spontaneous redox reactions. For any spontaneous process at constant T and P, ΔG must be negative.
- What is the fundamental difference in energy conversion between a Galvanic cell and an Electrolytic cell?
A Galvanic cell converts chemical energy into electrical energy. An electrolytic cell uses electrical energy to drive a non-spontaneous chemical reaction.
- State the polarity of the anode in both types of electrochemical cells and explain the reason.
Galvanic cell: Anode is negative — it is the source of electrons produced by spontaneous oxidation.
Electrolytic cell: Anode is positive — it is connected to the positive terminal of the external battery to forcefully strip electrons from ions.
- Discuss the role of ΔG° in defining the two main types of electrochemical cells. Give one real-world example of each.
Role of ΔG°: If ΔG° < 0, the reaction is spontaneous → Galvanic cell (energy is released as electrical work). If ΔG° > 0, the reaction is non-spontaneous → Electrolytic cell (external electrical energy must be supplied to force the reaction).
Galvanic example: AA alkaline battery in a flashlight — spontaneous reaction generates current.
Electrolytic example: Electroplating chrome onto a steel bumper — external power forces reduction of chromium ions onto the steel surface.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Oxidation always takes place at the anode, regardless of whether it is a Galvanic or electrolytic cell.
Reason (R): The anode is always the negative terminal in any electrochemical cell.Answer: (C) A is true, but R is false. While oxidation always happens at the anode, the anode is positive in an electrolytic cell and negative in a Galvanic cell.
4. Galvanic Cells
The Daniell Cell
The Daniell Cell is the classic example of a Galvanic cell. It consists of two half-cells:
- Anode (Zinc half-cell): Zinc rod in ZnSO₄ solution. Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻
- Cathode (Copper half-cell): Copper rod in CuSO₄ solution. Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall reaction: \(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\)
Cell Notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
[Diagram: Daniell Cell with Zn anode, Cu cathode, salt bridge, voltmeter, and electron flow direction]
Must Know
Salt Bridge — Purpose and Function
- Prevents direct mixing of the two electrolyte solutions
- Maintains electrical neutrality in both half-cells by allowing migration of ions (KCl, KNO₃, NH₄NO₃ in agar-agar gel)
- Without a salt bridge, the cell stops working quickly as charge builds up
Cell Notation Convention
Single vertical line (|) = phase boundary (e.g., solid | solution)
Double vertical line (||) = salt bridge
Anode written on LEFT; Cathode written on RIGHT
Example: Zn(s) | Zn²⁺(1M) || Cu²⁺(1M) | Cu(s) → E°cell = +1.10 V
Practice Worksheets
- In a standard Galvanic cell, electrons flow through the external circuit from:
Answer: (B) Electrons are produced at the anode (by oxidation) and flow through the external circuit to the cathode (where they are consumed by reduction).
- What is the primary function of the salt bridge in a Daniell cell?
Answer: (C) The salt bridge allows migration of ions (NOT electrons) to maintain electrical neutrality as ions are consumed/produced in each half-cell. Electrons flow through the external wire.
- Write the cell notation for the Daniell cell and identify the anode and cathode.
Cell notation: Zn(s) | Zn²⁺(aq, 1M) || Cu²⁺(aq, 1M) | Cu(s)
Anode (left): Zinc — undergoes oxidation (Zn → Zn²⁺ + 2e⁻)
Cathode (right): Copper — undergoes reduction (Cu²⁺ + 2e⁻ → Cu) - What happens to the mass of the zinc electrode and the blue colour of the copper sulphate solution in a Daniell cell as the cell operates?
The zinc electrode loses mass because Zn is oxidized and dissolves as Zn²⁺ ions into the solution. The blue colour of CuSO₄ fades because Cu²⁺ ions are reduced to Cu metal which deposits on the copper electrode, reducing the concentration of Cu²⁺ in solution.
- Describe the working of a Daniell cell. How is electrical energy generated from this device? Include all half-reactions and the overall cell reaction.
Construction: A zinc rod is dipped in ZnSO₄ solution (anode half-cell) and a copper rod is dipped in CuSO₄ solution (cathode half-cell). The two half-cells are connected by a salt bridge (KCl in agar-agar) and an external wire with a voltmeter.
Working: Zinc is more reactive than copper. It spontaneously oxidizes at the anode, releasing electrons. These electrons flow through the external circuit to the copper cathode, where Cu²⁺ ions gain electrons and are reduced to Cu metal.
Anode: Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = +1.10 V
The salt bridge allows K⁺ ions to migrate towards the cathode half-cell (where anions are depleted) and Cl⁻ ions towards the anode half-cell (where cations are increasing), maintaining electrical neutrality.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): The blue colour of CuSO₄ solution fades as the Daniell cell operates.
Reason (R): Cu²⁺ ions are deposited as copper metal at the cathode, reducing their concentration in solution.Answer: (A) Both A and R are true, and R correctly explains A. The blue colour of CuSO₄ is due to Cu²⁺ ions; as they are reduced at the cathode, their concentration falls and the solution becomes less blue.
5. Standard Electrode Potential
Concept Overview
When a metal electrode is dipped into a solution containing its own ions, a potential difference develops at the interface. This is the electrode potential.
If all species are at standard conditions (1 mol L⁻¹ concentration, 1 bar pressure, 298 K), the potential is called the Standard Electrode Potential (E°).
IUPAC Convention: Standard electrode potentials are always listed as Standard Reduction Potentials. Higher (more positive) E° = stronger tendency to be reduced = stronger oxidizing agent.
Core Reference
The Standard Hydrogen Electrode (SHE)
[Diagram: SHE — Pt electrode in 1M HCl with H₂ gas at 1 bar bubbling, E° = 0.00 V]
- Platinum foil coated with platinum black (catalyst), dipped in 1M H⁺ solution
- Pure H₂ gas at 1 bar bubbled through
- Assigned E° = 0.00 V at all temperatures by convention
- All other electrode potentials are measured relative to the SHE
Half-Reaction: H⁺(aq) + e⁻ → ½H₂(g), E° = 0.00 V
The Electrochemical Series
Electrodes arranged in order of increasing standard reduction potential:
- Li⁺/Li: −3.05 V — strongest reducing agent (most easily oxidized)
- Zn²⁺/Zn: −0.76 V
- H⁺/H₂ (SHE): 0.00 V
- Cu²⁺/Cu: +0.34 V
- Ag⁺/Ag: +0.80 V
- F₂/F⁻: +2.87 V — strongest oxidizing agent (most easily reduced)
Calculating E°cell: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)
Practice Worksheets
- Given E°: K⁺/K = −2.93V, Ag⁺/Ag = +0.80V, Hg²⁺/Hg = +0.79V, Mg²⁺/Mg = −2.37V. The strongest reducing agent is:
Answer: (D) K. The lowest (most negative) reduction potential means the highest tendency to be oxidized. The substance most easily oxidized is the strongest reducing agent.
- The standard electrode potential of the SHE is exactly zero:
Answer: (C) It is arbitrarily assigned 0.00 V at all temperatures as a universal baseline by convention.
- Standard conditions for measuring E° include all EXCEPT:
Answer: (B) Modern IUPAC standard pressure is 1 bar, not 1 atm (though very close).
- Why is platinum used in the Standard Hydrogen Electrode?
Platinum is chemically inert — it does not participate in the reaction. It provides a conducting surface for electron transfer and adsorbs hydrogen gas, facilitating the H⁺/H₂ equilibrium without contaminating the system.
- Can a copper vessel be used to store ZnSO₄ solution? (E°Cu²⁺/Cu = +0.34V, E°Zn²⁺/Zn = −0.76V)
Yes. Copper has a higher reduction potential (+0.34V) than zinc (−0.76V). This means copper prefers to remain as solid metal. Copper cannot displace zinc ions from ZnSO₄ solution, so no reaction occurs and the vessel is safe.
- What happens to E° if the half-reaction stoichiometry is multiplied by 2?
E° remains unchanged. Standard electrode potential is an intensive property — it depends on the nature of the species, not the quantity involved.
- Detail the experimental setup to determine E° for the Zn²⁺/Zn half-cell. Include cell notation.
Setup: Zinc rod dipped in 1M ZnSO₄ (298K) connected via salt bridge and external voltmeter to a SHE (Pt electrode in 1M HCl, H₂ at 1 bar).
Observation: Voltmeter reads 0.76V. Electrons flow from Zn to SHE → Zn is anode.
Calculation: E°cell = E°cathode − E°anode → 0.76 = 0.00 − E°Zn²⁺/Zn → E°Zn²⁺/Zn = −0.76V
Cell Notation: Zn(s) | Zn²⁺(aq, 1M) || H⁺(aq, 1M) | H₂(g, 1bar) | Pt(s)
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Absolute electrode potential of a single half-cell cannot be measured.
Reason (R): A half-cell reaction cannot occur independently; it needs another half-cell to complete the circuit.Answer: (A) Both A and R true; R correctly explains A. A voltmeter requires a complete circuit and measures the difference between two potentials. - Assertion (A): Standard reduction potential of F₂ is highest in the electrochemical series.
Reason (R): F₂ gas is the strongest reducing agent.Answer: (C) A is true, R is false. Because F₂ has the highest reduction potential, it is the strongest oxidizing agent, not reducing agent.
6. Nernst Equation
Why We Need the Nernst Equation
E°cell is measured at standard conditions (all concentrations = 1M, 298K). In reality, concentrations are rarely 1M. The Nernst equation allows us to calculate the actual cell potential (Ecell) at any concentration.
Core Formula
Nernst Equation at 298 K
\(E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q\)
Where:
- \(E_{cell}\) = actual cell potential at given conditions
- \(E^\circ_{cell}\) = standard cell potential
- \(n\) = number of electrons transferred in the balanced reaction
- \(Q\) = reaction quotient = [products]/[reactants] (omit pure solids and liquids)
General form (any temperature): \(E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q\)
Common exam trap: For the Daniell cell, \(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\). Students often invert this. Always write products over reactants.
What Happens to Ecell When Concentration Changes?
- Increasing [products] → Q increases → log Q increases → Ecell decreases
- Increasing [reactants] → Q decreases → log Q decreases → Ecell increases
- Diluting product → Ecell increases
- At equilibrium: Ecell = 0, Q = Kc
Practice Worksheets
- If concentration of products increases, cell potential:
Answer: (B) Increasing [products] increases Q. Since Ecell = E°cell − (0.0591/n)logQ, a larger Q means a larger value is subtracted, so Ecell decreases.
- For the cell Zn|Zn²⁺(0.01M)||Cu²⁺(1M)|Cu, the Ecell compared to E°cell will be:
Answer: (B) Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01 < 1. So logQ < 0, meaning the subtracted term is negative, making Ecell > E°cell.
- For the cell Mg(s)|Mg²⁺(0.1M)||Cu²⁺(0.001M)|Cu(s), calculate Ecell given E°cell = 2.70V at 298K.
Reaction: Mg + Cu²⁺ → Mg²⁺ + Cu, n = 2
Q = [Mg²⁺]/[Cu²⁺] = 0.1/0.001 = 100
Ecell = 2.70 − (0.0591/2)log(100) = 2.70 − (0.02955)(2) = 2.70 − 0.0591 = 2.641 V - Write the Nernst equation for the half-cell reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
\(E = E^\circ - \frac{0.0591}{6}\log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}\)
- Calculate the potential of a hydrogen electrode in contact with a solution of pH = 10.
Half-reaction: H⁺(aq) + e⁻ → ½H₂(g), E° = 0.00V, n = 1
pH = 10 → [H⁺] = 10⁻¹⁰ M
Q = 1/[H⁺] (H₂ at 1 bar = pure gas, omitted from Q)
E = 0 − (0.0591/1)log(1/10⁻¹⁰) = −0.0591 × 10 = −0.591 V
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): The Ecell of a Galvanic cell decreases over time as the cell operates.
Reason (R): As the reaction proceeds, reactants are consumed and products accumulate, increasing Q.Answer: (A) Both true and R correctly explains A via the Nernst equation: as Q increases, the subtracted term increases and Ecell falls.
7. Equilibrium Constant from Nernst Equation
Concept Overview
As a Galvanic cell operates, reactant concentration decreases and product concentration increases. The cell potential (Ecell) steadily decreases. At equilibrium, no net change occurs — the cell potential drops to zero and Q = Kc.
Core Derivation
Mathematical Relationship
At equilibrium: Ecell = 0 and Q = Kc. Substituting into the Nernst equation:
\(0 = E^\circ_{cell} - \frac{0.0591}{n}\log K_c\)
\(E^\circ_{cell} = \frac{0.0591}{n}\log K_c\)
Rearranging: \(\log K_c = \frac{n \times E^\circ_{cell}}{0.0591}\)
This equation is powerful — it lets you calculate equilibrium constants (often very difficult to measure directly) simply by measuring the standard cell voltage.
Practice Worksheets
- When a Galvanic cell reaches chemical equilibrium, which is true?
Answer: (B) The actual cell potential (Ecell) drops to zero. E°cell is a fixed constant and does not change.
- For n = 2 and E°cell = 0.0591V at 298K, the value of log Kc is:
Answer: (B) 2. logKc = nE°/(0.0591) = 2×0.0591/0.0591 = 2.
- Calculate Kc for the Daniell cell reaction, given E°cell = 1.10V.
n = 2. logKc = (2 × 1.10)/0.0591 = 37.22. Kc = 10³⁷·²² ≈ 1.6 × 10³⁷. (Very large — reaction is essentially irreversible.)
- Calculate E°cell and Kc for: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s). Given E°Fe³⁺/Fe²⁺ = 0.77V, E°I₂/I⁻ = 0.54V.
Anode: 2I⁻ → I₂ + 2e⁻ (n = 2)
Cathode: 2Fe³⁺ + 2e⁻ → 2Fe²⁺
E°cell = 0.77 − 0.54 = 0.23V
logKc = (2 × 0.23)/0.0591 = 7.78
Kc = 10⁷·⁷⁸ = 6.0 × 10⁷
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): When a cell reaction reaches equilibrium, the voltmeter reading drops to zero.
Reason (R): At equilibrium, the standard cell potential (E°cell) becomes zero.Answer: (C) A is true but R is false. At equilibrium, Ecell (actual voltage) = 0. E°cell is a thermodynamic constant and does not change.
8. Electrochemical Cell and Gibbs Energy
Concept Overview
The reversible work done by a Galvanic cell equals the decrease in its Gibbs energy (ΔG). If n moles of electrons are transferred, the total charge passed is nF (F = Faraday's constant ≈ 96487 C mol⁻¹).
Core Formulas
Mathematical Relationships
General Condition:
\(\Delta_r G = -nFE_{cell}\)
Standard Condition (all concentrations = 1M):
\(\Delta_r G^\circ = -nFE^\circ_{cell}\)
Relationship with Equilibrium Constant:
\(\Delta_r G^\circ = -RT\ln K_c\)
Crucial distinction: E°cell is intensive (independent of stoichiometry). ΔG° is extensive (depends on n). If you double the cell reaction, E°cell stays same but ΔG° doubles.
Practice Worksheets
- For a spontaneous electrochemical cell reaction, the signs of ΔG and Ecell are respectively:
Answer: (C) For spontaneous reaction, ΔG must be negative. Since ΔG = −nFEcell, Ecell must be positive.
- If the stoichiometry of a cell reaction is multiplied by 2:
Answer: (C) E°cell is intensive (stays same). ΔG° is extensive — doubling n doubles ΔG°.
- Calculate ΔG° for the Daniell cell: Zn + Cu²⁺ → Zn²⁺ + Cu. E°cell = 1.10V.
n = 2, F = 96487 C mol⁻¹
ΔG° = −nFE°cell = −2 × 96487 × 1.10 = −212,271 J mol⁻¹ = −212.27 kJ mol⁻¹
- Calculate E°cell and maximum work for: 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s). E°Cr³⁺/Cr = −0.74V, E°Cd²⁺/Cd = −0.40V.
Anode: 2Cr → 2Cr³⁺ + 6e⁻; Cathode: 3Cd²⁺ + 6e⁻ → 3Cd. So n = 6.
E°cell = −0.40 − (−0.74) = +0.34V
ΔG° = −6 × 96487 × 0.34 = −196,834 J mol⁻¹ = −196.83 kJ mol⁻¹
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): If half-reactions are multiplied by 2, E°cell doubles but ΔG° remains constant.
Reason (R): E°cell is an extensive property while ΔG° is intensive.Answer: (D) Both A and R are completely false. It is the exact opposite — E°cell is intensive (stays same) and ΔG° is extensive (doubles).
9. Conductance in Electrolytic Solutions
Fundamental Definitions
- Resistance (R): Hindrance to current flow. Measured in Ω. \(R = \rho\frac{l}{A}\)
- Conductance (G): Inverse of resistance. Measured in Siemens (S). \(G = \frac{1}{R}\)
- Conductivity/Specific Conductance (κ): Conductance of 1 cm³ of solution. Measured in S cm⁻¹. \(\kappa = \frac{1}{\rho} = G\frac{l}{A}\)
Core Measurement
Cell Constant & Molar Conductivity
[Diagram: Conductivity cell with two Pt electrodes, showing l (length) and A (area)]
Cell Constant (G*):
\(G^* = \frac{l}{A} = R\kappa\) (units: cm⁻¹)
G* is determined by calibrating with KCl of known conductivity.
Molar Conductivity (Λm):
\(\Lambda_m = \frac{\kappa \times 1000}{M}\)
Units: S cm² mol⁻¹ (when κ in S cm⁻¹ and M in mol L⁻¹)
Practice Worksheets
- The correct unit for molar conductivity (Λm) is:
Answer: (B) S cm² mol⁻¹
- The cell constant of a conductivity cell:
Answer: (C) G* = l/A is a geometric property of the physical electrodes and does not depend on the solution inside.
- Which expression correctly relates κ, G, and G*?
Answer: (A) κ = G × (l/A) and G* = l/A, so κ = G × G*.
- When a solution is diluted, its specific conductivity (κ):
Answer: (B) Specific conductivity is the conductance of 1 cm³. On dilution, ions per cm³ decrease, so κ decreases.
- A conductivity cell has R = 100Ω with a solution of κ = 1.29×10⁻² S cm⁻¹. Calculate the cell constant.
G* = R × κ = 100 × 1.29×10⁻² = 1.29 cm⁻¹
- Why does Λm increase on dilution even though κ decreases?
Λm = κ × V (where V = volume holding 1 mole of electrolyte). On dilution, κ decreases but V increases at a much faster rate. The net product κ × V increases, so Λm increases.
- Why is AC used instead of DC to measure resistance of electrolytic solutions?
DC causes electrolysis of the solution, changing its composition and altering the resistance being measured. AC reverses direction rapidly, preventing electrolysis and maintaining stable solution composition.
- A column of 0.05 mol L⁻¹ NaOH (diameter 1 cm, length 50 cm) has resistance 5.55×10³ Ω. Calculate resistivity, conductivity, and molar conductivity.
1. Area: A = π(0.5)² = 0.785 cm², l = 50 cm
2. Resistivity: ρ = R×A/l = 5550×0.785/50 = 87.14 Ω cm
3. Conductivity: κ = 1/ρ = 1/87.14 = 0.01148 S cm⁻¹
4. Molar Conductivity: Λm = (0.01148×1000)/0.05 = 229.6 S cm² mol⁻¹
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): The specific conductivity of all electrolytes decreases on dilution.
Reason (R): On dilution, the number of ions per unit volume decreases.Answer: (A) Both A and R true, R correctly explains A. - Assertion (A): The cell constant depends on the nature of the electrolyte filling it.
Reason (R): Cell constant is defined as l/A.Answer: (D) A is false — cell constant is a purely physical dimension of the electrodes independent of the solution. R is true.
10. Variation of Conductivity with Concentration
1. Specific Conductivity (κ) on Dilution
Specific conductivity always decreases with dilution for both strong and weak electrolytes. Reason: fewer ions per unit volume as solution expands.
Core Distinction
2. Molar Conductivity (Λm) on Dilution
Λm always increases with dilution, but the reasons differ:
[Graph: Λm vs √c — straight line for strong electrolytes (e.g., KCl, HCl) extrapolating to Λ°m; steep curve for weak electrolytes (e.g., CH₃COOH) that becomes asymptotic near zero]
A. Strong Electrolytes (e.g., KCl, HCl)
Already 100% dissociated at all concentrations. Increase in Λm is gradual — due to reduced interionic attractions (ions move faster with more separation). Follows the Debye-Hückel-Onsager equation:
\(\Lambda_m = \Lambda^\circ_m - A\sqrt{c}\)
The straight line can be extrapolated to find Λ°m.
B. Weak Electrolytes (e.g., CH₃COOH)
Only partially dissociated. On dilution, degree of dissociation (α) increases dramatically → massive increase in ion count → steep rise in Λm near infinite dilution. Cannot extrapolate to find Λ°m — use Kohlrausch Law instead.
Practice Worksheets
- Upon dilution, the specific conductivity of CH₃COOH:
Answer: (B) Specific conductivity always decreases on dilution for both strong and weak electrolytes — ions per cm³ decrease.
- For a strong electrolyte, the Λm vs √c plot is:
Answer: (C) The Debye-Hückel-Onsager equation Λm = Λ°m − A√c is a straight line with negative slope.
- Why does Λm of a strong electrolyte increase upon dilution?
Answer: (C) Strong electrolytes are already fully dissociated. Ions move faster as separation increases and interionic attraction decreases.
- State the Debye-Hückel-Onsager equation and define all terms.
\(\Lambda_m = \Lambda^\circ_m - A\sqrt{c}\)
Λm = molar conductivity at concentration c; Λ°m = limiting molar conductivity (y-intercept); A = constant depending on temperature, solvent nature, and charge on ions; c = concentration. - Can you find Λ°m of NH₄OH by plotting Λm vs √c? Give a reason.
No. NH₄OH is a weak electrolyte. Its Λm vs √c plot becomes steep and asymptotic to the y-axis near zero concentration — the curve never cleanly intersects the y-axis, so Λ°m cannot be extrapolated. Kohlrausch Law must be used.
- Λm = κ × 1000/M. We know κ decreases on dilution. Comprehensively explain why Λm still increases.
Λm = κ × V, where V = volume containing 1 mole of electrolyte. On dilution: (1) κ decreases (fewer ions per cm³), but (2) V increases much faster. The product κ×V increases overall. Physically: while each cm³ carries less current, you have vastly more cm³ per mole. For strong electrolytes, reduced interionic friction speeds up individual ions. For weak electrolytes, the much higher degree of dissociation generates far more ions per mole.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Λm of CH₃COOH increases steeply at high dilutions.
Reason (R): At high dilutions, degree of dissociation of weak electrolytes approaches 100%, generating many more ions.Answer: (A) Both A and R true; R is the exact chemical reason for the steep curve. - Assertion (A): Λ°m for strong electrolytes can be determined by extrapolating the Λm vs √c graph.
Reason (R): Strong electrolytes follow Λm = Λ°m − A√c at all concentrations.Answer: (C) A is true but R is false. The Debye-Hückel-Onsager equation holds only at low concentrations, not all concentrations.
11. Kohlrausch Law
Why We Need Kohlrausch Law
For weak electrolytes (like CH₃COOH), we cannot find Λ°m by extrapolation. Kohlrausch Law provides an algebraic method using data from strong electrolytes.
Core Law
Law of Independent Migration of Ions
At infinite dilution, each ion makes a definite, independent contribution to the molar conductivity regardless of the nature of the other ion.
\(\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-\)
Where ν₊ and ν₋ are the stoichiometric numbers of cations and anions, and λ° are the limiting molar conductivities of individual ions.
Applications of Kohlrausch Law
1. Finding Λ°m for CH₃COOH:
\(\Lambda^\circ_m(CH_3COOH) = \Lambda^\circ_m(HCl) + \Lambda^\circ_m(CH_3COONa) - \Lambda^\circ_m(NaCl)\)
2. Degree of dissociation: \(\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}\)
3. Dissociation constant: \(K_a = \frac{c\alpha^2}{1-\alpha}\)
Practice Worksheets
- Λ°m of electrolyte AxBy is given by:
Answer: (A) Kohlrausch law: each ion contributes proportionally to its stoichiometric number.
- If Λ°m for BaCl₂, H₂SO₄, and HCl are x₁, x₂, x₃ respectively, then Λ°m for BaSO₄ is:
Answer: (B) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl. So Λ°m(BaSO₄) = x₁ + x₂ − 2x₃.
- Degree of dissociation (α) of a weak electrolyte equals:
Answer: (B) α = Λm/Λ°m
- Given λ°(H⁺) = 350, λ°(CH₃COO⁻) = 40, λ°(Na⁺) = 50, λ°(Cl⁻) = 76 S cm² mol⁻¹. Calculate Λ°m(CH₃COOH).
Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 350 + 40 = 390 S cm² mol⁻¹
- Express Λ°m(Al₂(SO₄)₃) in terms of individual ionic conductivities.
\(\Lambda^\circ_m(Al_2(SO_4)_3) = 2\lambda^\circ_{Al^{3+}} + 3\lambda^\circ_{SO_4^{2-}}\)
- The conductivity of 0.001028M acetic acid is 4.95×10⁻⁵ S cm⁻¹. Calculate Ka. (Λ°m = 390.5 S cm² mol⁻¹)
1. Λm: (4.95×10⁻⁵ × 1000) / 0.001028 = 48.15 S cm² mol⁻¹
2. α: 48.15/390.5 = 0.1233
3. Ka: (0.001028 × 0.1233²)/(1 − 0.1233) = (0.001028 × 0.0152)/0.8767 = 1.78×10⁻⁵ mol L⁻¹
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Λ°m of CH₃COOH can be obtained by extrapolating Λm vs √c to zero concentration.
Reason (R): CH₃COOH is a weak electrolyte and its Λm increases steeply near infinite dilution.Answer: (D) A is false; extrapolation is impossible for weak electrolytes precisely because of the steep curve. R is true. - Assertion (A): Λ°m/Λm gives the degree of dissociation of a weak electrolyte.
Reason (R): At infinite dilution, dissociation of a weak electrolyte is 100%.Answer: (D) A is false (it should be Λm/Λ°m, not the inverse). R is true.
12. Electrolysis & Electrolytic Cells
Concept Overview
Unlike a Galvanic cell (spontaneous), an Electrolytic Cell uses electrical energy to drive a non-spontaneous chemical reaction. This process is called electrolysis.
[Diagram: Electrolytic cell — external battery connected to two inert Pt electrodes in electrolyte solution, showing cation migration to cathode and anion migration to anode]
- Anode (+): Oxidation occurs. Anions migrate here to lose electrons.
- Cathode (−): Reduction occurs. Cations migrate here to gain electrons.
Core Principle
Faraday's Laws of Electrolysis
First Law
The mass of substance produced at an electrode is directly proportional to the quantity of electricity passed.
\(m = \frac{M \times I \times t}{n \times F}\)
Where: m = mass (g), M = molar mass, I = current (A), t = time (s), n = electrons transferred per ion, F = 96487 C mol⁻¹
Second Law
[Diagram: Two cells in series — same charge through each → masses deposited proportional to equivalent weights]
When the same quantity of electricity passes through different electrolytes in series, masses deposited are proportional to their equivalent weights.
\(\frac{m_1}{m_2} = \frac{E_1}{E_2} = \frac{M_1/n_1}{M_2/n_2}\)
Products of Electrolysis
- Molten NaCl: Only Na⁺ and Cl⁻ present. Cathode: Na(s). Anode: Cl₂(g).
- Aqueous NaCl (brine): H₂O also present. Cathode: H₂(g) (water reduced, not Na⁺). Anode: Cl₂(g) due to overpotential.
- Aqueous CuSO₄ (Pt electrodes): Cathode: Cu(s). Anode: O₂(g).
- Aqueous AgNO₃ (Ag electrodes): Cathode: Ag(s) deposited. Anode: Ag dissolves (active electrode).
Practice Worksheets
- How many Faradays are required to reduce 1 mole of MnO₄⁻ to Mn²⁺?
Answer: (C) 5 F. Mn oxidation state changes from +7 in MnO₄⁻ to +2 in Mn²⁺ — a change of 5, requiring 5 moles of electrons.
- Three Faradays passed through molten Al₂O₃, aqueous CuSO₄, and molten NaCl. Molar ratio of Al:Cu:Na deposited is:
Answer: (A) 1 : 1.5 : 3. Al³⁺ needs 3e⁻ → 1 mol; Cu²⁺ needs 2e⁻ → 1.5 mol; Na⁺ needs 1e⁻ → 3 mol. All from 3 Faradays.
- Why is alternating current not used for electrolysis?
Answer: (B) AC reverses direction, alternating anode/cathode roles at each electrode — net reaction is zero.
- A current of 1.5A passes through CuSO₄ for 10 minutes. Calculate mass of copper deposited. (M(Cu) = 63.5 g/mol)
Q = 1.5 × 600 = 900 C. Cu²⁺ + 2e⁻ → Cu. m = (63.5 × 900)/(2 × 96487) = 0.296 g
- Predict products of electrolysis of aqueous CuCl₂ using Pt electrodes.
Cathode: Cu(s) deposited (Cu²⁺ has higher reduction potential than water). Anode: Cl₂(g) evolved (Cl⁻ oxidized preferentially due to overpotential of O₂ formation).
- Discuss products of electrolysis of aqueous NaCl (brine) and why they differ from molten NaCl. Explain the role of overpotential at the anode.
Molten NaCl: Only Na⁺ and Cl⁻ present. Na deposited at cathode; Cl₂ at anode. Simple and straightforward.
Aqueous NaCl: Water is also present and competes.
Cathode: E°(H₂O) = −0.83V vs E°(Na⁺) = −2.71V. Water is reduced preferentially → H₂ gas.
Anode: Thermodynamically, water should oxidize to O₂ (+1.23V) before Cl⁻ oxidizes to Cl₂ (+1.36V). However, O₂ formation requires extra activation energy (overpotential). Kinetically, Cl⁻ oxidation is easier → Cl₂ gas produced.
Result: NaOH forms in solution (Na⁺ + OH⁻ remaining after H₂ and Cl₂ evolve).
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Electrolysis of aqueous NaCl gives H₂ at cathode and Cl₂ at anode.
Reason (R): Na⁺ has higher standard reduction potential than water.Answer: (C) A is true but R is false. Na⁺ has LOWER reduction potential (−2.71V) than water (−0.83V), which is exactly why water is reduced to H₂ instead. - Assertion (A): When 1 Faraday passes through Al³⁺ solution, 1 mole of Al is deposited.
Reason (R): 1 Faraday is the charge of 1 mole of electrons.Answer: (D) A is false — Al³⁺ + 3e⁻ → Al requires 3F for 1 mole; 1F only gives ⅓ mol. R is true.
13. Batteries (Commercial Cells)
Concept Overview
A battery is one or more Galvanic cells connected in series. A useful commercial battery should be light, compact, and maintain stable voltage during use. Batteries are classified as Primary (non-rechargeable) or Secondary (rechargeable).
1. Primary Batteries (Non-Rechargeable)
A. Dry Cell (Leclanché Cell) — ~1.5V
[Diagram: Dry cell cross-section — Zn casing (anode), carbon rod + MnO₂ (cathode), NH₄Cl paste electrolyte]
- Anode: Zinc container
- Cathode: Graphite rod surrounded by MnO₂ + carbon powder
- Electrolyte: Moist paste of NH₄Cl + ZnCl₂
- Anode: Zn(s) → Zn²⁺ + 2e⁻
- Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃
- NH₃ forms [Zn(NH₃)₄]²⁺ complex to prevent pressure buildup
B. Mercury Cell — ~1.35V (constant)
- Anode: Zinc-mercury amalgam; Cathode: HgO + carbon; Electrolyte: KOH + ZnO paste
- Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
- Constant voltage because no ions in solution change concentration
2. Secondary Batteries (Rechargeable)
Lead Storage Battery — ~12V (6 cells)
[Diagram: Lead battery — Pb anode plates, PbO₂ cathode plates, H₂SO₄ electrolyte]
- Anode: Spongy Pb; Cathode: PbO₂; Electrolyte: 38% H₂SO₄(aq)
- Anode (discharge): Pb + SO₄²⁻ → PbSO₄ + 2e⁻
- Cathode (discharge): PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
- Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- On charging, reactions are reversed; H₂SO₄ is regenerated (density rises)
Practice Worksheets
- Which is true for a mercury cell?
Answer: (C) Overall reaction involves only pure solids and liquids (no aqueous ions whose concentrations change), so voltage is constant at 1.35V.
- During discharging of a lead storage battery:
Answer: (C) H₂SO₄ is consumed and H₂O is produced during discharge, diluting the electrolyte and decreasing its density.
- In a dry cell, Mn oxidation state changes from:
Answer: (A) MnO₂ (+4) → MnO(OH) (+3).
- Write the charging reactions for a lead storage battery.
During charging, cell acts as electrolytic cell (reactions reversed):
Cathode: PbSO₄ + 2e⁻ → Pb + SO₄²⁻
Anode: PbSO₄ + 2H₂O → PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ - Why does a dry cell become dead even when not used for a long time?
The electrolyte paste contains NH₄Cl, which is acidic. Over time, this acidic medium slowly corrodes the zinc container (anode) even without a complete circuit, eventually destroying it.
- Compare the Leclanché cell and the mercury cell with respect to construction and voltage profile.
Leclanché Cell: Anode=Zn, Cathode=C+MnO₂, Electrolyte=NH₄Cl+ZnCl₂ paste. Voltage starts at 1.5V but drops gradually because NH₄⁺ ion concentration decreases as reaction proceeds.
Mercury Cell: Anode=Zn-Hg amalgam, Cathode=HgO+C, Electrolyte=KOH+ZnO paste. Voltage remains constant at 1.35V throughout its life because the reaction (Zn+HgO→ZnO+Hg) involves only pure solids/liquids — no aqueous ions that change concentration.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Voltage of a mercury cell remains constant throughout its life.
Reason (R): The overall cell reaction does not involve any ions in solution whose concentration changes.Answer: (A) Both true; R is the perfect thermodynamic explanation (Q remains constant, so Ecell = E°cell − 0 = constant). - Assertion (A): When a lead storage battery is discharging, it acts as an electrolytic cell.
Reason (R): During discharge, sulfuric acid is consumed.Answer: (D) A is false — discharging is a spontaneous process, making it a Galvanic cell. R is true (H₂SO₄ is consumed during discharge).
14. Fuel Cells
Concept Overview
Fuel cells convert the energy of combustion of fuels (hydrogen, methane, methanol) directly into electrical energy. Unlike batteries, they produce electricity continuously as long as fuel and oxidant are supplied — they never become "dead."
Core Example
The Hydrogen-Oxygen (H₂-O₂) Fuel Cell
[Diagram: H₂-O₂ fuel cell — H₂ bubbled into anode, O₂ into cathode, porous carbon electrodes with Pt catalyst, KOH electrolyte, electrical output shown]
Construction: Porous carbon electrodes embedded with Pt/Pd catalyst, immersed in concentrated KOH or NaOH solution.
- Cathode (Reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
- Anode (Oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
- Overall: 2H₂(g) + O₂(g) → 2H₂O(l)
Used in Apollo space missions — water produced was condensed for drinking water.
Advantages of Fuel Cells
- High efficiency (~70%) vs. thermal plants (~40%) — direct conversion bypasses heat/mechanical losses
- Pollution-free — only byproduct is pure water
- Continuous operation — never becomes dead as long as fuel supplied
Practice Worksheets
- Only byproduct of the H₂-O₂ fuel cell is:
Answer: (C) Water. 2H₂ + O₂ → 2H₂O.
- Theoretical efficiency of a fuel cell is approximately:
Answer: (C) 70%. Much superior to thermal power plants (~40%).
- Fuel cells differ from standard batteries because:
Answer: (C) Batteries have fixed internal reactants; fuel cells consume externally supplied fuel continuously.
- Write electrode reactions for the H₂-O₂ fuel cell.
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ - State two major advantages of fuel cells over thermal power plants.
1. Higher efficiency (~70%) vs ~40% for thermal plants — no energy lost as heat in mechanical stages.
2. Environmentally friendly — only byproduct is water; no greenhouse gases produced.
- Describe the working of a H₂-O₂ fuel cell. Why was it chosen for the Apollo space programme?
Working: Porous carbon electrodes with Pt/Pd catalyst are immersed in concentrated KOH/NaOH. H₂ is continuously supplied to the anode compartment and O₂ to the cathode. H₂ is oxidized at the anode releasing electrons; O₂ is reduced at the cathode consuming electrons. Electrons flow through the external circuit generating electrical power. The overall reaction is simply H₂ combustion producing water.
Apollo Programme: Chosen for its extremely high energy-to-weight ratio (critical in space), high efficiency, and the fact that the only byproduct is pure water — which was condensed and used as the astronauts' drinking water supply, solving two problems at once.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Fuel cells produce electricity at ~70% efficiency.
Reason (R): Chemical energy is converted directly to electrical energy without thermal and mechanical intermediate stages.Answer: (A) Both true; R perfectly explains why fuel cells avoid the energy losses inherent in thermal → mechanical → electrical conversion chains. - Assertion (A): Fuel cells never become "dead" in the way primary batteries do.
Reason (R): Reactants are continuously supplied from an external reservoir.Answer: (A) Both true; R is the correct explanation. Unlike a dry cell with fixed internal reactants, fuel cells run indefinitely as long as they are fed.
15. Corrosion
What is Corrosion?
Corrosion is the slow eating away of metals due to attack by atmospheric gases (O₂, CO₂, H₂S, SO₂) and moisture. It is essentially an electrochemical process — the metal acts as a Galvanic cell with itself.
The most common example is rusting of iron.
Core Mechanism
Electrochemical Theory of Rusting
[Diagram: Iron surface with anode region (Zn-depleted, impure Fe), cathode region (pure Fe), water droplet, electron flow and ion migration shown]
Iron rusts in the presence of both water and oxygen. Impurities in iron (e.g., carbon particles) create small Galvanic cells on the iron surface:
- Anode (impure/stressed region): Fe(s) → Fe²⁺(aq) + 2e⁻ (iron oxidizes)
- Cathode (purer region/carbon impurity): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
- Fe²⁺ ions migrate through the moisture film to the cathode region
- Further oxidation: 4Fe²⁺ + O₂ + 8H⁺ → 4Fe³⁺
- Rust forms: Fe³⁺ + 3H₂O → Fe₂O₃·3H₂O (hydrated iron(III) oxide = rust)
Both water AND oxygen are necessary for rusting. Pure water or pure dry air alone will not cause rusting.
Prevention of Corrosion
- Barrier protection: Painting, oiling, greasing, electroplating with Ni/Cr
- Galvanizing: Coating iron with zinc. Even if zinc layer breaks, Zn is oxidized preferentially (sacrificial anode) — iron is protected
- Tinning: Coating with tin. If Sn layer breaks, Fe is oxidized (tin is NOT a sacrificial anode for iron — Fe is more reactive)
- Sacrificial anode: Connecting iron to a more reactive metal (Mg, Zn). The reactive metal oxidizes first, protecting iron (used in ship hulls)
- Anti-rust solutions: Adding inhibitors that coat the metal surface
- Alloying: Stainless steel (Fe + Cr + Ni) — Cr forms a protective Cr₂O₃ layer
Practice Worksheets
- Rusting of iron is enhanced in the presence of:
Answer: (C) Dissolved electrolytes (like NaCl in sea water) increase conductivity of the moisture layer, accelerating the electrochemical corrosion process dramatically.
- Galvanized iron is iron coated with zinc. If the zinc coating is scratched:
Answer: (B) Zn has lower reduction potential than Fe (E°Zn²⁺/Zn = −0.76V vs E°Fe²⁺/Fe = −0.44V). Zn oxidizes preferentially, acting as sacrificial anode and protecting Fe even where coating is broken.
- In the electrochemical corrosion of iron, the cathodic reaction is:
Answer: (B) Dissolved O₂ is reduced at the cathodic region.
- What is the chemical formula of rust? Write the overall equation for its formation.
Rust is Fe₂O₃·3H₂O (hydrated iron(III) oxide).
Overall: 4Fe + 3O₂ + 6H₂O → 2Fe₂O₃·3H₂O - Why does galvanizing protect iron better than tinning when the coating is scratched?
Zinc (E° = −0.76V) is more reactive than iron (E° = −0.44V). When the Zn coating scratches, Zn acts as sacrificial anode — it oxidizes instead of Fe, protecting it. Tin (E° = −0.14V) is less reactive than iron. When Sn is scratched, Fe becomes the anode and corrodes more rapidly. So galvanizing provides cathodic protection even when scratched; tinning does not.
- Explain the electrochemical theory of rusting. What conditions are necessary? How does the sacrificial anode method prevent corrosion in ship hulls?
Conditions: Both water (with dissolved electrolytes) and oxygen must be present. Presence of CO₂ or SO₂ (forming carbonic/sulfurous acid) also accelerates the process.
Mechanism: Impurities in iron create microscopic Galvanic cells. The impure/stressed region acts as anode — Fe oxidizes to Fe²⁺. The purer region (or carbon particles) acts as cathode — O₂ is reduced. Fe²⁺ ions migrate to cathode region, undergo further oxidation to Fe³⁺, and combine with OH⁻ to form Fe(OH)₃, which dehydrates to rust (Fe₂O₃·3H₂O).
Sacrificial anode in ships: Blocks of Mg or Zn are bolted to the steel hull below the waterline. These metals have lower reduction potential than iron, so they oxidize preferentially. The electrons generated keep the iron cathodically protected. The Mg/Zn blocks dissolve slowly and are periodically replaced.
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): Iron does not rust in dry air or pure water alone.
Reason (R): Both water and oxygen must be present simultaneously for the electrochemical corrosion process to occur.Answer: (A) Both true; R correctly explains A. Dry air has no water film (no electrolyte medium); pure water has no O₂ to act as cathodic reactant. - Assertion (A): A tin-coated iron can (food tin) corrodes faster than bare iron once the tin coating is broken.
Reason (R): Tin has a higher reduction potential than iron, making iron the anode in the galvanic couple formed.Answer: (A) Both true; R correctly explains A. E°Sn²⁺/Sn = −0.14V vs E°Fe²⁺/Fe = −0.44V. Fe has lower potential, acts as anode, and corrodes rapidly once Sn is broken.
16. Core In-Text Conceptual Questions
Concept Overview
This section focuses on fundamental conceptual questions from the NCERT chapter text — practical applications like chemical storage safety, single-electrode potential measurement, and the effect of pH on the Nernst equation.
Practice Worksheets
- Calculate the potential of a hydrogen electrode in contact with a solution of pH = 10.
Answer: (C) −0.591 V. pH = 10 → [H⁺] = 10⁻¹⁰ M. E = 0 − (0.0591/1)×log(1/10⁻¹⁰) = −0.0591 × 10 = −0.591 V.
- Which substance can oxidize Fe²⁺ to Fe³⁺ (E°Fe³⁺/Fe²⁺ = +0.77V)?
Answer: (C) Cl₂. To oxidize Fe²⁺, the oxidizing agent must have E° > +0.77V. Cl₂ (+1.36V) qualifies. Note: it's Ag⁺ ions (not Ag metal) that would also qualify (+0.80V).
- Can you store copper sulphate solution in a zinc pot? Explain thermodynamically.
No. E°Zn²⁺/Zn = −0.76V, E°Cu²⁺/Cu = +0.34V. Zinc is more reactive and will spontaneously oxidize, displacing copper: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). E°cell = +0.34 − (−0.76) = +1.10V > 0, confirming spontaneous reaction. The pot will dissolve.
- Why does conductivity of a solution decrease with dilution?
Conductivity (κ) is the conductance of exactly 1 cm³ of solution. On dilution, total volume increases — the number of current-carrying ions per cm³ decreases, lowering κ.
- Describe how you would determine E° for the Mg²⁺/Mg half-cell using a SHE. Include cell notation and calculation.
Setup: Mg rod in 1M MgSO₄ (298K) connected via salt bridge + voltmeter to SHE (Pt in 1M HCl, H₂ at 1 bar).
Observation: Voltmeter reads 2.36V; electrons flow from Mg to SHE → Mg is the anode.
Calculation: E°cell = E°cathode − E°anode → 2.36 = 0.00 − E°Mg²⁺/Mg → E°Mg²⁺/Mg = −2.36V
Cell Notation: Mg(s) | Mg²⁺(aq, 1M) || H⁺(aq, 1M) | H₂(g, 1bar) | Pt(s)
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): A silver vessel is safe for storing CuSO₄ solution.
Reason (R): E°Ag⁺/Ag (+0.80V) is higher than E°Cu²⁺/Cu (+0.34V).Answer: (A) Both true; R correctly explains A. Silver has higher reduction potential — it prefers to stay as solid metal and cannot displace copper from CuSO₄. - Assertion (A): The potential of a hydrogen electrode decreases when pH is increased.
Reason (R): Increasing pH means decreasing [H⁺], which decreases reduction potential per Nernst equation.Answer: (A) Both true; R correctly explains A. E = −0.0591 × pH, so higher pH → more negative voltage.
17. Comprehensive Solved Exercises
Concept Overview
Advanced problems often chain multiple formulas: use resistance → find conductivity → find molar conductivity → find degree of dissociation → find Ka. Practice these multi-step chains until they feel natural.
Practice Worksheets
- If E°cell = 0.591V for a reaction with n = 2, what is Kc?
Answer: (B) 10²⁰. logKc = nE°/0.0591 = (2×0.591)/0.0591 = 20. Kc = 10²⁰.
- 2.0A passed for 5 hours deposits 22.2g of metal (M = 177 g/mol). Oxidation state of metal?
Answer: (C) +3. Q = 2×5×3600 = 36000C. Moles of metal = 22.2/177 = 0.125. F used = 36000/96487 = 0.373. n = 0.373/0.125 ≈ 3.
- Calculate potential for: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Given [Cr₂O₇²⁻] = 0.10M, [Cr³⁺] = 0.20M, [H⁺] = 1.0×10⁻⁴M, E° = 1.33V.
n = 6. E = 1.33 − (0.0591/6)log{(0.20)²/[0.10×(10⁻⁴)¹⁴]} = 1.33 − 0.00985×log(4×10⁵⁵) ≈ 1.33 − 0.55 = 0.78V
- A conductivity cell with 0.001M KCl (κ = 0.146×10⁻³ S cm⁻¹) has R = 1500Ω. The same cell with 0.001M CH₃COOH has R = 3200Ω. Find: (a) cell constant, (b) degree of dissociation of acetic acid, (c) Ka. (Λ°m = 390.5 S cm² mol⁻¹)
(a) Cell constant: G* = κ × R = 0.146×10⁻³ × 1500 = 0.219 cm⁻¹
(b) Conductivity of CH₃COOH: κ = G*/R = 0.219/3200 = 6.84×10⁻⁵ S cm⁻¹
Λm = (6.84×10⁻⁵ × 1000)/0.001 = 68.4 S cm² mol⁻¹
α = Λm/Λ°m = 68.4/390.5 = 0.175
(c) Ka: Ka = cα²/(1−α) = (0.001 × 0.0306)/0.825 = 3.7×10⁻⁵ mol L⁻¹
Directions: (A) Both A and R true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
- Assertion (A): The cell constant calculated using KCl is independent of the volume of KCl solution in the cell.
Reason (R): Cell constant is a purely physical characteristic determined by electrode geometry (l/A).Answer: (A) Both true; R correctly explains A. As long as electrodes are fully submerged, l and A do not change with volume.
18. Standard Question Bank
Overview
This question bank is organized into three sections following the CBSE board exam pattern:
- Section A (Section 19): MCQ — 1 mark each. Tests rapid recall, conceptual clarity, and application of formulas.
- Section B (Section 20): Short Answer — 2 marks each. Tests concise explanation and single-step calculations.
- Section C (Section 21): Long Answer — 3–5 marks each. Tests multi-step numericals and detailed conceptual explanations.
Marking Scheme Insights
What Examiners Look For
- Numericals: Always write the formula first, then substitute values with units, then calculate. Even if the final answer is wrong, formula and method marks are awarded.
- Reactions: Write balanced equations — unbalanced equations lose marks even if the species are correct.
- Definitions: Use precise scientific language — e.g., "limiting molar conductivity" not just "maximum conductivity."
- 1-mark MCQs: No partial marks — ensure conceptual clarity.
- Assertion-Reasoning: Evaluate A and R independently first, then judge if R explains A.
Most Frequently Tested Topics (Last 10 Years)
- Nernst equation application (every year)
- Kohlrausch law + Ka calculation
- Faraday's law numericals
- Lead storage battery reactions
- Galvanic vs. Electrolytic cell comparison
- Corrosion mechanism
- Fuel cell electrode reactions
- ΔG° and Kc from E°cell
19. Section A: Multiple Choice Questions (1 Mark)
Instructions
1-mark MCQs covering the entire chapter. Answer before revealing the solution.
- Standard reduction potentials of metals A, B, C are +0.5V, −3.0V, −1.2V. Order of reducing power:
Answer: (B) B > C > A. Lower (more negative) reduction potential = stronger reducing agent.
- Unit of cell constant (G*) is:
Answer: (C) cm⁻¹. G* = l/A = cm/cm² = cm⁻¹.
- For Zn|Zn²⁺||Cu²⁺|Cu, if [Cu²⁺] is increased, Ecell will:
Answer: (B) Increase. Q = [Zn²⁺]/[Cu²⁺]. Increasing [Cu²⁺] decreases Q → smaller log Q → smaller subtracted term → higher Ecell.
- Λ°m(NaCl) = 126.4, Λ°m(HCl) = 425.9, Λ°m(CH₃COONa) = 91.0 S cm² mol⁻¹. Λ°m(CH₃COOH) =
Answer: (B) 390.5 S cm² mol⁻¹. 91.0 + 425.9 − 126.4 = 390.5.
- Faradays required to reduce 1 mole of Cr₂O₇²⁻ to Cr³⁺:
Answer: (D) 6 F. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. 6 moles of electrons required.
- Electrolysis of aqueous NaCl using Pt electrodes — product at anode:
Answer: (C) Cl₂ gas. Though thermodynamics favors O₂, the high overpotential of O₂ formation makes Cl₂ evolution kinetically preferred.
- For spontaneous electrochemical cell reaction, signs of Ecell and ΔG:
Answer: (C) ΔG = −nFEcell. For spontaneous process: ΔG < 0 → Ecell > 0.
- On diluting a strong electrolyte, κ and Λm change as:
Answer: (B) κ decreases (fewer ions per cm³); Λm increases (volume per mole increases faster).
- Electrolyte in lead storage battery is:
Answer: (C) 38% H₂SO₄(aq). Consumed during discharge, regenerated during charging.
- Primary advantage of H₂-O₂ fuel cell over conventional thermal power plant:
Answer: (C). Direct conversion bypasses thermal/mechanical losses; water is the only byproduct.
20. Section B: Short Answer Questions (2 Marks)
Instructions
For numerical problems, state the formula used and include units in the final answer.
- Q1: Explain why specific conductivity decreases with dilution but molar conductivity increases.
Specific conductivity (κ): = conductance of 1 cm³. On dilution, ions per cm³ decrease → κ decreases.
Molar conductivity (Λm): = κ × V (V = volume holding 1 mole). On dilution, V increases much faster than κ decreases → net product κV increases → Λm increases. - Q2: Calculate mass of copper deposited when 2.0A flows through CuSO₄ for 20 minutes. (M(Cu) = 63.5 g/mol, F = 96487 C mol⁻¹)
Q = 2.0 × (20×60) = 2400 C. Reaction: Cu²⁺ + 2e⁻ → Cu (n=2). m = (63.5 × 2400)/(2 × 96487) = 0.789 g
- Q3: State Kohlrausch's law. Write its expression for Λ°m of BaCl₂.
Law: At infinite dilution, each ion makes a definite, independent contribution to the limiting molar conductivity, irrespective of the other ion present.
\(\Lambda^\circ_m(BaCl_2) = \lambda^\circ_{Ba^{2+}} + 2\lambda^\circ_{Cl^-}\) - Q4: Why does a dry cell become dead even when not in use for a long time?
The NH₄Cl electrolyte paste is acidic. Even without a complete circuit, the acidic medium slowly corrodes the zinc container (anode) over time, eventually destroying it.
- Q5: Write the Nernst equation for the Daniell cell. How does Ecell change if [Zn²⁺] is increased?
\(E_{cell} = E^\circ_{cell} - \frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
Increasing [Zn²⁺] increases the numerator of log Q, making the subtracted term larger → Ecell decreases.
21. Section C: Long Answer Questions (3+ Marks)
Instructions
State your formula, show all steps, and include units. Multi-part questions — answer each part clearly labeled.
- Q1: Galvanic Cell Thermodynamics
Cell: Mg(s)|Mg²⁺(0.1M)||Cu²⁺(0.001M)|Cu(s). E°Mg²⁺/Mg = −2.36V, E°Cu²⁺/Cu = +0.34V, F = 96487 C mol⁻¹.
(a) Write cell reaction. (b) Calculate Ecell. (c) Calculate ΔG°.(a) Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), n = 2
(b) E°cell = 0.34 − (−2.36) = 2.70V
Q = [Mg²⁺]/[Cu²⁺] = 0.1/0.001 = 100
Ecell = 2.70 − (0.0591/2)×log100 = 2.70 − 0.0591 = 2.641V
(c) ΔG° = −nFE°cell = −2 × 96487 × 2.70 = −521,030 J mol⁻¹ = −521.03 kJ mol⁻¹ - Q2: Faraday's Laws (Series Cells)
Two cells in series — AgNO₃ (Cell A) and CuSO₄ (Cell B). Current = 1.5A until 1.45g Ag deposits in A.
(a) How long did current flow? (b) Mass of Cu deposited in B? (M(Ag)=108, M(Cu)=63.5 g/mol)(a) Ag⁺ + e⁻ → Ag (n=1). Moles Ag = 1.45/108 = 0.01343 mol. Q = 0.01343 × 96487 = 1295.8C. t = Q/I = 1295.8/1.5 = 863.9 s ≈ 14.4 min
(b) Same charge (1295.8C) through CuSO₄. Cu²⁺ + 2e⁻ → Cu (n=2). m = (63.5 × 1295.8)/(2 × 96487) = 0.426 g - Q3: Kohlrausch Law + Ka Calculation
Conductivity of 0.00241M CH₃COOH = 7.896×10⁻⁵ S cm⁻¹. Λ°m = 390.5 S cm² mol⁻¹. Find Λm, α, and Ka.Λm = (7.896×10⁻⁵ × 1000)/0.00241 = 32.76 S cm² mol⁻¹
α = 32.76/390.5 = 0.0839
Ka = (0.00241 × 0.0839²)/(1 − 0.0839) = (0.00241 × 0.00704)/0.9161 = 1.85×10⁻⁵ mol L⁻¹ - Q4: Lead Storage Battery
Describe construction, electrode reactions during discharge, and explain why electrolyte density drops with use.Construction: Alternating plates of spongy Pb (anode) and PbO₂ (cathode) immersed in 38% H₂SO₄(aq) electrolyte.
Discharge reactions:
Anode: Pb + SO₄²⁻ → PbSO₄(s) + 2e⁻
Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄(s) + 2H₂O
Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
Density drop: H₂SO₄ is actively consumed and water is produced during discharge, diluting the electrolyte. A hydrometer measures density; low density = low charge remaining.