**Definition**: An electrochemical cell is a device that either converts chemical energy into electrical energy (galvanic cell) or uses electrical energy to drive non-spontaneous chemical reactions (electrolytic cell).
**Key distinction**: In a galvanic (voltaic) cell, a spontaneous redox reaction produces electrical energy. In an electrolytic cell, external electrical energy drives a non-spontaneous reaction.
**Example**: The Daniell cell converts the spontaneous reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) into electrical energy, generating an emf of 1.1 V at standard conditions.
**Important exam points**:
**Cell components**:
**Half-cell reactions** (using Daniell cell example):
**Electrode definition**:
**Cell notation convention** (IUPAC):
**Cell potential (emf)**:
The potential difference between cathode and anode, measured in volts when no current flows:
**E°cell = E°cathode − E°anode = E°right − E°left**
**Exam-critical points**:
**Standard Electrode Potential (E°)**: The potential of a half-cell measured against the standard hydrogen electrode under standard conditions (all concentrations 1 M, gases 1 bar pressure, 298 K).
**Standard Hydrogen Electrode (SHE)**:
**Measuring standard electrode potential**:
When a half-cell is connected with SHE as anode and the half-cell as cathode, the measured emf equals the standard electrode potential of that half-cell:
**E°cell = E°half-cell − E°SHE = E°half-cell − 0 = E°half-cell**
**Examples from standard table**:
**Interpretation of electrode potential values**:
**Trends in standard electrode potentials table**:
**Real-life application**: Cu does not dissolve in dilute HCl because E°(H⁺/H₂) = 0.00 V > E°(Cu²⁺/Cu) = +0.34 V, meaning H⁺ cannot oxidize Cu. However, HNO₃ oxidizes Cu because NO₃⁻ is a stronger oxidizing agent than H⁺.
**Exam-important fact**: Standard electrode potentials are reduction potentials—they always represent the reduction half-reaction. To find cell emf: E°cell = E°(cathode, reduction) − E°(anode, oxidation)
**Definition**: The Nernst equation relates electrode potential under non-standard conditions to the standard electrode potential and the concentration of reactants and products.
**For a single electrode** (reduction half-reaction: Mⁿ⁺ + ne⁻ → M):
**E = E° − (RT/nF) ln(1/[Mⁿ⁺])**
Or equivalently:
**E = E° − (RT/nF) ln[M]/[Mⁿ⁺]** (concentration of solid M = 1)
**For a complete cell reaction**:
**Ecell = E°cell − (RT/nF) ln Q**
Where:
**At 298 K, converting to base-10 logarithm**:
**E = E° − (0.0592/n) log Q**
**Nernst equation for Daniell cell** (n = 2):
**Ecell = 1.10 − (0.0592/2) log([Zn²⁺]/[Cu²⁺])**
**Nernst equation for Ni|Ni²⁺||Ag⁺|Ag cell** (n = 2):
**Ecell = E°cell − (0.0592/2) log([Ni²⁺]/[Ag⁺]²)**
**Important observations**:
**Example calculation**: For Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s)
**Thermodynamic relationship**:
The Gibbs free energy change is directly related to the electrical work done by the cell:
**ΔG° = −nFE°cell**
Where:
**Derivation of relationship with Kc**:
At equilibrium, Ecell = 0 and Q = Kc, so:
0 = E°cell − (0.0592/n) log Kc
Rearranging:
**E°cell = (0.0592/n) log Kc**
Or in terms of natural logarithm:
**E°cell = (2.303RT/nF) ln Kc**
**Interpretation**:
**Example**: For Daniell cell reaction:
**Exam importance**: This relationship allows calculation of equilibrium constants from electrochemical data—useful when direct measurement is difficult.
**Definition**: An electrolytic cell uses external electrical energy to drive non-spontaneous chemical reactions.
**Comparison with galvanic cells**:
| Feature | Galvanic | Electrolytic |
|---------|----------|-------------|
| **Spontaneity** | Spontaneous (ΔG° < 0) | Non-spontaneous (ΔG° > 0) |
| **E°cell** | Positive | Negative |
| **Anode** | Negative (oxidation occurs) | Positive (oxidation occurs) |
| **Cathode** | Positive (reduction occurs) | Negative (reduction occurs) |
| **External voltage** | None needed | Eext > E°cell required |
| **Power source** | Produces electrical energy | Consumes electrical energy |
**Operating principle**: When external voltage applied equals cell emf, reaction stops (I = 0). When external voltage exceeds cell emf (Eext > E°cell), reaction reverses, and the cell now functions as an electrolytic cell.
**Electrolysis of ionic solutions**: The actual products depend on the relative positions of ions in the electrochemical series and their concentrations. Ions that are easier to oxidize/reduce (higher/lower E° values) preferentially react.
**Exam-critical concept**: In an electrolytic cell, the anode is connected to the positive terminal of the external source, and the cathode to the negative terminal—opposite to the galvanic cell configuration.
**First Law of Electrolysis**: The amount of substance deposited (or dissolved) at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the cell.
Mathematically: **m ∝ Q**, or **m = (M × Q)/(n × F)**
Where:
Also expressed in terms of current and time:
**m = (M × I × t)/(n × F)**
Where **I** = current (in amperes) and **t** = time (in seconds)
**Second Law of Electrolysis**: When the same amount of charge is passed through different electrolytes, the masses of substances deposited at the electrodes are proportional to their chemical equivalent weights (molar mass/number of electrons).
**m₁/m₂ = (M₁/n₁)/(M₂/n₂)**
Or equivalently: **m₁/m₂ = (Eq₁)/(Eq₂)** where Eq = equivalent weight
**Practical applications**:
**Example problem**: Calculate mass of Cu deposited when 5 A current passes through CuSO₄ solution for 2 hours.
**Electrical conductivity (κ)**: The ability of a solution to conduct electric current; measured in siemens per meter (S m⁻¹) or siemens per centimeter (S cm⁻¹).
**Resistance (R)** and **Resistivity (ρ)** relationship:
**R = ρ × (l/A)**
Where:
**Conductance (G)**: Reciprocal of resistance:
**G = 1/R** (measured in siemens, S)
**Conductivity (κ)**: Reciprocal of resistivity:
**κ = 1/ρ** (measured in S cm⁻¹)
Therefore: **κ = G × (l/A)** or **G = κ × (A/l)**
**Molar conductivity (Λₘ)**: The conductivity of solution containing 1 mole of electrolyte; measured in S cm² mol⁻¹.
**Λₘ = κ × (1000/c)**
Where:
Alternative expression:
**Λₘ = (κ/c) × 1000** (when c is in mol dm⁻³)
**Measurement of conductivity**:
**Variation with concentration**:
**Exam-important distinctions**:
**Statement**: The molar conductivity of an electrolyte at infinite dilution is equal to the sum of molar conductivities of its constituent ions at infinite dilution.
**Mathematical form**: **Λ°ₘ = λ°₊ + λ°₋**
Where:
**For compounds with polyatomic ions**:
**Λ°ₘ(NaCl) = λ°(Na⁺) + λ°(Cl⁻)**
**Λ°ₘ(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻)**
**Important applications**:
1. **Determination of degree of dissociation (α)**:
**α = Λₘ/Λ°ₘ** (at any concentration)
Where Λₘ is molar conductivity at concentration c, and Λ°ₘ is at infinite dilution
2. **Calculation of unknown molar conductivity**:
If Λ°ₘ of an unknown electrolyte cannot be directly measured, it can be calculated from known ionic conductivities.
3. **Determination of Ka for weak electrolytes**:
For weak electrolyte HA:
**Ka = (α² × c)/(1 − α)** where **α = Λₘ/Λ°ₘ**
Or: **Ka = (Λₘ)² × c / (Λ°ₘ × (Λ°ₘ − Λₘ))**
**Example**: For acetic acid (weak acid):
**Exam-critical points**:
**Definition**: A battery is a galvanic cell (or combination of cells) that converts chemical energy directly into electrical energy through spontaneous redox reactions.
**Construction**:
**Electrode reactions during discharge**:
**At anode (oxidation)**:
Pb(s) + SO₄²⁻ → PbSO₄(s) + 2e⁻
**At cathode (reduction)**:
PbO₂(s) + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄(s) + 2H₂O(l)
**Overall cell reaction**:
Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
**Standard cell potential**: E°cell = 2.1 V per cell
**Charging process**: When connected to external source, reactions reverse (electrolytic mode):
**Advantages**:
**Disadvantages**:
**Exam importance**: Lead storage battery is the most common secondary battery (rechargeable cell) in vehicles.
**Definition**: A fuel cell is a galvanic cell in which chemical energy of fuel combustion is directly converted to electrical energy without heat generation.
**Construction of H₂-O₂ fuel cell**:
**Electrode reactions in acidic medium**:
**At anode (oxidation)**:
H₂(g) → 2H⁺(aq) + 2e⁻ (E° = 0.00 V)
**At cathode (reduction)**:
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) (E° = 1.23 V)
**Overall cell reaction**:
2H₂(g) + O₂(g) → 2H₂O(l)
**Standard cell potential**:
E°cell = 1.23 − 0.00 = 1.23 V
**Electrode reactions in alkaline medium**:
**At anode**:
2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
**At cathode**:
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
**Overall reaction**: Same as acidic medium: 2H₂ + O₂ → 2H₂O
**Advantages of fuel cells**:
**Disadvantages**:
**Real-world applications**:
**Exam-critical concept**: Fuel cells are non-rechargeable galvanic cells that consume reactants and cannot be regenerated by reversing the reaction with external voltage.
**Definition**: Corrosion is the spontaneous electrochemical degradation of metals and their alloys when exposed to environmental conditions (moisture, oxygen, etc.).
**Types of corrosion**:
1. **Electrochemical corrosion** (most common)
2. **Direct chemical attack**
**Mechanism of electrochemical corrosion** (rusting of iron example):
Iron corrodes in presence of water and oxygen:
**4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)** (overall)
Or: 4Fe(s) + 3O₂(g) + 2H₂O(l) → 4FeO·OH(s) (hydrated iron oxide)
This involves two half-reactions:
**Anode reaction (oxidation)**:
2Fe(s) → 2Fe²⁺(aq) + 4e⁻
**Cathode reaction (reduction)**:
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
The Fe²⁺ ions react with OH⁻ ions and oxygen to form rust (Fe₂O₃·nH₂O or FeO·OH).
**Electrochemical cell formation**:
**Galvanic corrosion** (when two different metals in contact):
**Factors affecting corrosion rate**:
**Method 1: Coating/Barrier protection**
**Method 2: Sacrificial anode protection (Cathodic protection)**
**Method 3: Impressed current cathodic protection**
**Method 4: Alloying**
**Method 5: Corrosion inhibitors**
**Method 6: Environmental control**
**Real-life example**: Protection of underground pipelines uses combination of:
**Exam-important points**:
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**Summary of Key Formulae for Board Exam**:
1. **Cell potential**: E°cell = E°cathode − E°anode
2. **Nernst equation**: Ecell = E°cell − (0.0592/n) log Q (at 298 K)
3. **Gibbs energy relation**: ΔG° = −nFE°cell
4. **Equilibrium constant**: E°cell = (0.0592/n) log Kc
5. **Conductivity**: κ = G × (l/A)
6. **Molar conductivity**: Λₘ = κ × (1000/c)
7. **Kohlrausch's law**: Λ°ₘ = λ°₊ + λ°₋
8. **Degree of dissociation**: α = Λₘ/Λ°ₘ
9. **Faraday's law**: m = (M × I × t)/(n × F)
10. **Ka from conductivity**: Ka = (Λₘ)² × c / (Λ°ₘ × (Λ°ₘ − Λₘ))
Q1. In a Daniell cell, the standard cell potential E° = 1.1 V at 25°C. What is ΔG° for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)? (F = 96500 C/mol)
Answer: A — ΔG° = −nFE° = −2 × 96500 × 1.1 = −212,300 J/mol = −212.3 kJ/mol; negative value indicates spontaneity.
Q2. A 0.1 M NaCl solution has conductivity κ = 1.2 × 10⁻² S/cm. What is its molar conductivity?
Answer: A — Molar conductivity Λm = (κ/c) × 1000 = (1.2 × 10⁻²/0.1) × 1000 = 120 S·cm²/mol (factor 1000 converts concentration from mol/cm³ to mol/L).
Q3. According to Kohlrausch's law, the molar conductivity at infinite dilution Λ°m for an electrolyte depends on:
Answer: C — Kohlrausch's law states Λ°m = Σ(λ° cation) + Σ(λ° anion); it is independent of concentration and depends on ionic mobilities.
Q4. In the electrolysis of molten NaCl, 0.5 mol of electrons is passed through the cell. How many moles of Cl₂ gas are produced at the anode?
Answer: A — At the anode: 2Cl⁻ → Cl₂ + 2e⁻; 0.5 mol electrons produces 0.5/2 = 0.25 mol Cl₂.
Q5. Which of the following statements about the Nernst equation is NOT correct? (I) Q represents the reaction quotient with all species at their instantaneous concentrations. (II) As Q increases, Ecell decreases. (III) At equilibrium, Ecell = E° because Q = K.
Answer: C — At equilibrium, Ecell = 0 (not E°) because E° = (0.0592/n)log(K), so Ecell = E° − (0.0592/n)log(K) = 0; statements (I) and (II) are correct.
Q6. A galvanic cell has E° = +0.34 V. Which statement is correct? (A) The cell reaction is non-spontaneous and ΔG° > 0. (B) The cell reaction is spontaneous and ΔG° < 0. (C) The cell reaction is at equilibrium with ΔG° = 0. (D) The cell potential will increase with temperature.
Answer: B — Since E° is positive, ΔG° = −nFE° is negative, indicating the reaction is spontaneous; options (A) and (C) contradict this.
Q7. Calculate the mass of copper deposited at the cathode when 2 A current passes through a copper sulfate solution for 30 minutes. (Atomic mass Cu = 64 g/mol, F = 96500 C/mol)
Answer: A — Charge Q = I × t = 2 × (30 × 60) = 3600 C; moles of Cu = Q/(nF) = 3600/(2 × 96500) = 0.01865 mol; mass = 0.01865 × 64 = 1.19 g.
Q8. In a galvanic cell, why does molar conductivity increase as the solution becomes more dilute?
Answer: B — Upon dilution, ions are more dispersed, interionic attractions decrease, and ions move faster with higher mobility, increasing molar conductivity despite lower total ion concentration.
Q9. HOTS: A galvanic cell is constructed using Zn and Cu electrodes. Initially, [Zn²⁺] = 0.1 M and [Cu²⁺] = 1.0 M. Given E°Zn²⁺|Zn = −0.76 V and E°Cu²⁺|Cu = +0.34 V, calculate Ecell using the Nernst equation at 25°C. (R = 8.314 J/(mol·K), F = 96500 C/mol)
Answer: A — E°cell = 0.34 − (−0.76) = 1.10 V; Q = [Zn²⁺]/[Cu²⁺] = 0.1/1.0 = 0.1; Ecell = 1.10 − (0.0592/2)log(0.1) = 1.10 − (0.0296 × −1) = 1.10 + 0.0296 ≈ 1.03 V.
Q10. During the electrolysis of acidified water using inert electrodes, oxygen is produced at the anode and hydrogen at the cathode. If 500 cm³ of O₂ (at STP) is collected, what volume of H₂ (at STP) is produced? (Assume 2 mol H₂O → 2 mol H₂ + 1 mol O₂)
Answer: C — Stoichiometry: 1 mol O₂ : 2 mol H₂; moles O₂ = 500/22400 mol; moles H₂ = 2 × (500/22400) = 1000/22400; volume H₂ = 1000 cm³ at STP.
What is the definition of a galvanic cell?
A galvanic cell is an electrochemical device that converts the chemical energy of a spontaneous redox reaction into electrical energy.
In a galvanic cell, which electrode is negative and where does oxidation occur?
The anode is negative and oxidation occurs there; electrons flow from anode to cathode through the external circuit.
State the Nernst equation and define each symbol.
Ecell = E° − (RT/nF)ln(Q), where R = gas constant, T = temperature, n = moles of electrons, F = Faraday constant (96500 C/mol), Q = reaction quotient.
What is molar conductivity and how does it vary with dilution?
Molar conductivity Λm = κ/c (conductivity ÷ concentration) and it increases with dilution because ions move faster when dispersed.
State Kohlrausch's law of independent ion migration.
At infinite dilution, the molar conductivity of an electrolyte is the sum of limiting molar conductivities of its constituent cations and anions: Λ°m = Σ(λ° cation) + Σ(λ° anion).
What is Faraday's first law of electrolysis?
The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of charge passed: moles = Q/(nF) where Q is charge and n is electrons transferred.
How are ΔG° and E° cell related?
ΔG° = −nFE°cell; if E° is positive, ΔG° is negative and the reaction is spontaneous.
In an electrolytic cell, which electrode is the anode and what happens there?
The anode is positive and oxidation occurs there; species lose electrons and are oxidized.
What is the relationship between conductivity κ and resistivity ρ?
Conductivity κ is the reciprocal of resistivity: κ = 1/ρ; conductivity measures how well a solution conducts electricity.
How does applying an external voltage greater than the cell potential change the direction of a galvanic cell reaction?
When external voltage exceeds cell potential, the reaction reverses and the cell functions as an electrolytic cell, converting electrical energy into chemical energy.
Define electrode potential and standard electrode potential. Give one example. [2 marks]
Electrode potential is the potential difference at electrode-electrolyte interface; standard electrode potential is when all species are at unit concentration (1 M) or unit activity. Example: E°Cu²⁺|Cu = +0.34 V means Cu²⁺ + 2e⁻ → Cu at standard conditions.
Derive the relationship between ΔG°, E° and equilibrium constant K. Show all steps and explain the significance when E° > 0, E° = 0, and E° < 0. [5 marks]
Start with ΔG° = −nFE° and relate to ΔG° = −RT ln(K); set up equilibrium condition Ecell = 0 = E° − (RT/nF)ln(K); analyse sign of E° against ΔG° and K to interpret spontaneity and direction.
A silver coulometer (containing AgNO₃ solution with silver electrodes) is connected in series with a copper sulfate cell during electrolysis. When 1.08 g of silver is deposited at the cathode of the coulometer, calculate: (i) the total charge passed through the circuit in coulombs, (ii) the mass of copper deposited at the cathode of the copper sulfate cell. (Atomic masses: Ag = 108 g/mol, Cu = 64 g/mol; F = 96500 C/mol) [6 marks]
For (i): Ag⁺ + e⁻ → Ag; moles Ag = 1.08/108 = 0.01 mol; charge = 0.01 × F = 965 C. For (ii): In CuSO₄ cell, Cu²⁺ + 2e⁻ → Cu; same charge flows so moles Cu = 965/(2 × 96500) = 0.005 mol; mass Cu = 0.005 × 64 = 0.32 g (show stoichiometric ratio 1 mol e⁻ : Ag but 2 mol e⁻ : Cu).
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