⚡ Electrochemistry - Complete Notes
📘 1. Introduction to Electrochemistry
Electrochemistry deals with the relationship between electrical energy and chemical energy — specifically, the study of redox reactions that involve this interconversion.
Key Concepts:
- Electrolysis = Chemical energy is created from electrical energy (using electricity to cause a redox reaction)
- Electrochemical Cell = Electrical energy is created by converting chemical energy (a redox process produces electrical energy)
📗 2. Electrolytes
An electrolyte is a substance that conducts electricity when dissolved in water (or in molten state) by producing ions.
Classification of Electrolytes
| Type | Description | Examples |
|---|---|---|
| Strong Electrolytes | Completely dissociates into ions in water | NaCl, KCl, HCl, HNO₃, H₂SO₄ |
| Weak Electrolytes | Partially dissociates/ionizes into ions | CH₃COOH, NH₄OH, NH₃, H₂CO₃ |
| Non-Electrolytes | Does not conduct electric current in molten state or aqueous solution | Urea, Glucose, Ethanol |
Dissociation Example:
NaCl (aq) → Na⁺(aq) + Cl⁻(aq)
H₂O → H⁺(aq) + OH⁻(aq)
📙 3. Difference Between Electrolyte and Electrochemistry
| Property | Electrolyte | Electrochemistry |
|---|---|---|
| Definition | A substance that conducts electricity when dissolved in water | A system/reaction involving the relationship between electrical and chemical energy |
| Energy Conversion | Chemical energy is created from electric energy | Electric energy is created by converting chemical energy |
| Process Type | Using electricity causes a redox reaction | A redox process produces electrical energy |
| Focus | The conducting medium itself | The entire system of electrical-chemical interconversion |
📕 4. Electrodes
Electrodes are metallic rods dipped into the electrolyte through which electricity enters or leaves the electrolyte.
Two Types of Electrodes:
| Electrode | Type of Reaction | Charge | Function |
|---|---|---|---|
| Anode | Oxidation | Positive (+) | Electrons leave; loss of electrons |
| Cathode | Reduction | Negative (−) | Electrons enter; gain of electrons |
🔑 Memory Trick:
- Oxidation = Loss of electrons (Anode)
- Reduction = Gain of electrons (Cathode)
- "An Ox, Red Cat" → Anode Oxidation, Reduction Cathode
📓 5. Electrolysis Reactions
A) Electrolysis of Water (H₂O)
At Cathode (Reduction): $\(2H^+ + 2e^- \rightarrow H_2 \uparrow\)$
At Anode (Oxidation): $\(2OH^- \rightarrow H_2O + \tfrac{1}{2}O_2 + 2e^-\)$
B) Electrolysis of Aqueous NaCl Solution
Dissociation: $\(NaCl \rightarrow Na^+ + Cl^-\)$
At Cathode (Reduction): $\(Na^+ + e^- \rightarrow Na \text{ (Sodium deposited at cathode)}\)$
At Anode (Oxidation): $\(2Cl^- \rightarrow Cl_2 \uparrow + 2e^- \text{ (Cl₂ gas released at anode)}\)$
C) Electrolysis of Molten NaCl
Dissociation: $\(NaCl (l) \rightarrow Na^+ + Cl^-\)$
At Cathode: $\(Na^+ + e^- \rightarrow Na\)$
At Anode:$\(2Cl^- \rightarrow Cl_2 + 2e^-\)$
📔 6. Faraday's Laws of Electrolysis
🔬 Faraday's First Law
The mass of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematical Form: $\(W \propto Q\)$ $\(W = ZQ\)$
Where: - W = Mass of substance deposited (grams) - Q = Quantity of electricity (Coulombs) - Z = Electrochemical equivalent (g/C)
Also: $\(Q = I \times t\)$
Where: - I = Current (Amperes) - t = Time (seconds)
Relation: $\(Z = \frac{W}{Q} = \frac{E}{F}\)$
Where: - E = Equivalent weight - F = Faraday constant = 96,500 C
🔬 Faraday's Second Law
When the same quantity of electricity is passed through different electrolytic solutions connected in series, the ratio of the weights of different substances deposited at different electrodes is equal to the ratio of their equivalent weights.
Mathematical Form: $\(\frac{W_1}{W_2} = \frac{E_1}{E_2}\)$
Where: - W₁, W₂ = Weights of substances deposited - E₁, E₂ = Equivalent weights of the substances
Derivation: $\(\because W_1 = Z_1 \times Q \quad \text{and} \quad W_2 = Z_2 \times Q\)$
🔋 7. Faraday Constant (F)
Definition: The charge carried by 1 mole of electrons is called 1 Faraday.
Calculation:
- Charge of 1 electron = \(1.6 \times 10^{-19}\) Coulombs
- 1 mole of electrons = \(6.022 \times 10^{23}\) electrons
Charge of 1 mole of electrons: $\(F = 6.022 \times 10^{23} \times 1.6 \times 10^{-19}\)$ $\(F = 96{,}368 \text{ C} \approx 96{,}500 \text{ C}\)$
⚖️ 8. Equivalent Weight Calculations
Formula:
Examples:
| Compound | Atomic/Molecular Weight | Valency (n) | Equivalent Weight (E) |
|---|---|---|---|
| NaCl | 23 (Na) | 1 | 23 |
| Na⁺ | 23 | 1 | 23 |
| Cu | 64 | 2 | 32 |
| Al | 27 | 3 | 9 |
| Ag | 108 | 1 | 108 |
| Mg(NO₃)₂ | 148 | 2 | 74 |
| NaNO₃ | 85 | 1 | 85 |
| BaSO₄ | 233 | 2 | 116.5 |
| Fe₂SO₄ | 304 | 2 | 152 |
| Zn(NO₃)₂ | 189 | 2 | 94.5 |
| FeCl₃ | 162.5 | 3 | 54.17 |
✏️ 9. Numerical Problems
📌 Problem 1: Electrochemical Equivalent of Ag
Question: When 3 amperes of electric current is passed for 20 minutes through a AgNO₃ solution, 4 g of Ag is deposited. Calculate the electrochemical equivalent (Z) of Ag.
Given: - Current (I) = 3 A - Time (t) = 20 min = 1200 s - Mass deposited (W) = 4 g
Solution:
Using Q = I × t: $\(Q = 3 \times 1200 = 3600 \text{ C}\)$
Using Faraday's First Law: W = ZQ $\(4 = Z \times 3600\)$ $\(Z = \frac{4}{3600} = \frac{1}{900} \text{ g/C}\)$
📌 Problem 2: Mass of Silver Deposited
Question: What amount of silver (Ag) will be deposited when an electric current of 2.1 A is passed through a AgNO₃ solution for 20 minutes? (Atomic weight of Ag = 108 a.m.u.)
Given: - Current (I) = 2.1 A - Time (t) = 20 min = 1200 s - Atomic weight of Ag = 108 - Valency = 1 → Equivalent weight (E) = 108 - Faraday constant (F) = 96,500 C
Solution:
Step 1: Calculate the charge (Q) $\(Q = I \times t = 2.1 \times 1200 = 2520 \text{ C}\)$
Step 2: Use the formula $\(W = \frac{1}{F} \times E \times Q\)$ $\(W = \frac{1}{96500} \times 108 \times 2520\)$ $\(\boxed{W = 2.82 \text{ g of Ag}}\)$
📌 Problem 3: Charge Required for Conversion
Question: How many coulombs of charge are required for the following conversion: $\(Cr^{2+}(aq) \rightarrow Cr(s) \text{ for 2 moles of electrons}\)$
Solution:
The reaction is: $\(Cr^{2+} + 2e^- \rightarrow Cr(s)\)$
For 2 moles of electrons, the charge required is: $\(Q = n \times F = 2 \times 96{,}500\)$
🧠 10. Quick Revision - Key Formulas
| Formula | Description |
|---|---|
| \(Q = I \times t\) | Charge = Current × Time |
| \(W = ZQ\) | Faraday's 1st Law |
| \(Z = \frac{E}{F}\) | Electrochemical equivalent |
| \(\frac{W_1}{W_2} = \frac{E_1}{E_2}\) | Faraday's 2nd Law |
| \(E = \frac{\text{Atomic Weight}}{n\text{-factor}}\) | Equivalent weight |
| \(F = 96{,}500 \text{ C}\) | Faraday's constant |
| \(W = \frac{E \times Q}{96500}\) | Mass deposited |
📌 11. Important Constants
| Constant | Value |
|---|---|
| Charge of 1 electron | $$ 1.6 \times 10^{-19} C$$ |
| Avogadro's number (Nₐ) | $\(6.022 \times 10^{23}\)$ |
| 1 Faraday (F) | 96,500 C ≈ 96,368 C |
💡 Tip: Remember the mnemonic "OIL RIG" — Oxidation Is Loss, Reduction Is Gain (of electrons), and "An Ox, Red Cat" for anode (oxidation) and cathode (reduction).