Electrochemistry – Study Notes for AIIMS Chemistry
(Prepared for AIIMS‑admission candidates – concise, exam‑focused)
1. Simple Explanation
Electrochemistry studies the relationship between electricity and chemical change.
- Redox reactions (oxidation + reduction) can occur when electrons move through an external circuit.
- A galvanic (voltaic) cell generates electricity spontaneously from a redox reaction.
- An electrolytic cell uses electricity to drive a non‑spontaneous reaction.
- Electrolysis (e.g., water splitting, metal deposition) and cell potentials (voltage) are the two main pillars.
Remember: “Electro‑” means “electric”, “chemistry” means “chemical change”.
2. Key Points
| Concept | What It Means | Important Equations / Facts |
|---|---|---|
| Standard Electrode Potential (E°) | Potential of a half‑cell vs. SHE at 1 M, 25 °C, 1 atm. | (E = E^0 - \frac{0.0592}{n}\log Q) (Nernst) |
| Half‑Reaction | Either oxidation or reduction part of a redox pair. | (\text{Ox} + e^- \rightarrow \text{Red}) |
| Cell Potential (Ecell) | Net voltage of a galvanic cell. | (E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}) |
| Nernst Equation | Adjusts E° for non‑standard conditions. | (E = E^0 - \frac{RT}{nF}\ln Q) |
| Faraday’s Laws of Electrolysis | Quantitative relation between charge and mass. | 1. (m = \dfrac{QzM}{F}) 2. (Q = It) |
| Electrolytic Cell | Uses external power; products are at electrodes. | Anode: oxidation, Cathode: reduction |
| Electroplating/Corrosion | Practical uses of electrochemistry. | Electrodeposition vs. corrosion (unwanted oxidation) |
| Battery Chemistry | Common AIIMS topics: Alkaline, Lead‑acid, Lithium‑ion, Ni‑Cd. | Energy density, self‑discharge, safety |
| Electrolysis of Water | (2H_2O \rightarrow 2H_2 + O_2) | Overpotential, electrode materials |
Tip: Always write the full cell notation:
Anode || Cathode(e.g.,Zn | Zn²⁺ (0.01 M) || Cu²⁺ (0.01 M) | Cu)
3. Worked Example
Problem
Calculate the cell potential for the following galvanic cell at 25 °C:
[ \text{Zn | Zn}^{2+}(0.01,\text{M}) ;||; \text{Cu}^{2+}(0.01,\text{M}) | \text{Cu} ]
Solution
-
Identify Standard Potentials
[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \quad (n=2)\ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \quad (n=2) ] -
Write Half‑Reactions
[ \text{Anode (oxidation)}: \quad \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \ \text{Cathode (reduction)}: \quad \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} ] -
Calculate Ecell (Standard) [ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\ \text{V} ]
-
Apply Nernst Equation (non‑standard concentrations)
For each half‑cell: [ E_{\text{Zn}} = -0.76 - \frac{0.0592}{2}\log\frac{[Zn^{2+}]}{1} = -0.76 - 0.0296\log(0.01) = -0.76 - 0.0296(-2) = -0.76 + 0.0592 = -0.7008\ \text{V} ][ E_{\text{Cu}} = 0.34 - \frac{0.0592}{2}\log\frac{[Cu^{2+}]}{1} = 0.34 - 0.0296(-2) = 0.34 + 0.0592 = 0.3992\ \text{V} ]
-
Find Ecell (actual)
[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.3992 - (-0.7008) = 1.10\ \text{V} ]Result: The cell potential remains 1.10 V (concentrations were equal, so Q = 1).
Check
- Positive Ecell → spontaneous (galvanic).
- 1.10 V is a typical value for a Zn–Cu cell.
4. Quick Revision Summary
- Redox: oxidation = loss of e⁻, reduction = gain