Electrochemistry is a crucial and highly scoring topic in JEE Chemistry. It deals with the relationship between electrical energy and chemical changes. This guide provides a thorough understanding of electrochemical cells, electrode potentials, the Nernst equation, conductance, electrolysis, and important formulas and concepts for exam preparation.
Electrochemistry studies chemical processes that cause electrons to move, producing electric current or involving electrical energy to cause chemical changes. It involves two major types of cells:
A galvanic cell consists of two electrodes immersed in electrolyte solutions connected externally by a wire allowing electron flow.
The half-cell reactions involve oxidation at the anode and reduction at the cathode.
Standard electrode potential is the measure of the ability of a half-cell to gain or lose electrons compared to the standard hydrogen electrode (SHE), defined as 0 V.
\text{Oxidation half reaction: } \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \quad E^\circ = -0.76 \text{ V}
\text{Reduction half reaction: } \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \quad E^\circ = +0.34 \text{ V}
Cell potential is calculated as:
E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}
The Daniell cell consists of a zinc electrode in ZnSO₄ solution and a copper electrode in CuSO₄ solution. Oxidation at Zn anode and reduction at Cu cathode produce cell voltage:
E^\circ_\text{cell} = 0.34 - (-0.76) = 1.10 \text{ V}
The Nernst equation relates the cell potential to the concentration (or pressure) of the reactants and products under non-standard conditions.
E = E^\circ - \frac{RT}{nF} \ln Q
Where:
At 25°C (298 K), the equation simplifies to:
E = E^\circ - \frac{0.0591}{n} \log Q
The electrochemical series lists elements and their standard electrode potentials in order. Applications include:
Conductance measures how well an electrolyte solution conducts electricity.
Conductance is the reciprocal of resistance (\( R \)):
G = \frac{1}{R}
Measured in Siemens (S) or mho.
Specific conductance or conductivity is conductance of a solution of 1 cm length and 1 cm² cross-sectional area:
\kappa = \frac{G \times l}{A}
Where \( l \) is distance between electrodes and \( A \) is electrode area.
Molar conductance is conductance of all ions produced by one mole of electrolyte in solution:
\Lambda_m = \frac{\kappa \times 1000}{C}
Where \( C \) = concentration in mol/L.
Molar conductance increases with dilution due to decreased inter-ionic attraction.
At infinite dilution, each ion contributes a definite value to the molar conductance, independent of other ions.
\Lambda_m^\infty = \lambda^+ + \lambda^-
Electrolysis involves passing electric current to bring about a chemical change.
Mass of substance liberated at an electrode is proportional to the quantity of electricity passed.
m \propto Q = I \times t
or
m = Z \times Q
Where \( Z \) = electrochemical equivalent.
Masses of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights.
Using Faraday’s laws:
Q = I \times t
Moles of electrons = \( \frac{Q}{F} \)
Mass of substance = \( \frac{(E \times Q)}{F} \), where \( E \) is equivalent weight.
Primary Cells: Non-rechargeable cells, e.g., Dry cell.
Secondary Cells: Rechargeable cells, e.g., Lead-acid battery.
Fuel Cells: Generate electricity by continuous fuel supply, e.g., Hydrogen-oxygen fuel cell.
| Concept | Formula | Notes |
|---|---|---|
| Cell potential | \( E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \) | Standard conditions |
| Nernst Equation | \( E = E^\circ - \frac{0.0591}{n} \log Q \) | At 25°C |
| Conductance | \( G = \frac{1}{R} \) | Siemens |
| Specific Conductance | \( \kappa = \frac{G l}{A} \) | |
| Molar Conductance | \( \Lambda_m = \frac{\kappa \times 1000}{C} \) | mol/L concentration |
| Faraday’s First Law | \( m = Z Q \) | Electrochemical equivalent |
| Faraday’s Second Law | \( m \propto \text{Equivalent weight} \) | Mass proportionality |
Calculate the standard cell potential for the following cell:
\(\mathrm{Zn(s)}|\mathrm{Zn^{2+}(aq)}||\mathrm{Cu^{2+}(aq)}|\mathrm{Cu(s)}\)
Given, \(E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76 \, V\), \(E^\circ_{\mathrm{Cu^{2+}/Cu}} = +0.34 \, V\).
Solution:
\( E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = 0.34 - (-0.76) = 1.10 \, V \)
Calculate the emf of a cell at 25°C if the concentration of \(\mathrm{Cu^{2+}}\) is 0.01 M and \(\mathrm{Zn^{2+}}\) is 1 M. (Use Nernst equation)
Solution:
Reaction quotient, \( Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} = \frac{1}{0.01} = 100 \)
Number of electrons, \( n = 2 \)
\( E = E^\circ - \frac{0.0591}{n} \log Q = 1.10 - \frac{0.0591}{2} \log 100 = 1.10 - 0.0591 = 1.0409\, V \)
Electrochemistry is an interesting yet challenging topic, but with consistent practice and concept clarity, mastering it is definitely achievable and rewarding for your JEE exams.