Why does molar conductivity (\Lambda_m) increase upon dilution for both strong and weak electrolytes?

For strong electrolytes, dilution increases ionic separation, drastically reducing interionic attractive forces and boosting ionic velocity. For weak electrolytes, dilution increases the degree of dissociation (\alpha) rapidly according to Ostwald's Dilution Law ($K_a = C\alpha^2$), generating significantly more charge carriers per mole of solute.

How do you accurately predict product formation at electrodes during the electrolysis of aqueous salts?

Compare reduction potentials at cathode and oxidation potentials at anode. At the cathode, the species with higher $E^\circ_{red}$ is reduced first (e.g., $Cu^{2+}$ over $H^+$). At the anode, the species with higher $E^\circ_{ox}$ is oxidized first. However, for gas evolution like $O_2$ vs $Cl_2$, kinetic overpotential can make $Cl_2$ oxidation kinetically preferred despite lower thermodynamic preference.

Why does a salt bridge prevent liquid junction potential in a galvanic cell?

The salt bridge contains inert ions like $K^+$ and $NO_3^-$ with nearly identical ionic transport numbers ($u_+ \approx u_-$). They migrate into half-cells at identical speeds to neutralize charge accumulation without letting solution boundaries mix, thereby completely eliminating liquid junction potential.

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Question #1NEET 2024

Calculate the cell potential $E_{cell}$ of the following cell at $298\text{ K}$ : $Zn(s) | Zn^{2+}(aq, 0.1\text{ M}) || Cu^{2+}(aq, 0.01\text{ M}) | Cu(s)$ . Given standard potentials $E^\circ_{Zn^{2+}/Zn} = -0.76\text{ V}$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate standard cell potential $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = +0.34\text{ V} - (-0.76\text{ V}) = +1.10\text{ V}$ with $n = 2$ .

Step 2

Write the Nernst equation for the cell reaction: $E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log_{10} \left(\frac{[Zn^{2+}]}{[Cu^{2+]}}\right)$ .

Step 3

Substitute given concentrations: $E_{cell} = 1.10 - 0.02955 \log_{10}\left(\frac{0.1}{0.01}\right) = 1.10 - 0.02955(1) \approx 1.070\text{ V}$ .

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Question #2JEE Main 2023

The molar conductivity of a $0.01\text{ M}$ weak monobasic acid $HA$ is $19.6\text{ S cm}^2\text{ mol}^{-1}$ . If limiting molar conductivities are $\lambda^\circ_{H^+} = 349.6\text{ S cm}^2\text{ mol}^{-1}$ and $\lambda^\circ_{A^-} = 42.4\text{ S cm}^2\text{ mol}^{-1}$ , calculate the acid dissociation constant $K_a$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Compute limiting molar conductivity using Kohlrausch's law: $\Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{A^-} = 349.6 + 42.4 = 392.0\text{ S cm}^2\text{ mol}^{-1}$ .

Step 2

Calculate degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{19.6}{392.0} = 0.05$ .

Step 3

Evaluate dissociation constant: $K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 = (0.01)(0.05)^2 = 2.5 \times 10^{-5}\text{ M}$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Nernst Equation for Cell EMF

Gibbs Free Energy and Electromotive Force

Kohlrausch's Law of Independent Migration

Molar Conductivity and Specific Conductance

Faraday's First Law of Electrolysis

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Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

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