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Electrochemistry - Conductance of Electrolytic Solutions

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Resistance (RR): The opposition to the flow of current. It is directly proportional to length (ll) and inversely proportional to the area of cross-section (AA): R=ρlAR = \rho \frac{l}{A}, where ρ\rho is resistivity.

Conductance (GG): The ease with which current flows through a conductor. It is the reciprocal of resistance: G=1RG = \frac{1}{R}. Unit: SS (Siemens) or Ω1\Omega^{-1}.

Conductivity (Specific Conductance, κ\kappa): The inverse of resistivity: κ=1ρ\kappa = \frac{1}{\rho}. It represents the conductance of a solution of 1 cm1\text{ cm} length and 1 cm21\text{ cm}^2 area of cross-section.

Cell Constant (GG^*): For a given cell, it is the ratio of the distance between electrodes (ll) to the area of the electrodes (AA): G=lAG^* = \frac{l}{A}. Also, κ=G×G\kappa = G \times G^*.

Molar Conductivity (Λm\Lambda_m): The conducting power of all the ions produced by dissolving one mole of an electrolyte in solution: Λm=κC\Lambda_m = \frac{\kappa}{C}.

Variation with Concentration: κ\kappa always decreases with a decrease in concentration (dilution) because the number of ions per unit volume decreases. However, Λm\Lambda_m increases with dilution as the total volume containing one mole of electrolyte increases.

Kohlrausch's Law of Independent Migration of Ions: At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions: Λm=ν+λ++νλ\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ.

Degree of Dissociation (α\alpha): For weak electrolytes, the ratio of molar conductivity at concentration CC to the limiting molar conductivity: α=ΛmΛm\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}.

📐Formulae

R=ρlAR = \rho \frac{l}{A}

G=1R=κAlG = \frac{1}{R} = \kappa \frac{A}{l}

κ=GR\kappa = \frac{G^*}{R}

Λm=κ×1000M (where κ is in S cm1 and M is Molarity)\Lambda_m = \frac{\kappa \times 1000}{M} \text{ (where } \kappa \text{ is in } S \text{ cm}^{-1} \text{ and } M \text{ is Molarity)}

Λm=ΛmAc (Debye-Huckel-Onsager equation for strong electrolytes)\Lambda_m = \Lambda_m^\circ - A\sqrt{c} \text{ (Debye-Huckel-Onsager equation for strong electrolytes)}

Λm(AxBy)=xλ++yλ\Lambda_m^\circ (A_xB_y) = x\lambda_+^\circ + y\lambda_-^\circ

α=ΛmΛm\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

Ka=Cα21αK_a = \frac{C\alpha^2}{1-\alpha}

💡Examples

Problem 1:

The resistance of a conductivity cell containing 0.001M0.001 M KClKCl solution at 298K298 K is 1500Ω1500 \Omega. What is the cell constant if conductivity of 0.001M0.001 M KClKCl solution at 298K298 K is 0.146×103S cm10.146 \times 10^{-3} S \text{ cm}^{-1}?

Solution:

G=κ×RG^* = \kappa \times R G=0.146×103S cm1×1500ΩG^* = 0.146 \times 10^{-3} S \text{ cm}^{-1} \times 1500 \Omega G=0.219 cm1G^* = 0.219 \text{ cm}^{-1}

Explanation:

The cell constant (GG^*) is the product of conductivity (κ\kappa) and resistance (RR). The units of Ω\Omega and SS (1/Ω1/\Omega) cancel out, leaving the unit cm1\text{cm}^{-1}.

Problem 2:

Calculate Λm\Lambda_m^\circ for CaCl2CaCl_2 and MgSO4MgSO_4 from the following data: λ(Ca2+)=119.0\lambda^\circ(Ca^{2+}) = 119.0, λ(Cl)=76.3\lambda^\circ(Cl^-) = 76.3, λ(Mg2+)=106.0\lambda^\circ(Mg^{2+}) = 106.0, λ(SO42)=160.0\lambda^\circ(SO_4^{2-}) = 160.0 (all units in S cm2 mol1S \text{ cm}^2 \text{ mol}^{-1}).

Solution:

For CaCl2CaCl_2: Λm(CaCl2)=λ(Ca2+)+2λ(Cl)\Lambda_m^\circ(CaCl_2) = \lambda^\circ(Ca^{2+}) + 2\lambda^\circ(Cl^-) Λm=119.0+2(76.3)=119.0+152.6=271.6S cm2 mol1\Lambda_m^\circ = 119.0 + 2(76.3) = 119.0 + 152.6 = 271.6 S \text{ cm}^2 \text{ mol}^{-1} For MgSO4MgSO_4: Λm(MgSO4)=λ(Mg2+)+λ(SO42)\Lambda_m^\circ(MgSO_4) = \lambda^\circ(Mg^{2+}) + \lambda^\circ(SO_4^{2-}) Λm=106.0+160.0=266.0S cm2 mol1\Lambda_m^\circ = 106.0 + 160.0 = 266.0 S \text{ cm}^2 \text{ mol}^{-1}

Explanation:

According to Kohlrausch's Law, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its ions, multiplied by their stoichiometric coefficients.

Conductance of Electrolytic Solutions Revision - Class 12 Chemistry CBSE