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Electrochemistry - Conductance in electrolytic solutions

Grade 12ICSEChemistry

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

🔑Concepts

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Ohm's Law states that the current flowing through a conductor is directly proportional to the potential difference across it: V=IRV = IR.

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Resistance (RR) is given by R=ρlAR = \rho \frac{l}{A}, where ρ\rho is the resistivity, ll is the length, and AA is the area of cross-section. The unit is Ohm (Ω\Omega).

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Conductance (GG) is the reciprocal of resistance: G=1RG = \frac{1}{R}. Its unit is SiemensSiemens (SS) or Ω−1\Omega^{-1}.

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Conductivity (Specific Conductance, κ\kappa) is the reciprocal of resistivity: κ=1ρ=G×lA\kappa = \frac{1}{\rho} = G \times \frac{l}{A}. It represents the conductance of 11 cm3cm^3 of the electrolytic solution.

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Cell Constant (G∗G^*) is defined as the ratio of distance between electrodes (ll) to the area of cross-section (AA): G∗=lAG^* = \frac{l}{A}. Thus, κ=G×G∗\kappa = G \times G^*.

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Molar Conductivity (Λm\Lambda_m) is the conducting power of all the ions produced by dissolving 11 mole of an electrolyte in solution: Λm=κ×1000M\Lambda_m = \frac{\kappa \times 1000}{M}, where MM is Molarity.

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Equivalent Conductivity (Λeq\Lambda_{eq}) is the conducting power of all the ions produced by dissolving 11 gram equivalent of an electrolyte: Λeq=κ×1000N\Lambda_{eq} = \frac{\kappa \times 1000}{N}, where NN is Normality.

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Variation with Dilution: Conductivity (κ\kappa) decreases with dilution because the number of ions per unit volume decreases. However, Molar Conductivity (Λm\Lambda_m) and Equivalent Conductivity (Λeq\Lambda_{eq}) increase with dilution.

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Kohlrausch's Law states that at infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions: Λm∘=ν+λ+∘+ν−λ−∘\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ.

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Degree of Dissociation (α\alpha) for weak electrolytes can be calculated as α=ΛmcΛm∘\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}.

📐Formulae

R=ρlAR = \rho \frac{l}{A}

G=1RG = \frac{1}{R}

κ=G×lA\kappa = G \times \frac{l}{A}

G∗=lAG^* = \frac{l}{A}

Λm=κ×1000M\Lambda_m = \frac{\kappa \times 1000}{M}

Λeq=κ×1000N\Lambda_{eq} = \frac{\kappa \times 1000}{N}

Λm=Λm∘−Ac (Debye-Huckel-Onsager Equation)\Lambda_m = \Lambda_m^\circ - A \sqrt{c} \text{ (Debye-Huckel-Onsager Equation)}

Λm∘(AxBy)=xλm∘(Ay+)+yλm∘(Bx−)\Lambda_m^\circ (A_xB_y) = x \lambda_m^\circ (A^{y+}) + y \lambda_m^\circ (B^{x-})

α=ΛmcΛm∘\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}

Ka=cα21−αK_a = \frac{c \alpha^2}{1 - \alpha}

💡Examples

Problem 1:

The resistance of a 0.010.01 MM KClKCl solution is 15001500 Ω\Omega at 298298 KK. Calculate the conductivity and molar conductivity if the cell constant is 0.1460.146 cm−1cm^{-1}.

Solution:

Given: R=1500R = 1500 Ω\Omega, M=0.01M = 0.01 MM, G∗=0.146G^* = 0.146 cm−1cm^{-1}.

  1. Conductivity κ=G∗R=0.1461500=9.73×10−5\kappa = \frac{G^*}{R} = \frac{0.146}{1500} = 9.73 \times 10^{-5} S cm−1S\,cm^{-1}.
  2. Molar Conductivity Λm=κ×1000M=9.73×10−5×10000.01=9.73\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{9.73 \times 10^{-5} \times 1000}{0.01} = 9.73 S cm2 mol−1S\,cm^2\,mol^{-1}.

Explanation:

We first use the relationship between cell constant and resistance to find conductivity, then use the molarity to find molar conductivity.

Problem 2:

Calculate Λm∘\Lambda_m^\circ for CH3COOHCH_3COOH given Λm∘(HCl)=426\Lambda_m^\circ(HCl) = 426 S cm2 mol−1S\,cm^2\,mol^{-1}, Λm∘(NaCl)=126\Lambda_m^\circ(NaCl) = 126 S cm2 mol−1S\,cm^2\,mol^{-1}, and Λm∘(CH3COONa)=91\Lambda_m^\circ(CH_3COONa) = 91 S cm2 mol−1S\,cm^2\,mol^{-1}.

Solution:

Using Kohlrausch's Law: Λm∘(CH3COOH)=Λm∘(CH3COONa)+Λm∘(HCl)−Λm∘(NaCl)\Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(CH_3COONa) + \Lambda_m^\circ(HCl) - \Lambda_m^\circ(NaCl) Λm∘(CH3COOH)=91+426−126=391\Lambda_m^\circ(CH_3COOH) = 91 + 426 - 126 = 391 S cm2 mol−1S\,cm^2\,mol^{-1}.

Explanation:

To find the limiting molar conductivity of a weak electrolyte, we combine the limiting molar conductivities of strong electrolytes such that the unwanted ions cancel out.