CBSE Class 12 Chemistry 2023: question 33(a)

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Step-by-step solution

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1. (i) Kohlrausch's law: The limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion: ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$.

   1 mark

2. For acetic acid: ${\Lambda }_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})={\lambda }_{\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}}^{\circ }+{\lambda }_{{\mathrm{H}}^{+}}^{\circ }$
   (This can also be found as ${\Lambda }_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COONa})+{\Lambda }_m^{\circ }(\mathrm{HCl})-{\Lambda }_m^{\circ }(\mathrm{NaCl})$.)

   1 mark

3. (ii) $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0\cdot 80-(-0\cdot 25)=1\cdot 05$ V, with $n=2$.

   ½ mark

4. Maximum work $=-{\Delta }_r{G}^{\circ }=nF{E}_{cell}^{\circ }$.

   ½ mark

5. $=2\times 96500{\text{C mol}}^{-1}\times 1\cdot 05\text{V}=202650{\text{J mol}}^{-1}=202\cdot 65$ kJ mol⁻¹.

   1 mark

6. $\log K_c=\frac{n{E}_{cell}^{\circ }}{0\cdot 059}=\frac{2\times 1\cdot 05}{0\cdot 059}=35\cdot 59$

   1 mark

Answer: (i) Λ°m(CH₃COOH) = λ°(CH₃COO⁻) + λ°(H⁺); (ii) maximum work = 202.65 kJ mol⁻¹, log Kc ≈ 35.59

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
• Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant