CBSE 2023 · Region 1 · Set 1 · Q33 · 5 marks
(i)State Kohlrausch's law of independent migration of ions. Write an expression for the limiting molar conductivity of acetic acid according to Kohlrausch's law.(ii)Calculate the maximum work and $\displaystyle \log \mathrm{K}_{\mathrm{c}}$ for the given reaction at $\displaystyle 298$ K: \[\mathrm{Ni}(\mathrm{~s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{~s}) \] Given: $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}$ \[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1} \](i)State Faraday's first law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of $\displaystyle 1 \mathrm{~mol} \mathrm{Cu}^{2+}$ to Cu?(ii)Calculate emf of the following cell at $\displaystyle 298$ K for \[\begin{aligned} & \mathrm{Mg}(\mathrm{~s})\left|\mathrm{Mg}^{2+}(0 \cdot 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(0 \cdot 01 \mathrm{M})\right| \mathrm{Cu}(\mathrm{~s}) \\ & {\left[\mathrm{E}_{\text {cell }}^{\circ}=+2 \cdot 71 \mathrm{~V}, \quad 1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}, \quad \log 10=1\right]} \end{aligned} \]
(i)
State Kohlrausch's law of independent migration of ions. Write an expression for the limiting molar conductivity of acetic acid according to Kohlrausch's law.
(ii)
Calculate the maximum work and $\displaystyle \log \mathrm{K}_{\mathrm{c}}$ for the given reaction at $\displaystyle 298$ K: \[\mathrm{Ni}(\mathrm{~s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{~s}) \] Given: $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}$ \[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1} \]
(i)
State Faraday's first law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of $\displaystyle 1 \mathrm{~mol} \mathrm{Cu}^{2+}$ to Cu?
(ii)
Calculate emf of the following cell at $\displaystyle 298$ K for \[\begin{aligned} & \mathrm{Mg}(\mathrm{~s})\left|\mathrm{Mg}^{2+}(0 \cdot 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(0 \cdot 01 \mathrm{M})\right| \mathrm{Cu}(\mathrm{~s}) \\ & {\left[\mathrm{E}_{\text {cell }}^{\circ}=+2 \cdot 71 \mathrm{~V}, \quad 1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}, \quad \log 10=1\right]} \end{aligned} \]
Marking-scheme solution
(i)
Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
\[\mathbf{\Lambda}_{\mathrm{m}}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=\lambda^{\circ} \mathrm{CH}_{3} \mathrm{COO}^{-}+\lambda^{\circ} \mathrm{H}^{+}
\]
ElectrochemistryNernst EquationApplylong_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.