CBSE 2023 · Region 4 · Set 1 · Q33 · 5 marks
(i)Calculate the emf of the following cell at $\displaystyle 298$ K : \[\begin{aligned} & \mathrm{Al}(\mathrm{~s})\left|\mathrm{Al}^{3+}(0.001 \mathrm{M}) \| \mathrm{Ni}^{2+}(0.1 \mathrm{M})\right| \mathrm{Ni}(\mathrm{~s}) \\ & {\left[\text { Given }: \mathrm{E}_{\mathrm{Al}^{\circ}}{ }^{3+} / \mathrm{Al}=-1.66 \mathrm{~V}, \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{~V}, \log 10=1\right]} \end{aligned} \](ii)With the help of a graph explain why it is not possible to determine $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ}$ for a weak electrolyte by extrapolating the molar conductivity $\displaystyle \left(\mathbf{\Lambda}_{\mathrm{m}}\right)$ versus $\displaystyle \mathrm{C}^{1 / 2}$ curve as for strong electrolyte.(i)The molar conductivities of $\displaystyle \mathrm{NH}_{4}^{+}$and $\displaystyle \mathrm{Cl}^{-}$ion are $\displaystyle 73.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\displaystyle 76.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The conductivity of $\displaystyle 0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$ is $\displaystyle 1.29 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation.(ii)Calculate the half-cell potential at $\displaystyle 298$ K for the reaction \[\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} \] if $\displaystyle \left[\mathrm{Zn}^{2+}\right]=0.1 \mathrm{M}$ and $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$.
(i)
Calculate the emf of the following cell at $\displaystyle 298$ K : \[\begin{aligned} & \mathrm{Al}(\mathrm{~s})\left|\mathrm{Al}^{3+}(0.001 \mathrm{M}) \| \mathrm{Ni}^{2+}(0.1 \mathrm{M})\right| \mathrm{Ni}(\mathrm{~s}) \\ & {\left[\text { Given }: \mathrm{E}_{\mathrm{Al}^{\circ}}{ }^{3+} / \mathrm{Al}=-1.66 \mathrm{~V}, \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{~V}, \log 10=1\right]} \end{aligned} \]
(ii)
With the help of a graph explain why it is not possible to determine $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ}$ for a weak electrolyte by extrapolating the molar conductivity $\displaystyle \left(\mathbf{\Lambda}_{\mathrm{m}}\right)$ versus $\displaystyle \mathrm{C}^{1 / 2}$ curve as for strong electrolyte.
(i)
The molar conductivities of $\displaystyle \mathrm{NH}_{4}^{+}$and $\displaystyle \mathrm{Cl}^{-}$ion are $\displaystyle 73.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\displaystyle 76.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The conductivity of $\displaystyle 0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$ is $\displaystyle 1.29 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation.
(ii)
Calculate the half-cell potential at $\displaystyle 298$ K for the reaction \[\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} \] if $\displaystyle \left[\mathrm{Zn}^{2+}\right]=0.1 \mathrm{M}$ and $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$.
Marking-scheme solution
(a) (i)
\[\begin{aligned}
\mathrm{E}_{\text {cell }}= & \mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{6} \log \frac{\left[\mathrm{Al}^{3+}\right]^{2}}{\left[\mathrm{Ni}^{2+}\right]^{3}} \\
\mathrm{E}_{\text {cell }}= & {[-0.25+1.66]-\frac{0.059}{6} \log \frac{[0.001]^{2}}{[0.1]^{3}} } \\
& =1.41-\frac{0.059}{6} \log 10^{-6+3} \\
& =1.41+\frac{0.059}{6} \times 3 \\
& =1.41+0.0295 \\
& =1.4395 \mathrm{~V}
\end{aligned}
\]
(b) (i)
(ii) As seen from the curve, it runs parallel to the y-axis. So, even on extrapolation, it will not intercept, hence \(\displaystyle \mathbf{\Lambda}^{\circ}{ }_{\mathrm{m}}\) cannot be obtained.
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.