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Chemistry · 2023 · 5 marks
CBSE 2023 · Region 3 · Set 1 · Q35
(a) Calculate the emf of the following cell at 25°C: Zn ( s )| Zn^2+(0 · 1 M ) ∥ H^+(0 · 01 M )| H_2( g )(1 bar ), Pt ( s ) [Given: E_Zn^2+ /…
(a)
Calculate the emf of the following cell at $\displaystyle 25$°C : \[\mathrm{Zn}(\mathrm{~s})\left|\mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M}) \| \mathrm{H}^{+}(0 \cdot 01 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \text { bar }), \operatorname{Pt}(\mathrm{s}) \] [Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{\circ}=0.00 \mathrm{~V}, \log 10=1$ ]
(b)
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution ?
Marking-scheme solution
(a)
\[\begin{aligned}
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \\
& =0.76 \mathrm{~V}-\frac{0.059}{2} \log \frac{[0.1]}{[0.01]^{2}} \\
& =0.76 \mathrm{~V}-\frac{0.059}{2} \log 10^{3} \\
& =0.76 \mathrm{~V}-\frac{0.059 \times 3}{2} \\
& \mathrm{E}_{\text {cell }}=0.671 \mathrm{~V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.