JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
For an electrochemical cell \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})\) the ratio \(\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\) when this cell attains equilibrium is (Given \(\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}\) \(\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)\)
- A \(2.15\)
- B \(3.33\)
- C \(1.67\)
- D \(4.33\)
Answer & Solution
Correct Answer
(A) \(2.15\)
Step-by-step Solution
Detailed explanation
Cell reaction is \(\mathrm{Sn}(\mathrm{s})+\mathrm{Pb}^{+2}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{+2}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s})\dots (1)\) Apply Nernst equation:…
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