JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
For the cell \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.1 \mathrm{M}) \| \mathrm{Ag}^{+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})\) the cell potential \(\mathrm{E}_{1}=0.3095\, \mathrm{~V}\) For the cell \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(\mathrm{aq})(0.001 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})\) the cell potential \(=.....\times 10^{-2} \mathrm{~V}\) (Round off to the Nearest Integer). [Use: \(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\) ]
- A \(4\)
- B \(14\)
- C \(28\)
- D \(36\)
Answer & Solution
Correct Answer
(C) \(28\)
Step-by-step Solution
Detailed explanation
Cell reaction is: \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}(\mathrm{s})\) Now, \(\mathrm{E}_{\text {cell }}-\frac{0.059}{2} \log \frac{\left[\mathrm{cu}^{+}+\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}...(1)\)…
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