JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Assume a cell with the following reaction \(\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}^{+}\left(1 \times 10^{-3} \,\mathrm{M}\right) \rightarrow \mathrm{Cu}^{2+}(0.250\, \mathrm{M})+2 \mathrm{Ag}_{(\mathrm{s})}\) \(\mathrm{E}_{\mathrm{Cell}}^{\ominus}=2.97\, \mathrm{~V}\) \(\mathrm{E}_{\text {cell }}\) for the above reaction is \(....\,V.\) (Nearest integer) [Given : \(\log 2.5=0.3979, T=298\, \mathrm{~K}]\)
- A \(5\)
- B \(2\)
- C \(3\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}\) \(=2.97-\frac{0.059}{2} \log \frac{0.25}{\left(10^{-3}\right)^{2}}=2.81\,\mathrm{~V}\)
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