AP EAMCET · Chemistry · Electrochemistry
Calculate the equilibrium constant of the reaction, \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\), given that for the reaction \(E_{\text {cell }}=0.46 \mathrm{~V}\).
- A \(4.2 \times 10^8\)
- B \(6.23 \times 10^9\)
- C \(3.92 \times 10^{15}\)
- D \(4.54 \times 10^{20}\)
Answer & Solution
Correct Answer
(C) \(3.92 \times 10^{15}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} \text { } \Delta G^{\circ} & =-n F E_{\text {cell }}^{\circ} \\ \Delta G^{\circ} & =-2.303 R T \log K \\ \log K & =\frac{n F E_{\text {cell }}^{\circ}}{2303 R T}=\frac{2 \times 96500 \times 0.46}{2.303 \times 8.31 \times 298}=15.56 \\ \log K & =15.56 \Rightarrow…
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