Search any question & find its solution
Question:
Answered & Verified by Expert
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 $E_{\text {cell }}=0.46 \mathrm{~V}$.
Options:
Solution:
2542 Upvotes
Verified Answer
The correct answer is:
$3.92 \times 10^{15}$
$$
\begin{aligned}
\text { } \Delta G^{\circ} & =-n F E_{\text {cell }}^{\circ} \\
\Delta G^{\circ} & =-2.303 R T \log K \\
\log K & =\frac{n F E_{\text {cell }}^{\circ}}{2303 R T}=\frac{2 \times 96500 \times 0.46}{2.303 \times 8.31 \times 298}=15.56 \\
\log K & =15.56 \Rightarrow K=3.92 \times 10^{15}
\end{aligned}
$$
Hence, the correct option is (3).
\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 K=3.92 \times 10^{15}
\end{aligned}
$$
Hence, the correct option is (3).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.