Question

Calculate the equilibrium constant of the following reaction:
\[ \mathrm{Cu_{(s)} + 2Ag^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}}, \qquad E^\circ_{\text{cell}} = 0.46\,V \]

• A. \(3.92 \times 10^{14}\)
• B. \(4.92 \times 10^{13}\)
• C. \(4.92 \times 10^{16}\)
• D. \(3.92 \times 10^{15}\)

Hint:

For electrochemical cells at \(298\,K\), always remember the formula \(\displaystyle E^\circ_{\text{cell}}=\frac{0.0591}{n}\log K\). A positive and large \(E^\circ_{\text{cell}}\) means the value of \(K\) will also be very large.

Answer 1

The Correct Option is D

Step 1: Use the relation between \(E^\circ_{\text{cell}}\) and equilibrium constant.
At \(298\,K\), the standard cell potential and equilibrium constant are related by the formula:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n}\log K \] where \(n\) is the number of electrons transferred in the balanced redox reaction.
Step 2: Find the value of \(n\).
From the reaction, copper is oxidised as:
\[ \mathrm{Cu \rightarrow Cu^{2+} + 2e^-} \] and silver ions are reduced as:
\[ \mathrm{2Ag^+ + 2e^- \rightarrow 2Ag} \] Hence, the number of electrons transferred is:
\[ n = 2 \] Step 3: Substitute the given values.
Given,
\[ E^\circ_{\text{cell}} = 0.46\,V \] Using the formula,
\[ 0.46 = \frac{0.0591}{2}\log K \] So,
\[ \log K = \frac{2 \times 0.46}{0.0591} \] \[ \log K = \frac{0.92}{0.0591} \approx 15.57 \] Step 4: Calculate \(K\).
\[ K = 10^{15.57} \] \[ K \approx 3.92 \times 10^{15} \] Step 5: Conclusion.
Therefore, the equilibrium constant of the given reaction is \(3.92 \times 10^{15}\). Hence, the correct option is \((D)\).
Final Answer:\(3.92 \times 10^{15}\).