Question

The equilibrium constant for the disproportionation reaction
\( 2\text{Cu}^+ (aq) \rightleftharpoons \text{Cu} (s) + \text{Cu}^{2+} (aq) \)
at \( 25^\circ \text{C} \) is given that
\( E^\circ_{\text{Cu}^+/\text{Cu}} = 0.52\,\text{V}, \quad E^\circ_{\text{Cu}^{2+}/\text{Cu}^+} = 0.16\,\text{V} \).

• A. \(6\times10^4\)
• B. \(6\times10^6\)
• C. \(1.2\times10^6\)
• D. 1.2×10⁸

Hint:

Use log K=(nE^∘)/(0.0591) at 25^∘C.

Answer 1

The Correct Option is C

Step 1: Standard EMF:
\( E^\circ = 0.52 - 0.16 = 0.36\,\text{V} \) 

Step 2: Relation between \( E^\circ \) and \( K \):
\( \log K = \dfrac{nE^\circ}{0.0591} \) 

Step 3: Substituting values \( (n = 1) \):
\( \log K = \dfrac{0.36}{0.0591} \approx 6.1 \)
\( \Rightarrow K \approx 1.2 \times 10^6 \)