Question

Calculate log \(K_c\) for the following reaction at 298 K: Zn(s) + Cu$^{2+}$(aq) $\rightleftharpoons$ Zn$^{2+}$(aq) + Cu(s)
Given:
$E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}$, \quad $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V}$

Hint:

Use this shortcut: \(\log K_c = \frac{n E^\circ_{\text{cell}} \times 96500}{2.303RT}\)
At 298 K, the denominator simplifies to approx. 5700.

Answer 1

Step 1: Calculate standard EMF:
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10 \, \text{V}\]
Step 2: Use relation between EMF and equilibrium constant:
\[\Delta G^\circ = -nFE^\circ_{\text{cell}} = -RT \ln K_c\]
Or,
\[\log K_c = \frac{n E^\circ_{\text{cell}} \times 96500}{2.303 \times 8.314 \times 298}\]
Step 3: Substituting:
\[n = 2, \quad E^\circ = 1.10 \, \text{V} \]
\[\log K_c = \frac{2 \times 1.10 \times 96500}{2.303 \times 8.314 \times 298} \approx 37.29\]