Question

Given that
\[ \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s); \quad E^\circ_{\text{red}} = -0.76\,\text{V} \] \[ \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s); \quad E^\circ_{\text{red}} = -0.74\,\text{V} \] Calculate the equilibrium constant K at 25°C for the following balanced reaction,
\[ 3\text{Zn}(s) + 2\text{Cr}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Cr}(s) \]

• A. $K = e^{-0.02}$
• B. $K = e^{0.02}$
• C. $K = e^{4.7}$
• D. $K = e^{2.0}$

Hint:

To calculate equilibrium constant from standard EMF, use $\ln K = \dfrac{nFE^\circ}{RT}$ at 25°C.

Answer 1

The Correct Option is C

The standard EMF of the cell is: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.74) - (-0.76) = 0.02\,V \] For the reaction: \[ 3\text{Zn} \rightarrow 3\text{Zn}^{2+} + 6e^-, \quad 2\text{Cr}^{3+} + 6e^- \rightarrow 2\text{Cr} \Rightarrow n = 6 \] Using the relation: \[ \ln K = \frac{nFE^\circ_{\text{cell}}}{RT} = \frac{6 \cdot 96500 \cdot 0.02}{8.314 \cdot 298} \approx 4.7 \Rightarrow K = e^{4.7} \]