Question

The standard Gibbs energy for the reaction is \(-212.27\,\mathrm{kJ\,mol^{-1}}\). \[ Zn(s)+Cu^{2+}(aq.) \rightarrow Zn^{2+}(aq.)+Cu(s) \] The standard electrode potential \((E^\circ_{\text{cell}})\) for the Daniell cell is:

• A. \(1.1\text{ V}\)
• B. \(2.2\text{ V}\)
• C. \(0.002\text{ V}\)
• D. \(0.001\text{ V}\)

Hint:

Use the relation \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] A negative \(\Delta G^\circ\) corresponds to a positive cell potential and a spontaneous reaction.

Answer 1

The Correct Option is A

Concept: The relationship between Gibbs free energy change and cell potential is \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] where \[ n=\text{number of electrons transferred} \] \[ F=96500\ \mathrm{C\,mol^{-1}} \]

Step 1: Determine the number of electrons transferred.
The reaction is \[ Zn \rightarrow Zn^{2+}+2e^- \] \[ Cu^{2+}+2e^- \rightarrow Cu \] Hence, \[ n=2 \]

Step 2: Substitute the values in the equation.
\[ \Delta G^\circ=-212.27\times10^3\ \mathrm{J\,mol^{-1}} \] Using \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] \[ 212.27\times10^3 = 2\times96500\times E^\circ_{\text{cell}} \] \[ E^\circ_{\text{cell}} = \frac{212270}{193000} \] \[ E^\circ_{\text{cell}} \approx1.10\text{ V} \]

Step 3: Final answer.
\[ {E^\circ_{\text{cell}}=1.1\text{ V}} \]