Question

The value of standard electrode potential of Daniell cell is 1.1 V. Calculate the value of standard Gibbs energy for the following reaction. \[ \text{Zn (s) + Cu}^{2+}\text{(aq)} \longrightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu (s)} \]

Hint:

Always remember: A negative Gibbs free energy (\(\Delta G^\circ<0\)) indicates a spontaneous reaction under standard conditions.

Answer 1

Step 1: Recall the relationship between Gibbs free energy and cell potential.
The formula is: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] where, - \(n\) = number of electrons transferred in the reaction = 2
- \(F\) = Faraday constant = \(96500 \, \text{C mol}^{-1}\)
- \(E^\circ_{\text{cell}}\) = standard electrode potential = \(1.1 \, \text{V}\) Step 2: Substitute the values.
\[ \Delta G^\circ = - (2)(96500)(1.1) \] \[ \Delta G^\circ = -212300 \, \text{J mol}^{-1} \] \[ \Delta G^\circ = -212.3 \, \text{kJ mol}^{-1} \] Conclusion:
The standard Gibbs energy for the Daniell cell reaction is: \[ \boxed{-212.3 \, \text{kJ mol}^{-1}} \]