Question

The standard electrode potential of Daniel cell is 1.1 volt. Calculate the standard Gibbs energy for the following reaction: \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]

Hint:

The Gibbs free energy of a cell reaction is calculated using \( \Delta G^0 = -nFE^0_{\text{cell}} \).

Answer 1

Step 1: Gibbs free energy is related to e.m.f. using: \[ \Delta G^0 = -n F E^0_{\text{cell}} \] where \( n = 2 \) (electrons transferred), \( F = 96500 \) C/mol, and \( E^0_{\text{cell}} = 1.1V \). 

Step 2: Calculate: \[ \Delta G^0 = - (2 \times 96500 \times 1.1) \] \[ \Delta G^0 = -212300 \text{ J/mol} = -212.3 \text{ kJ/mol} \] Thus, the Gibbs free energy is **-212.3 kJ/mol**.