Following reaction takes place in Daniell cell:

\(Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+} (aq) + Cu(s); \) \(E^o_{cell} = 1.1 V\)

Standard Gibbs energy in this case will be:

\(-212.3 kJ/mol\)

The correct answer is option 2. \(-212.3 kJ/mol\).

The given reaction is

\(Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+} (aq) + Cu(s)\)

From the reaction, \(n = 2\)

Given,

\(E^o_{cell} = 1.1 V\)

\(F = 96500 C\)

Gibbs Free Energy,

\(\Delta G^o = -nFE^o_{cell}\)

or, \(\Delta G^o = - 2 × 96500 × 1.1\)

or, \(\Delta G^o = -212300 J\)

or, \(\Delta G^o = -212.3 kJ/mol\)