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:
Answer & explanation
Correct answer: option 2
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, \(\Delta G^o = -212.3 kJ/mol\)