The standard electrode potential for Daniell cell is 1.1 V, calculate the standard Gibbs energy for the reaction,

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

It is found to be

\(-212.27 \text{kJ mol}^{-1}\)

The correct answer is option 1. \(-212.27 \text{kJ mol}^{-1}\).

To calculate the standard Gibbs free energy (\(\Delta G^\circ\)) for the reaction using the standard electrode potential (\(E^\circ\)), you can use the following relationship:

\(\Delta G^\circ = -nFE^\circ\)

where:

\(n\) is the number of moles of electrons transferred in the reaction,

\(F\) is the Faraday constant (\(96,485 \, \text{C/mol}\)),

\(E^\circ\) is the standard electrode potential of the cell.

For the Daniell cell reaction:

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

The number of moles of electrons transferred (\(n\)) is 2 (since \(Zn\) is oxidized from \(0\) to \(+2\) and \(Cu^{2+}\) is reduced from \(+2\) to \(0\)).

The standard electrode potential (\(E^\circ\)) is \(1.1 \, \text{V}\).

Now, calculating \(\Delta G^\circ\):

\(\Delta G^\circ = -nFE^\circ\)

\(\Delta G^\circ = -2 \times 96,485 \, \text{C/mol} \times 1.1 \, \text{V}\)

\(\Delta G^\circ = -2 \times 106,133.5 \, \text{J/mol}\)

\(\Delta G^\circ = -212,267 \, \text{J/mol}\)

\(\Delta G^\circ = -212.27 \, \text{kJ/mol}\)

So the correct answer is:1. \(-212.27 \text{kJ mol}^{-1}\)