The standard electrode potential for the Daniell cell is $1.1 \, V$. Calculate the standard Gibbs energy for the reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$

(N/A) The formula for standard Gibbs energy is ${\Delta _r}{G^\Theta } = - nFE_{(cell)}^\Theta $.
In this reaction,the number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is $96487 \, C \, mol^{-1}$ and the standard cell potential $(E_{(cell)}^\Theta )$ is $1.1 \, V$.
Substituting these values into the equation:
${\Delta _r}{G^\Theta } = - 2 \times 96487 \, C \, mol^{-1} \times 1.1 \, V$
${\Delta _r}{G^\Theta } = - 212271 \, J \, mol^{-1}$
Converting to kilojoules:
${\Delta _r}{G^\Theta } = - 212.27 \, kJ \, mol^{-1}$