\(\Delta G^\circ\) (in kJ mol\(^{-1}\)) for the cell reaction given below is
\[
2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)
\]
\[
\text{Given: }E^\circ_{Al^{3+}/Al}=-1.66V,\quad E^\circ_{Cu^{2+}/Cu}=+0.34V,\quad F=96500\;C\,mol^{-1}
\]
Show Hint
Always use
\[
\Delta G^\circ=-nFE^\circ_{cell}
\]
and remember that \(n\) is the number of electrons exchanged in the balanced cell reaction.
Step 1: Recall the Gibbs energy link.
The standard free energy change of a cell reaction is related to its cell potential by \[ \Delta G^\circ = -nFE^\circ_{cell}, \] where $n$ is the electrons transferred and $F$ is the Faraday constant.
Step 2: Identify the half-reactions.
Aluminium is oxidised: $Al \rightarrow Al^{3+} + 3e^-$. Copper ion is reduced: $Cu^{2+} + 2e^- \rightarrow Cu$.
Step 3: Find the cell potential.
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (+0.34) - (-1.66) = 2.00 \ \text{V}. \]
Step 4: Count the electrons.
To balance the reaction $2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu$, six electrons are exchanged (two Al lose $3e^-$ each). So $n = 6$.
Step 5: Plug into the formula.
\[ \Delta G^\circ = -(6)(96500)(2.00) = -1158000 \ \text{J mol}^{-1}. \]
Step 6: Convert and conclude.
Dividing by $1000$ gives kilojoules: \[ \boxed{\Delta G^\circ = -1158 \ \text{kJ mol}^{-1}} \]