Question

\(\Delta G^\circ\) (in kJ mol\(^{-1}\)) for the cell reaction given below is \[ 2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s) \] \[ \text{Given: }E^\circ_{Al^{3+}/Al}=-1.66V,\quad E^\circ_{Cu^{2+}/Cu}=+0.34V,\quad F=96500\;C\,mol^{-1} \]

• A. \(-1158\)
• B. \(-579\)
• C. \(-386\)
• D. \(-772\)

Hint:

Always use \[ \Delta G^\circ=-nFE^\circ_{cell} \] and remember that \(n\) is the number of electrons exchanged in the balanced cell reaction.

Answer 1

The Correct Option is A

Concept: The relationship between standard Gibbs free energy change and standard cell potential is \[ \Delta G^\circ=-nFE^\circ_{cell} \] where

• \(n\) = number of electrons transferred

• \(F\) = Faraday constant

• \(E^\circ_{cell}\) = standard cell potential

Step 1: Identify oxidation and reduction half-reactions. Oxidation: \[ Al\rightarrow Al^{3+}+3e^- \] Reduction: \[ Cu^{2+}+2e^-\rightarrow Cu \]

Step 2: Calculate standard cell potential. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ = (+0.34)-(-1.66) \] \[ =2.00V \]

Step 3: Determine number of electrons transferred. Balanced reaction: \[ 2Al+3Cu^{2+} \rightarrow 2Al^{3+}+3Cu \] Electrons exchanged: \[ n=6 \]

Step 4: Calculate \(\Delta G^\circ\). \[ \Delta G^\circ = -nFE^\circ_{cell} \] \[ =-(6)(96500)(2.00) \] \[ =-1158000J\,mol^{-1} \] \[ =-1158kJ\,mol^{-1} \]

Step 5: Final conclusion. \[ \boxed{\Delta G^\circ=-1158kJ\,mol^{-1}} \] Hence, \[ \boxed{\text{Option (A)}} \]