Question

The following reaction takes place in a galvanic cell \[ 2Cr(s)+3Cd^{2+}(aq)\rightarrow2Cr^{3+}(aq)+3Cd(s) \] What is \(\Delta_rG^\circ\) of this cell? Given: \[ F=96500~Cmol^{-1} \] \[ E_{Cd^{2+}/Cd}^{\circ}=-0.4V \] \[ E_{Cr^{3+}/Cr}^{\circ}=-0.74V \]

• A. -196.86
• B. -1968.6
• C. -32.81
• D. -19.686

Hint:

Always use formula: \(\Delta G^\circ=-nFE^\circ_{cell}\). Positive cell potential means negative Gibbs free energy.

Answer 1

The Correct Option is A

Concept: The Gibbs free energy change in electrochemical cells is related to cell potential by equation \[ \Delta G^\circ=-nFE^\circ_{cell} \] where \[ n=\text{electrons transferred} \] \[ F=\text{Faraday constant} \] \[ E^\circ_{cell}=\text{standard cell potential} \]

Step 1: Determine oxidation and reduction half reactions. Chromium loses electrons. Oxidation: \[ Cr\rightarrow Cr^{3+}+3e^- \] Cadmium ion gains electrons. Reduction: \[ Cd^{2+}+2e^-\rightarrow Cd \]

Step 2: Calculate cell potential. Formula: \[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \] Cathode: \[ Cd^{2+}/Cd=-0.4V \] Anode: \[ Cr^{3+}/Cr=-0.74V \] Thus \[ E^\circ_{cell}=(-0.4)-(-0.74) \] \[ E^\circ_{cell}=0.34V \]

Step 3: Determine electron transfer number. Balanced reaction shows \[ n=6 \]

Step 4: Apply Gibbs free energy equation. \[ \Delta G^\circ=-nFE^\circ \] \[ =-6\times96500\times0.34 \] \[ =-196860J \] Convert to kilojoules. \[ =-196.86kJ \] Hence \[ \boxed{-196.86kJ} \]