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

Step 1: Identify the two half reactions.
Chromium is oxidised: $Cr \rightarrow Cr^{3+}+3e^-$. Cadmium ion is reduced: $Cd^{2+}+2e^- \rightarrow Cd$. So $Cr$ is the anode and $Cd^{2+}/Cd$ is the cathode.
Step 2: Find the standard cell potential.
$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=(-0.4)-(-0.74)=0.34\ V$.
Step 3: Count the electrons in the balanced equation.
Balancing $2Cr + 3Cd^{2+} \rightarrow 2Cr^{3+}+3Cd$, chromium gives $2\times3=6$ electrons and cadmium takes $3\times2=6$ electrons. So $n=6$.
Step 4: Write the Gibbs energy relation.
$\Delta_rG^\circ = -nFE^\circ_{cell}$, with $F=96500\ C\,mol^{-1}$.
Step 5: Substitute the numbers.
$\Delta_rG^\circ = -6\times96500\times0.34 = -196860\ J$.
Step 6: Convert to kilojoules and choose the option.
$-196860\ J = -196.86\ kJ$, which matches option (1).
\[ \boxed{\Delta_rG^\circ=-196.86\ kJ\ \ \text{(Option 1)}} \]