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Q2.4 (i):
Calculate the standard cell potentials of galvanic cell in which the following reaction takes place: (i) (2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd) Calculate the (\Delta_{r}G^{o}) and equilibrium constant of the reaction.

Solution :

Step 1: Identification of Half-Cell Reactions and Standard Potentials

To calculate the standard cell potential ($E^o_{cell}$), we identify the oxidation and reduction half-reactions from the given global equation:

  • Oxidation (Anode): $2Cr(s) \rightarrow 2Cr^{3+}(aq) + 6e^-$
  • Reduction (Cathode): $3Cd^{2+}(aq) + 6e^- \rightarrow 3Cd(s)$

Referring to standard electrode potential tables [$E^o_{Cr^{3+}/Cr} = -0.74\text{ V}$ and $E^o_{Cd^{2+}/Cd} = -0.40\text{ V}$]:

Step 2: Calculation of Standard Cell Potential ($E^o_{cell}$)

The standard cell potential is defined as the difference between the reduction potential of the cathode and the anode [Per Nernst Equation principles at standard conditions]:

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

$E^o_{cell} = E^o_{Cd^{2+}/Cd} - E^o_{Cr^{3+}/Cr}$

$E^o_{cell} = -0.40\text{ V} - (-0.74\text{ V}) = +0.34\text{ V}$

Step 3: Calculation of Standard Gibbs Free Energy Change ($\Delta_r G^o$)

The relationship between the standard Gibbs free energy and the cell potential is given by the formula [Per Faraday’s Laws of Electrolysis and Thermodynamic Relations]:

$\Delta_r G^o = -nFE^o_{cell}$

Where:

  • $n = 6$ (moles of electrons transferred)
  • $F = 96487\text{ C mol}^{-1}$ (Faraday's Constant)
  • $E^o_{cell} = 0.34\text{ V}$

$\Delta_r G^o = -(6) \times (96487\text{ C mol}^{-1}) \times (0.34\text{ J C}^{-1})$

$\Delta_r G^o = -196833.48\text{ J mol}^{-1} \approx -196.83\text{ kJ mol}^{-1}$

Step 4: Calculation of Equilibrium Constant ($K_c$)

The equilibrium constant is related to the standard cell potential by the expression [Derived from the Nernst Equation at equilibrium, where $E_{cell} = 0$]:

$\ln K_c = \frac{nFE^o_{cell}}{RT}$ or $\log K_c = \frac{nE^o_{cell}}{0.0591}$

Using the value $0.0591$ for $2.303RT/F$ at $298\text{ K}$:

$\log K_c = \frac{6 \times 0.34}{0.0591} = \frac{2.04}{0.0591} \approx 34.5177$

$K_c = \text{antilog}(34.5177) = 3.29 \times 10^{34}$

Anode Cathode e- Flow

Final Solution: The standard cell potential $E^o_{cell} = 0.34\text{ V}$, the standard Gibbs free energy $\Delta_r G^o = -196.83\text{ kJ mol}^{-1}$, and the equilibrium constant $K_c = 3.29 \times 10^{34}$.


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