Question

Consider the following cell reaction: Mg + Cd$^{2+}$ ⇌ Mg$^{2+}$ + Cd The standard Gibbs free energy change for the reaction is ........... kJ (rounded off to an integer). Given: Standard oxidation potentials with respect to standard hydrogen electrode are: Mg ⇌ Mg$^{2+}$ + 2e$^-$, E° = 2.37 V
Cd ⇌ Cd$^{2+}$ + 2e$^-$, E° = 0.403 V
Faraday's constant = 96500 C mol$^{-1}$

Hint:

Always convert oxidation potentials to reduction potentials if needed, then apply $E°_{cell} = E°_{cathode} - E°_{anode}$. Use ΔG° = −nFE to get energy change.

Answer 1

Step 1: Determine half-cell potentials.
Oxidation potential of Mg is given as 2.37 V. Hence the reduction potential is −2.37 V.
Oxidation potential of Cd is given as 0.403 V. Hence the reduction potential is −0.403 V. Step 2: Identify anode and cathode.
In the cell, Mg is oxidized and Cd$^{2+}$ is reduced.
Therefore: - Anode (oxidation): Mg → Mg$^{2+}$ + 2e$^-$ (E° oxidation = +2.37 V)
- Cathode (reduction): Cd$^{2+}$ + 2e$^-$ → Cd (E° reduction = −0.403 V) Step 3: Cell potential.
\[ E°_{cell} = E°_{cathode} - E°_{anode(reduction)} \] But since anode data is given in oxidation form, simply add: \[ E°_{cell} = E°_{oxidation, Mg} + E°_{reduction, Cd} = 2.37 + (−0.403) = 1.967 V \] Step 4: Gibbs free energy change.
\[ ΔG° = −n F E°_{cell} \] where n = 2 electrons, F = 96500 C mol$^{-1}$. \[ ΔG° = −2 \times 96500 \times 1.967 \] \[ = −379963 J \approx −380 kJ \] \[ \boxed{ΔG° = −380 \, kJ} \]