The value of (kJ mol) for the given cell is : — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cell Reaction and the Number of Transferred Electrons ($n$) The given electrochemical cell is represented as: $$\ce{M(s) | M^2+(saturated solution of MX2) || M^2+(0.001 mol dm^-3) | M(s)}$$ This is a metal-ion concentration cell. The individual half-reactions taking place at the electrodes are: * **Anode (Oxidation):** $$\ce{M(s) -> M^2+(anode) + 2e^-}$$ * **Cathode (Reduction):** $$\ce{M^2+(cathode) + 2e^- -> M(s)}$$ Adding these two half-reactions gives the net overall cell reaction: $$\ce{M^2+(cathode) -> M^2+(anode)}$$ From this balanced equation, the number of moles of electrons transferred in the cell reaction is: $$n = 2$$ Step 2 - Apply the Thermodynamic Equation for Gibbs Free Energy Change ($\Delta G$) The change in Gibbs free energy ($\Delta G$) is related to the electromotive force ($E_{\text{cell}}$) of the cell by the fundamental equation: $$\Delta G = -n F E_{\text{cell}}$$ Where: * $n = 2$ (moles of electrons transferred). * $F$ is Faraday's constant = $96,500\text{ C mol}^{-1}$. * $E_{\text{cell}}$ is the cell potential = $0.059\text{ V}$. Step 3 - Substitute the Values and Calculate $\Delta G$ in Joules Substituting the given values into the equation: $$\Delta G = -2 \times 96,500\text{ C mol}^{-1} \times 0.059\text{ V}$$ Since $1\text{ V} = 1\text{ J C}^{-1}$: $$\Delta G = -2 \times 96,500 \times 0.059\text{ J mol}^{-1}$$ $$\Delta G = -11,387\text{ J mol}^{-1}$$ Step 4 - Convert the Answer to Kilojoules and Match with the Options To convert the Gibbs free energy change from Joules per mole ($\text{J mol}^{-1}$) to kilojoules per mole ($\text{kJ mol}^{-1}$), we divide by $1000$: $$\Delta G = \frac{-11,387\text{ J mol}^{-1}}{1000\text{ J kJ}^{-1}}$$ $$\Delta G \approx -11.4\text{ kJ mol}^{-1}$$ This matches Option (D) from the multiple-choice selection. $$\text{Correct Option: } \boxed{D}$$