The given electrochemical cell is a concentration cell where both the electrodes are of the same metal but immersed in solutions of different concentrations of the same metal ion. The emf (E) generated by this cell can be calculated using the Nernst equation, which for this cell at 298 K (25°C) is given by:

$ E = E^\circ - \frac{0.059}{n} \log \frac{[C_1]}{[C_2]} $

Where:

• $E^\circ$ is the standard electrode potential which is zero in a concentration cell because both electrodes are same.


• $n$ is the number of moles of electrons transferred in the redox reaction (2 in this case, as it involves $M^{2+}$ ions).


• $[C_1]$ and $[C_2]$ are the concentrations of $M^{2+}$ at the two electrodes.


• $E$ is given to be 0.059 V.

Assuming the more concentrated solution is at the left hand electrode and the less concentrated solution (0.001 M) is at the right hand electrode, the Nernst equation simplifies to:

$ E = - \frac{0.059}{2} \log \frac{0.001}{[C_1]} $

To find $[C_1]$, we solve for the argument of log such that the calculated emf matches the given emf (0.059 V):

$ 0.059 = - \frac{0.059}{2} \log \frac{0.001}{[C_1]} $

$ -2 = \log \frac{0.001}{[C_1]} $

$ 10^{-2} = \frac{0.001}{[C_1]} $

$ [C_1] = 0.1 \text{ M} $

This $[C_1]$ value supports the direction of the redox reactions assumed. Now, to find the change in Gibbs free energy $\Delta G$ for this cell, we use the relationship:

$ \Delta G = -nFE $

But $E$ should be positive for the spontaneous reaction, hence:

$ \Delta G = -2 \times 96500 \times 0.059 \, \text{V} = -11381 \, \text{J/mol} = -11.381 \, \text{kJ/mol} $

Therefore, $\Delta G$ for the cell is approximately -11.4 kJ/mol. The correct answer is Option D: -11.4.