To find the standard Gibbs energy change for the reaction, we can use the relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the standard cell potential ($E^\circ$) given by the following equation:

$$\Delta G^\circ = -nFE^\circ$$

Where:

• $n$ is the number of moles of electrons transferred in the reaction.
• $F$ is the Faraday constant ($$1 \mathrm{~F} = 96487 \mathrm{~C \cdot mol^{-1}}$$).
• $E^\circ$ is the standard cell potential.

From the given reaction:

$$\mathrm{Zn}(\mathrm{s}) + \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{Fe}(\mathrm{s})$$

It can be observed that $n = 2$ because two electrons are transferred from zinc to iron.

Given data:

• $$E^\circ = 0.32 \mathrm{~V}$$
• $$1 \mathrm{~F} = 96487 \mathrm{C/mol}$$

Now, substitute the values into the equation:

$$\Delta G^\circ = -nFE^\circ$$

$$\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}$$

Calculate the value:

$$\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}$$

$$\Delta G^\circ = -2 \times 30875.84 \mathrm{~J/mol}$$

$$\Delta G^\circ = -61751.68 \mathrm{~J/mol}$$

Converting the units to kJ/mol:

$$\Delta G^\circ = -61.75168 \mathrm{~kJ/mol}$$

The closest answer, when rounded to two decimal places, is:

$$\Delta G^\circ = -61.75 \mathrm{~kJ/mol}$$

Therefore, the correct option is:

Option A: $$-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$$