Question

Calculate the standard free-energy change at 25 ∘C for the following reaction:

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)
Express your answer to three significant figures and include the appropriate units.

Answer 1

To calculate the standard free-energy change for a reaction, the standard reduction potentials are needed to determine the standard cell potential, which can then be used in the formula ΔG° = -nFE°.

Step-by-step explanation:

To calculate the standard free-energy change (ΔG°) for the reaction Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s) at 25 ℃, you would need standard reduction potentials for magnesium and iron. Unfortunately, the provided references do not directly give the ΔG° for this reaction, but you can use the formula ΔG° = -nFE°, where ΔG° is the standard free-energy change, n is the number of moles of electrons exchanged, F is the Faraday constant (96485 C/mol), and E° is the standard cell potential. The standard cell potential can be calculated from the standard reduction potentials of the involved half-reactions, which can be found in standard tables.

Note that to obtain the full answer, consult a resource like Appendix L (as suggested in your textbook examples) or another reliable source for the standard reduction potentials of Mg²⁺/Mg and Fe²⁺/Fe. Then, following the steps outlined above will yield the ΔG° value for this reaction.

Answer 2

`-3.72* 10^5\ J`

Step-by-step explanation:

The given reaction is:

`Mg(s)+Fe^(2+)(aq)\rightarrow Mg^(2+)(aq)Fe(s)`

The two half reactions and their half potentials are as follows:

`Mg^(2+)(aq)+2e^(-)\rightarrow Mg(s)` E° = -2.37 V

`Fe^(2+)(aq)+2e^(-)\rightarrow Fe(s)` E° = -0.44 V

Half cell with more negative potential will act as anode and half cell with less negative potential will act as cathode.

Calculate cell potential as follows:

`E\°_(cell)=E\°_(cathode)-E\°_(anode)\\=-0.44\ V-(-2.37\ V)\\=+1.93\ V`

Formula for the calculation of standard free energy is as follows:

`\Delta G\°=-nFE\°_(cell)`

F = 96500 c/mol

n for the given reaction is 2.

`\Delta G\°=-nFE\°_(cell)\\=-2 * 96500\ C/mol * 1.93\ V\\=-3.72* 10^5\ J`