What is the standard free-energy change for the following reaction at 25°C? (F = 96,500 C • mol^-1) 2Al(s) + 3Sn⁴⁺(aq,1 M) → 2Al³⁺(aq,1 M) + 3Sn²⁺(aq,1 M)

Question

Quizplus · Accepted Answer

The standard free-energy change (∆G°) can be calculated using the equation:

∆G° = -nFE°

where n is the number of moles of electrons transferred in the reaction, F is Faraday's constant, and E° is the standard cell potential.

First, we need to balance the equation by multiplying SnS1U1P14+ and SnS1U1P12+ by 3 and 2, respectively:

2Al(s) + 3SnS1U1P14+S1S1P0(aq,1 M) → 2AlS1U1P13+S1S1P0(aq,1 M) + 9SnS1U1P12+S1S1P0(aq,1 M)

The overall reaction involves the transfer of 6 moles of electrons (2 from Al and 3x2 from SnS1U1P14+). The standard reduction potentials for Al3+ and SnS1U1P14+ are -1.66 V and -0.14 V, respectively. The standard cell potential (E°) can be calculated using the equation:

E° = E°(cathode) - E°(anode)

E° = E°(AlS1U1P13+S1S1P0) - E°(SnS1U1P14+S1S1P0)

E° = (-0.83 V) - (-0.14 V)

E° = -0.69 V

Substituting n, F, and E° into the previous equation, we get:

∆G° = -nFE°

∆G° = -(6 mol)(96,500 C • molS1U1P1-1S1S1P0)(-0.69 V)

∆G° = -1.04 × 10S1U1P13S1S1P0 kJ/mol

Therefore, the best choice is C.

What Is the Standard Free-Energy Change for the Following Reaction