Answer to Question #348452 in General Chemistry for ngoma

Question

3. Calculate the maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V. Al(s) |Al3+ (aq) || H+ (aq) | O2(g) |Pt . Note that O2 is reduced to H2O. Use appropriate standard reduction potentials.

Answer 1

Answer:

W = 615.91 kJ

Explanation:

We need to use the following expression:

W = ΔG° * mol (1)

But in order to determine the ΔG° we need the following expression:

ΔG° = -n * F * E° (2)

Where F is a constant and is 96,500 J/V mol

n is the number of transferred electrons in the reaction. As we are passing from Al to Al³⁺ we can say that the number of electrons are 3.

Finally to get the moles, we need the the atomic weight of aluminum which is 26.98 g/mol, so the moles:

moles = m/MM (3)

Let's calculate the moles of aluminum:

moles = 50 / 26.98 = 1.85 moles of aluminum

Now let's calculate the gibbs energy using (2):

ΔG° = -3 * 96,500 * 1.15

ΔG° = -332,925 J or simply -332.925 kJ/mol

Finally, using (1) we can determine the work done:

W = 332.925 * 1.85

W = 615.91 kJ