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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.​

Respuesta :

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

Hope this helps

shristiparmar1221

The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ

What is cell potential ?

The cell potential, Ecell, is the measure of the potential difference between two half cells in an electrochemical cell.

Expression for work done ;

W = ΔG° x mol   (1)

But, to determine the ΔG°

ΔG° = -n x F x 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 there is variation from Al to Al³⁺ the change in number of electrons are 3)

To get the moles,

moles = m/MM  (3)

moles = 50 / 26.98 = 1.85 moles of aluminum

Now,

ΔG° = -3 x 96,500 x 1.15

ΔG° = -332,925 J  

       = -332.925 kJ/mol

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

W = 332.925 x 1.85

W = 615.91 kJ

Hence, The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ

Learn more about work done here ;

https://brainly.com/question/26622905

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