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) |Al

Question

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.​

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Nho 5 years 2021-08-16T07:53:12+00:00 1 Answers 563 views 2

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  1. niczorrrr
    2
    2021-08-16T07:54:40+00:00
    Reply

    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

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