Question

12. Calculate the maximum work to be done in the cell: Al (s) + Ni^2+ (ac) --> Al^3+ (ac) + Ni (s), if 15.8g of Al (26.98g) are used / mol) (F = 96,485 C / mol)

Answer 1

The given cell is   Al (s) + Ni^2+ (ac) --> Al^3+ (ac) + Ni (s)

Now it is a redox cell where both Reduction (i.e gain of electron) and Oxidation (i.e loss of electron) takes place

Now in this cell,

• Oxidation half : Al (s) --------------------> Al^3+ (ac) + 3e E⁰red = -1.66 V
• Reduction half   Ni^2+ (ac) + 2e --------------------> Ni (s) E⁰red = -0.23 V

Now to calculate E⁰cell i.e standard cell potential, we have to balance the electrons in the half cells

So

• Oxidation half : Al (s) --------------------> Al^3+ (ac) + 3e ]*2 E⁰red = -1.66 V * 2
• Reduction half   Ni^2+ (ac) + 2e --------------------> Ni (s) ]*3 E⁰red = -0.23 V * 3

Or

• Oxidation half : 2Al (s) --------------------> 2Al^3+ (ac) + 6e E⁰red = -3.32 V
• Reduction half 3Ni^2+ (ac) + 6e --------------------> 3Ni (s) E⁰red = -0.69 V

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Overall cell 2Al (s) + 3Ni^2+ (ac) -----------------> 2Al^3+ (ac) + 3Ni (s)  

So here-

E⁰cell = E⁰red (reduction half) - E⁰red (reduction half)

= -0.69 V - (-3.32 V)

= -0.69 V + 3.32 V

= 2.63‬ V

Now the workdone (W) is actually the actual amount of free energy (ΔG^∘) available from the cell

So W = ΔG^∘ (Here only the +ve value of ΔG^∘ will be taken)

Now the formula to calculate ΔG^∘ is

ΔG^∘ = -nFE⁰cell

where n = number of electrons transfered = Here it s 6

F = Faraday's constant = 96,485 C

Calculated E⁰cell = 2.63‬ V

Now putting these values-

ΔG^∘ = -nFE⁰cell

W = -6 * 96,485 C * 2.63‬ V

= 15,22,533 J (Here we only need the value of energy available. So -ve sign will be omitted)

This indicates that when in the overall reaction, we have taken 2 moles of Al, then final workdone = 15,22,533 J

Now we are given with Al = 15.8g

So mols of Al given = mass / molar mass

= 15.8g / 26.98 g /mol

= 0.585 mols

So final workdone by 0.585 mols of Al is = 15,22,533 J /  2 moles * 0.585 mols

= 4,45,812 J

= 445.8 kJ

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