I\'m lost on how to solve this word problem: Aluminum-air batteries produce elec

Question

I'm lost on how to solve this word problem:

Aluminum-air batteries produce electricity from the reaction of aluminum with oxygen in the air. Such batteryies have a very small mass for the energy delivered because aluminum has a low density and is the only reactant that must be carried when the battery is used in the air.

The anode oxidation half-reaction is:     Al + 3OH- --> Al(OH)3 + 3e-        E_ox = +2.31V

The cathode reduction half-reaction is:   O2 + 2H2O + 4e- --> 4OH-           E_red = +0.40V

                 The total reaction is: 4Al + 3O2 + 6H2O --> 4Al(OH)3      E_cell = +2.71V

However, due to non-standard cell conditions, the potential produced by this cell is only about 1.2V.

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a) If you assume that the standard cell potential can be obtained with improved technology, what would be the total output of an aluminum-air battery, in kiloJoules, if 1.00 kg of aluminum metal were consumed in its discharge?

b) How long would such a battery keep a 60W light bulb lit?

Explanation / Answer

A)

moles of Al = 1000/27 = 37.04 moles

4moles of Al give 2.71V

so 4*27g gives 2.71 V

so 1000g gives = 2.71*1000/(4*27) = 25.09 V

4 moles involve trans fer of 12 moles e- = 12*96500 coulomb

so 37.04 moles give transfer of 12*37.04/4 = 10723080 coulomb

so in joules = 10723080*25.09 = 268398.69 KJ

B)

now time = 268398.69/60 = 1242.586 hrs

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