Calculate the quantity of charge required to obtain one mole of aluminum from al2o3?
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The balance half reaction will be as follows:
Al2O3 + 6H+ + 6e- --> 2Al + 3H2O
Since it requires 6 moles e- to produce 2mole of Al
It will require 3moles e- to produce 1mole of Al.
Thus quantity of charge will be Q=nF = 3*96500 = 289500C or simply 3F
Al2O3 + 6H+ + 6e- --> 2Al + 3H2O
Since it requires 6 moles e- to produce 2mole of Al
It will require 3moles e- to produce 1mole of Al.
Thus quantity of charge will be Q=nF = 3*96500 = 289500C or simply 3F
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