Question

Calculate the amount of Al deposited by passing 3 amperes of current for a duration of 6 hours through molten AlCl3

Answer 1

Answer Text

Solution :

Molten Al2O3−→−−−−Eletrolysis2Al3+3O2−

Number of Faradays =I×ts96500C

=9.65A×10s96500C=10−3F

First method

Reduction of Al3+ at cathode

Al3++3e−→Al

3e−=3F=1molAl=27gAl

∴10−3F=273×10−3g

=0.009gAl deposited

Second method

1F=1 equivalent of Al=1molCharge=27g3=9g

∴10−3F=9×10−3gAl deposited

Answer 2

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