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3.12     How much charge is required for the following reductions:

              (i) 1\ mol\ of\ Al^{3+}\ to\ Al\ ?

Answers (1)

best_answer

The equation becomes :-

                                  Al^{+3}\ + 3e^-\ =\ Al

So required charge is 3F.

Q = n*96500

Q  =   3*96500  =  289500 C

Posted by

Devendra Khairwa

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