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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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