Question

. If 36 g. of Al is deposited at the cathode, the
number of moles of electrons used is​

Answer 1

Given :

Mass of Al deposited at the cathode = 36 g

To Find :

No of moles of electrons used = ?

Solution :

The reaction at the cathode will be given as :

$Al^{3+} + 3 e^- \rightarrow Al$

No of electrons needed to reduce 1  atom of Al = 3

So, for reducing 1 mole of Al atoms  = $3 \times 6 .022 \times 10^{23} e^-$

Here the no of moles in 36 gm Al =$\frac{36 g }{molar \hspace3 mass \hspace3 of Al}$

                                                        =$\frac{36}{27}$ moles

So, the no of electrons required = $3 \times \frac{36}{27} \times 6.022 \times 10^{23}$

                                                     = $2.4 \times 10^{24}$

So, the no of electrons used is $2.4 \times 10^{24}$ electrons .