Question

The number of faradays required to produce 0.5 mol of free metal from al3+ is

Answer 1

this is how it's done

Answer 2

The number of faradays required to produce 0.5 mol of free metal from $Al^{3+}$ is 1.5

Explanation:

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C=1F$ (96500C=1Faraday)

Deposition of aluminium is given by:

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

1 mole of aluminium is deposited by = $3\times 1Farady=3Faraday$

0.5 mole of aluminium is deposited by =$\frac{3F}{1}\times 0.5=1.5F$

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