Question

How many moles of electrons are transferred when 2.0 moles of aluminum metal react with excess copper(II) nitrate in aqueous solution to form Al(NO3)3(aq) and Cu(s)

Answer 1

Moles of electrons:

The moles of electrons that are transferred are 12F

A balanced equation:

2 moles of Aluminium metal react with excess copper(II) nitrate.

$2Al + 3Cu{(NO_3)}_2 \rightarrow 2Al{(NO_3)}_3 +3 Cu$

Given:

Moles of Aluminium = 2

As Aluminium goes from 0 to +3 oxidation state

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

And copper goes from +2 to 0

$Cu^{2+} + 2e^-\rightarrow Cu$

On balancing the number of electrons we get:

For 1 mole of Al $6e^-$ is required.

Therefore for 2 moles of Al,

Total $(2\times6)$F mole of electrons

Where F= Faraday's constant= 96500 C

So, 12F moles of electrons are transferred.

Learn more about Faraday's Law here,

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