Question

The amount of electricity required to deposit 1mole of al on electrode frome a solution of alcl3 will be

Answer 1

Answer: The amount of electricity required to deposit 1 mole of Aluminium is 3 faraday.

Explanation:  $AlCl_{3}\rightarrow Al^{3+}+ 3Cl^{-}$

As shown in the above equation when 1 mole of $AlCl_{3}$ dissociates , it gives 1 mole of  $Al^{3+}$ and 3 moles of  $Cl^{-}$. The positively charged aluminium ions moves towards negative electrode or cathode and negatively charged chloride ions will move towards positively charged electrode or anode.

Reduction (on cathode): $Al^{3+}+ 3e^{-}\rightarrow Al$

1 mole of Aluminium ions will gain 3 mole of electrons to produce 1 mole of aluminium.

1 electron carries charge = $1.6\times 10^{-19}C$

1 mole of electrons ($6.023\times 10^2^3 electrons)$ carries charge = $1.6\times 10^{-19}\times 6.023\times 10^2^3C$ = $96500 C$

Also $96500 C$ = $1 Faraday$

As 1 mole of electrons contains 1 faraday so 3 mole of electrons carries 3 Faraday of electricity.