Question

How much electric charge is required to oxidise:
(a) 1 mole of ${H_2}O$ to ${O_2}$
(b) 1mole of $FeO$ to $F{e_2}{O_3}$

Answer 1

Hint: We can solve this problem by Faraday's Laws of Electrolysis that which depicts the relationship between the quantity of electricity measured in coulombs (q), moles of electron used (n), Faraday constant (F) which is equals to 96487 $Cmo{l^{ - 1}}$, thus $q = n \times F$ this will help approach the solution.

Complete step by step solution:

(a) We know that 1 mole of $H_2O$ has one atom of oxygen.
${H_2}O \to {H_2} + \dfrac{1}{2}{O_2}$
Charge on oxygen is -2
${O_{{2^ - }}} \to \dfrac{1}{2}{O_2} + 2{e^ - }$
difference in charge (n) = 2

Now with the help of Faraday's Laws of Electrolysis we will find out that the charges required to oxidise 1 mole of ${H_2}O$ to ${O_2}$ $= n \times F$
(And here n = 2)
$= 2 \times 96487$ C
$= 192974$C
$= 1.93 \times {10^5}$C
so $1.93 \times {10^5}$Coulombs charges are required to oxidise 1 mole of ${H_2}O$ to ${O_2}$

(b) 1mole of $FeO$ to $F{e_2}{O_3}$
Now again with the help of Faraday's Laws of Electrolysis we will find out the charges required to oxidise 1 mole of $FeO$ to $F{e_2}{O_3}$
Here we know that the charge on Fe is +2
and charge on Fe in compound $F{e_2}{O_3}$ is +3
$F{e^{2 + }} \to F{e^{3 + }} + {e^{ - 1}}$
difference in charge (n) = 1
electric charge required to oxidise 1 mole of ${H_2}O$ to ${O_2}$ $= n \times F$
(And here n = 1)
$= 96487$C
$= 9.65 \times {10^4}$C
So here $9.65 \times {10^4}$ Coulombs charges are required to oxidise 1 mole of $FeO$ to $F{e_2}{O_3}$

Note: For finding the no of charges to oxidise any compound first we need write the chemical reaction so that we can get the difference of charge (n) and we know the value of faraday constant $(96487Cmo{l^{ - 1}})$ after multiplying difference of charge to difference of charge we get electric charge. So we have used the same approach here to solve the problem.