if an ionic oxide has both x amount of fe2 and y amount fe3 , where x and y do not have to be integers, the formula would be (fe2 )x(fe3 )yoz. calculate the amount of oxide ions, z, in terms of x and y (an algebra problem).

Answer :

Fe^2 + → Fe^3 + + e^- is an example of oxidation

How many oxidations do Fe3+ and Fe2+ ions have?

For instance, the oxygen ion O2+ has an oxidation number of 2 because it can contribute two electrons, whereas the iron ion Fe3+ has an oxidation number of +3 because it can take on three electrons to establish a chemical bond.

[tex]$\mathrm{Fe}_3 \mathrm{O}_4$[/tex] contains [tex]$\mathrm{F}$ e[/tex] atoms of both[tex]$+2$ and $+3$[/tex]oxidation number.

It is a stoichiometric mixture of[tex]Ferrous $(\mathrm{FeO})$[/tex]

and Ferric [tex]$\left(\mathrm{Fe}_2 \mathrm{O}_3\right)$ oxides combined as $\mathrm{FeO} \cdot \mathrm{Fe}_2 \mathrm{O}_3$.[/tex]

[tex]Oxidation state of oxygen in $\mathrm{FeO}$ and $\mathrm{Fe}_2 \mathrm{O}_3$ is $-2$.[/tex]Let the oxidation state of [tex]$\mathrm{Fe}$ in $\mathrm{FeO}$ and $\mathrm{Fe}_2 \mathrm{O}_3$ be $\mathrm{x}$ and $y$[/tex]

As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

ThereforeOxidation state of [tex]$\mathrm{F}$ e in $\mathrm{F}$ eO-[/tex]

&[tex]x+(-2)=0 \\& \Rightarrow x=2[/tex]

Oxidation state of[tex]$\mathrm{F}$ e in $\mathrm{Fe}_2 \mathrm{O}_3-$[/tex]

&[tex]2(\mathrm{y})+3(-2)=0 \\& \Rightarrow 2 \mathrm{y}=6 \\& \Rightarrow \mathrm{y}=3\end{aligned}[/tex]

Hence the oxidation number of [tex]$\mathrm{Fe}$ in $\mathrm{Fe}_3 \mathrm{O}_4$ are $+2$ and $+3$[/tex]respectively.

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