Question

Element `A` in a compound ABD has oxidation number `A^(n-)` .It is oxidised by `Cr_2O_7^(2-)` in acid medium. In an experiment `1.68*10^-3` moles of `K_2Cr_2O_7` were used for `3.36*10^(-3)` moles of ABD. the new oxidation number of A after oxidation is?

Answer 1

`3-n`

Explanation:

For starters, you should know that chromium has an oxidation number equal to `+6` in the dichromate anion.

The two most common oxidation states for chromium are `+6` and `+3`, so you can assume that in this reaction, the dichromate anion will be reduced to chromium(III), `"Cr"^(3+)`.

This means that a single atom of chromium will gain `3` electrons. Since the dichromate anion contains `2` atoms of chromium, it follows that a total of `6` electrons will be gained in the reduction half-reaction.

As you know, in a redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.

This implies that a total of `6` electrons must be lost by `stackrel(color(blue)(-n))("A")`in the oxidation half-reaction.

Now, notice that the reaction consumes `2` moles of `stackrel(color(blue)(-n))("A")` for every `1` mole of dichromate anions

`(3.36 * color(red)(cancel(color(black)(10^(-3))))color(white)(.)"moles"color(white)(.) stackrel(color(blue)(-n))("A"))/(1.68 * color(red)(cancel(color(black)(10^(-3)))) color(white)(.)"moles Cr"_2"O"_7^(2-)) = ("2 moles"color(white)(.) stackrel(color(blue)(-n))("A"))/("1 mole Cr"_2"O"_7^(2-))`

This means that the total of `6` electrons needed for the reduction half-reaction will be delivered by `2` `stackrel(color(blue)(-n))("A")` atoms, which implies that every `stackrel(color(blue)(-n))("A")` atom will lose `3` electrons.

Consequently, you can say that the oxidation number of `"A"` will increase by `3` and go from `-n` on the reactants' side to

`color(darkgreen)(ul(color(black)("new oxidation state" = -n + 3 = 3 - n)))`

on the products' side.