Question

`AO_2` dispropotionates into `AO_4^-` and `A^(n+)` ion. If the mole ratio of `AO_2` undergoing oxidation and reduction is `2:3`, determine the value of n?

If you check in `AO_4^-`, The oxidation state of `A` turns out to be positive. So I don't understand why is this being referred to as disproportionation reaction?

Answer 1

`n = 2`

Explanation:

The equation for the disproportionation is

`"AO"_2 → "AO"_4^"-" + "A"^(n"+")`

The oxidation numbers that we can calculate are

`stackrelcolor(blue)("+4")("A")"O"_2 → stackrelcolor(blue)("+7")("A")"O"_4^"-" + "A"^(n"+")`

The half-reactions are

Oxidation: `stackrelcolor(blue)("+4")("A")"O"_2 + 2"H"_2"O" → stackrelcolor(blue)("+7")("A")"O"_4^"-" + 4"H"^"+" + 3"e"^"-"`

Reduction: `stackrelcolor(blue)("+4")("A")"O"_2 + 4"H"^"+" + ("4-"n)"e"^"-" → "A"^(n"+") +2"H"_2"O"`

We also know that

`2 × 3color(red)(cancel(color(black)("e"^"-"))) = 3 × ("4-"n)color(red)(cancel(color(black)("e"^"-")))`

`6 = "12 - 3"n`

`3n = "12 - 6 = 6"`

`color(red)(bb(n = 2)`

This is a disproportion because `"A"` is simultaneously oxidized from +4 to +7 and reduced from +4 to +2.

Extra:

The balanced equation is

`2×["AO"_2 + 2"H"_2"O" → "AO"_4^"-" + 4"H"^"+" + 3"e"^"-"]`
`ul(3×["AO"_2 + 4"H"^"+" + "2e"^"-" → "A"^"2+" +2"H"_2"O"]color(white)(mm))`
`color(white)(mmll)"5AO"_2 + 4 "H"^"+" → "2AO"_4^"-" + "3A"^"2+" + 2"H"_2"O"`