Question

Define auto redox reaction. Balance the following equation by ion-electron or oxidation number method: `"K"_2"Cr"_2"O"_7(aq) + "FeSO"_4(aq) + "H"_2"SO"_4(aq) -> "K"_2"SO"_4(aq) + "Cr"_2("SO"_4)_3(s) + "Fe"_2("SO"_4)_3(s) + "H"_2"O"(l)`?

Answer 1

An Auto-redox reaction [1] is a redox reaction in which a substance act as both the oxidizing agent and the reducing agent.

Balanced chemical equation:
`color(black)("K"_2)"Cr"_2"O"_7 color(white)(l)(aq)+color(black)(7)"H"_2 color(black)("SO"_4)(aq) +color(black)(6)"Fe" color(black)("SO"_4)(aq)`
`to "Cr"_2color(black)("SO"_4)(aq)+7"H"_2"O"(l)+color(black)(3)"Fe"_2 color(black)(("SO"_4)_3)(aq)`

Explanation:

Start by identifying reducing and oxidizing agents.

Chromium atoms are reduced; the oxidation state of this element decrease from `+6` in `"K"_2stackrel(+6)"Cr"_2"O"_7` to `+3` in `stackrel(+3)"Cr"_2("SO"_4)_3`. Therefore chromate ions acts as oxidizing agents and undergo the following reaction:

Reduction half:
`stackrel(+6)"Cr"_2"O"_7 color(white)(l)^(2-)(aq)+ ul(6e^(-)) +14color(green)("H"^(+))(aq) to 2 stackrel(+3)"Cr"color(white)(l)^(3+)(aq)+7color(green)("H"_2"O")(l)`

Add water to the product side and protons (i.e., `"H"^(+)`) on the reactant side to balance the number of oxygen atoms.

Iron (II) sulfate is oxidized to iron (III) sulfate; therefore iron (II) sulfate undergoes oxidation and acts as the reduction agent.

Oxidation half:
`stackrel(+2) ("Fe") color(white)(l) ^(2+)(aq) to stackrel(+3)("Fe") color(white)(l) ^(3+)(aq)+ul(e^(-))`

Electrons should cancel out in the net reaction; each mole of the reduction half reaction consumes six moles of electrons while each mole of the oxidation half produces one moles; electrons will cancel out in the sum if coefficients of the reduction half reaction are expanded by a factor of six:

`color(green)(6) *`Oxidation half:
`color(green)(6)stackrel(+2) ("Fe") color(white)(l) ^(2+)(aq) to color(green)(6)stackrel(+3)("Fe") color(white)(l) ^(3+)(aq)+ul(color(green)(6)e^(-))`

Combining reactions of the reduction half and six times that of the oxidation half gives:

`stackrel(+6)"Cr"_2"O"_7 color(white)(l)^(2-)(aq)+ ul(cancel(6e^(-))) +14"H"^(+)(aq) +color(green)(6)stackrel(+2) ("Fe") color(white)(l) ^(2+)(aq) to 2 stackrel(+3)"Cr"color(white)(l)^(3+)(aq)+7"H"_2"O"(l)+color(green)(6)stackrel(+3)("Fe") color(white)(l) ^(3+)(aq)+ul(cancel(color(green)(6)e^(-)))`

That is:

`"Cr"_2"O"_7 color(white)(l)^(2-)(aq)+14"H"^(+)(aq) +color(green)(6)"Fe" color(white)(l) ^(2+)(aq)`
`to 2"Cr"color(white)(l)^(3+)(aq)+7"H"_2"O"(l)+color(green)(6)"Fe" color(white)(l) ^(3+)(aq)`

Pair ions with spectator ions of the respective opposite charges as seen in the unbalanced reaction given:

Anion reactants pair with potassium ions `"K"^(+)(aq)` whereas
Cation reactants pair with sulfate ions `"SO"_4^(2-)(aq)`.

Hence the net chemical equation would be

`color(blue)("K"_2)"Cr"_2"O"_7 color(white)(l)(aq)+color(blue)(7)"H"_2 color(blue)("SO"_4)(aq) +color(blue)(6)"Fe" color(blue)("SO"_4)(aq)`
`to "Cr"_2color(blue)("SO"_4)(aq)+7"H"_2"O"(l)+color(blue)(3)"Fe"_2 color(blue)(("SO"_4)_3)(aq)`

References
[1] "AUTO OXIDATION REDUCTION REACTION," City Collegiate, http://www.citycollegiate.com/auto_oxidation.htm

Answer 2

Jacob T. has answered the second half of the question; however, I think that the example provided from the question isn't really an auto-redox reaction since different species oxidize and reduce in the given reaction.

Instead, I'll give an example of a disproportionation reaction, which would follow the definition of an "auto-redox" reaction, where the same species gets oxidized and reduced.

The result is:

`2stackrel(color(blue)(+3))("Mn")_2"O"_3(s) + 4"H"^(+)(aq) -> 2stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 2"H"_2"O"(l)`

---

Consider the diagram below:

Manganese(III) oxide can get oxidized to `"MnO"_2(s)` in acidic pH (or basic pH).

`"Mn"_2"O"_3(s) -> "MnO"_2(s)`

First, balance the non-oxygen atoms.

`"Mn"_2"O"_3(s) -> color(blue)(2)"MnO"_2(s)`

Then balance the oxygen atoms with water.

`color(blue)("H"_2"O"(l)) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s)`

Balance the hydrogen atoms with `"H"^(+)`.

`"H"_2"O"(l) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s) + color(blue)(2"H"^(+)(aq))`

Balance the charge with `e^(-)` on the more positive side.

`color(green)("H"_2"O"(l) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s) + 2"H"^(+)(aq) + 2e^(-))`

Now, the other half-reaction is for the reduction, also of `"Mn"_2"O"_3(s)`, to `"Mn"^(2+)` in acidic pH. This can only happen below pH 7.5, and we would get interfering `"Mn"_3"O"_4(s)` unless we are below pH 6.

Let's say we are below pH 6. Then:

`"Mn"_2"O"_3(s) -> "Mn"^(2+)(aq)`

Repeat the same steps as before to get the reduction half-reaction:

`color(green)(2e^(-) + 6"H"^(+)(aq) + "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O"(l))`

Now, add these together to cancel out the electrons.

`"H"_2"O"(l) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s) + 2"H"^(+)(aq) + cancel(2e^(-))`
`ul(cancel(2e^(-)) + 6"H"^(+)(aq) + "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O"(l))`
`cancel("H"_2"O"(l)) + "Mn"_2"O"_3(s) + cancel(6)^(4)"H"^(+)(aq) + "Mn"_2"O"_3(s)`

`-> 2"MnO"_2(s) + cancel(2"H"^(+)(aq)) + 2"Mn"^(2+)(aq) + cancel(3)^(2)"H"_2"O"(l)`

Cancel out the `"H"^(+)` and `"H"_2"O"` in common on the left and right sides, and combine the duplicate species on the left side. The result is:

`bb(2stackrel(color(blue)(+3))("Mn")_2"O"_3(s) + 4"H"^(+)(aq) -> 2stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 2"H"_2"O"(l))`