# 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)?

Apr 21, 2018

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:
$\textcolor{b l a c k}{{\text{K"_2)"Cr"_2"O"_7 color(white)(l)(aq)+color(black)(7)"H"_2 color(black)("SO"_4)(aq) +color(black)(6)"Fe" color(black)("SO}}_{4}} \left(a q\right)$
 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 ${\text{K"_2stackrel(+6)"Cr"_2"O}}_{7}$ 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., ${\text{H}}^{+}$) 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:
$\stackrel{+ 2}{\text{Fe") color(white)(l) ^(2+)(aq) to stackrel(+3)("Fe}} {\textcolor{w h i t e}{l}}^{3 +} \left(a q\right) + \underline{{e}^{-}}$

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:

$\textcolor{g r e e n}{6} \cdot$Oxidation half:
$\textcolor{g r e e n}{6} \stackrel{+ 2}{\text{Fe") color(white)(l) ^(2+)(aq) to color(green)(6)stackrel(+3)("Fe}} {\textcolor{w h i t e}{l}}^{3 +} \left(a q\right) + \underline{\textcolor{g r e e n}{6} {e}^{-}}$

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:

$\text{Cr"_2"O"_7 color(white)(l)^(2-)(aq)+14"H"^(+)(aq) +color(green)(6)"Fe} {\textcolor{w h i t e}{l}}^{2 +} \left(a q\right)$
$\to 2 \text{Cr"color(white)(l)^(3+)(aq)+7"H"_2"O"(l)+color(green)(6)"Fe} {\textcolor{w h i t e}{l}}^{3 +} \left(a q\right)$

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

Anion reactants pair with potassium ions ${\text{K}}^{+} \left(a q\right)$ whereas
Cation reactants pair with sulfate ions ${\text{SO}}_{4}^{2 -} \left(a q\right)$.

Hence the net chemical equation would be

$\textcolor{b l u e}{{\text{K"_2)"Cr"_2"O"_7 color(white)(l)(aq)+color(blue)(7)"H"_2 color(blue)("SO"_4)(aq) +color(blue)(6)"Fe" color(blue)("SO}}_{4}} \left(a q\right)$
 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

Apr 22, 2018

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 ${\text{MnO}}_{2} \left(s\right)$ in acidic pH (or basic pH).

${\text{Mn"_2"O"_3(s) -> "MnO}}_{2} \left(s\right)$

First, balance the non-oxygen atoms.

${\text{Mn"_2"O"_3(s) -> color(blue)(2)"MnO}}_{2} \left(s\right)$

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 ${\text{H}}^{+}$.

"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.

$\textcolor{g r e e n}{{\text{H"_2"O"(l) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s) + 2"H}}^{+} \left(a q\right) + 2 {e}^{-}}$

Now, the other half-reaction is for the reduction, also of ${\text{Mn"_2"O}}_{3} \left(s\right)$, to ${\text{Mn}}^{2 +}$ in acidic pH. This can only happen below pH 7.5, and we would get interfering ${\text{Mn"_3"O}}_{4} \left(s\right)$ unless we are below pH 6.

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

${\text{Mn"_2"O"_3(s) -> "Mn}}^{2 +} \left(a q\right)$

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

$\textcolor{g r e e n}{2 {e}^{-} + 6 \text{H"^(+)(aq) + "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O} \left(l\right)}$

Now, add these together to cancel out the electrons.

${\text{H"_2"O"(l) + "Mn"_2"O"_3(s) -> 2"MnO"_2(s) + 2"H}}^{+} \left(a q\right) + \cancel{2 {e}^{-}}$
$\underline{\cancel{2 {e}^{-}} + 6 \text{H"^(+)(aq) + "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O} \left(l\right)}$
cancel("H"_2"O"(l)) + "Mn"_2"O"_3(s) + cancel(6)^(4)"H"^(+)(aq) + "Mn"_2"O"_3(s)

$\to 2 \text{MnO"_2(s) + cancel(2"H"^(+)(aq)) + 2"Mn"^(2+)(aq) + cancel(3)^(2)"H"_2"O} \left(l\right)$

Cancel out the ${\text{H}}^{+}$ and $\text{H"_2"O}$ 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))