# Balance the chemical equation FeSO4 - Fe2O3+SO2+SO3 ?

May 29, 2018

$2 {\text{FeSO"_4 to "Fe"_2"O"_3 + "SO"_2 + "SO}}_{3}$

#### Explanation:

Start by identifying the oxidation state for each of the element:

$\stackrel{\textcolor{n a v y}{\boldsymbol{+ 2}}}{\text{Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("NOT BALANCED}}$

Only three of the four chemicals- ${\text{FeSO}}_{4}$, ${\text{Fe"_2"O}}_{3}$, and ${\text{SO}}_{2}$ are directly involved in the redox reaction.

The oxidation state of iron $\text{Fe}$ has increased from

• $\textcolor{n a v y}{+ 2}$ in stackrel(color(navy)(bb(+2)))(bb("Fe"))stackrel(color(purple)(+6))("S")stackrel(-2)("O")_4 to
• $\textcolor{n a v y}{+ 3}$ in stackrel(color(navy)(bb(+3)))(bb("Fe")_2)stackrel(-2)("O")_3 by one

and therefore $\text{Fe}$ is oxidized.

The oxidation state of sulfur $\text{S}$ has declined from

• $\textcolor{p u r p \le}{+ 6}$ in ${\stackrel{\textcolor{n a v y}{+ 2}}{\text{Fe")stackrel(color(purple)(bb(+6)))(bb("S"))stackrel(-2)("O}}}_{4}$ to
• $\textcolor{p u r p \le}{+ 4}$ in stackrel(color(purple)(bb(+4)))(bb("S"))stackrel(-2)("O")_2 by two

and therefore some of the sulfur atoms have been reduced.

• The total increases in oxidation numbers shall be the same as the sum of decreases in oxidation numbers in a balanced redox reaction.

• The oxidation number increases by $1$ for each mole of $\text{Fe}$ atom oxidized and decreases by $2$ for each mole of $\text{S}$ oxidized.

• Therefore for each mole of $\text{S}$ reduced, two moles of $\text{Fe}$ atoms shall be oxidized.

Note that sulfur dioxide, stackrel(color(purple)(bb(+4)))(bb("S")) stackrel(-2)("O")_2, is the only species containing sulfur atoms of oxidation state $\textcolor{p u r p \le}{+ 4}$. Thus all of the reduced sulfur atoms would end up in ${\text{SO}}_{2}$.

The number of moles of sulfur atoms reduced shall therefore equal to the number of ${\text{SO}}_{2}$ molecules produced. This number would be slightly smaller than that of ${\stackrel{\textcolor{n a v y}{+ 2}}{\text{Fe")stackrel(color(purple)(bb(+6)))(bb("S"))stackrel(-2)("O}}}_{4}$ given that sulfur atoms that were not reduced got eventually into ${\stackrel{\textcolor{p u r p \le}{+ 6}}{\text{S")stackrel(-2)("O}}}_{3}$.

Add coefficients $2$ and $1$ in front of ${\text{FeSO}}_{4}$ and ${\text{SO}}_{2}$, respectively:

$\textcolor{g r e e n}{2} \textcolor{w h i t e}{l} \stackrel{\textcolor{n a v y}{\boldsymbol{+ 2}}}{\text{Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + color(green)(1)color(white)(l)stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("NOT YET BALANCED}}$

Deduce coefficients for the rest of the species based on the conservation of iron and sulfur atoms.

$\textcolor{\mathrm{da} r k g r e e n}{2} \textcolor{w h i t e}{l} \stackrel{\textcolor{n a v y}{\boldsymbol{+ 2}}}{\text{Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to color(green)(1)color(white)(l)stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + color(darkgreen)(1)color(white)(l)stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+color(green)(1)color(white)(l)stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("BALANCED}}$

Take coefficients "$1$" out of the expression:

$2 \textcolor{w h i t e}{l} {\text{FeSO"_4 to "Fe"_2"O"_3 + "SO"_2 + "SO}}_{3}$