Question

Balance the following equation by oxidation reduction method FeSO4

Answer 1

Answer:

$10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O$.

Explanation:

Identify the elements with oxidation state changes:

Oxidation states of iron, $\rm Fe$:

• $+2$ in $\rm FeSO_4$ among the reactants.
• $+3$ in $\rm Fe_2(SO_4)_3$ among the products.
• Change to the oxidation state: $+1$ (oxidation) for each $\rm Fe$ atom.

Oxidation state of manganese, $\rm Mn$:

• $+7$ in $\rm KMnO_4$ among the reactants.
• $+2$ in $\rm MnSO_4$ among the products.
• Change to the oxidation state: $(-5)$ (reduction) for each $\rm Mn$ atom.

The change in the oxidation state of $\rm Mn$ is five times the opposite of the change to the oxidation state of $\rm Fe$. If there are one mole of $\rm Mn\!$ atoms in each mole of this reaction, there would be five times as many $\rm Fe\!$ atoms per mole reaction. In other words:

$\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O$.

(Notice that each mole of this reaction would include five times as many $\rm Fe$ atoms as $\rm Mn$ atoms.)

Multiply the coefficients by $2$ to eliminate the fraction:

$\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O$.

Find the unknown coefficients using the conservation of atoms.

Reactants:

• $2$ potassium $\rm K$ atoms in two $\rm K_2SO_4$ formula units.

Therefore, among the products:

• $2$ potassium $\rm K$ atoms in one $\rm K_2SO_4$ formula unit.

$\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O$.

Products:

• $5 \times 3 + 2 + 1 = 18$ sulfur $\rm S$ atoms in five $\rm Fe_2(SO_4)_3$ formula units, two $\rm K_2 SO_4$ formula units, and one $\rm MnSO_4$ formula unit.

Reactants:

• There are already ten $\rm S$ atoms in that ten $\rm Fe(SO_4)_2$ formula units. The other $18 - 10 = 8$ formula units would correspond to eight $\rm H_2SO_4$ molecules among the reactants of this reaction.

$\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O$.

Products:

• There are $8 \times 2 = 16$ hydrogen $\rm H$ atoms in that eight $\rm H_2SO_4$ molecules.

Therefore, among the products:

• There would be $16 / 2 = 8$ molecules of $\rm H_2O$, with two $\rm H$ atoms in each $\rm H_2O\!$ molecule.

$\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O$.