Question

One mole of ferrous oxalate is oxidized by x mole of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ in acidic medium. X is:
A.0.6
B.0.1
C.0.3
D.1.0

Answer 1

Hint:
In a redox reaction, transfer of electrons from one substance (atom, ion, or molecule) to another substance takes place. It is also called an oxidation – reduction reaction as it involves both oxidation half reaction and reduction half reaction.
Oxidation is the process which involves the loss of electrons by an atom, ion or molecule while reduction is the process which involves the gain of electrons by an atom, ion or molecule. There are two methods to balance redox reactions. One is the oxidation number method and the other is the ion-electron or half reaction method.

Complete step by step answer:
Given that one mole of ferrous oxalate is oxidized by x mole of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ in acidic medium. We need to find out the value of x, i.e., the number of moles of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ reduced.
Let us first balance the equation by the use of ion-electron or the half-reaction method. Here, ferrous oxalate is undergoing oxidation and ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is undergoing reduction.
Therefore, we will at first write the two partial reactions, one representing oxidation and the other representing reduction.
Oxidation half reaction: \[{\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}\] (oxidation number of iron increases from +2 to +3)
Reduction half reaction: ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}$ (ox oxidation number of manganese decreases from +7 to +2)
Now, balance the two half reactions by balancing the elements including the number of oxygen and hydrogen atoms. For acidic medium, the hydrogen atoms can be balanced by adding a hydrogen ion for each hydrogen atom. Then add the required number of electrons to balance the charges. So, we will have:
${\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{ + 2C}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{e}}^{\text{ - }}}$
And ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 8}}{{\text{H}}^{\text{ + }}}{\text{ + 5}}{{\text{e}}^{\text{ - }}} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, multiply the two half reactions by suitable numbers to equate the number of electrons.

$\left. {{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{ + 2C}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{e}}^{\text{ - }}}} \right] \times 5$
$\left. {{\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 8}}{{\text{H}}^{\text{ + }}}{\text{ + 5}}{{\text{e}}^{\text{ - }}} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}}} \right] \times 3$

Add the two reactions to get the balanced equation.
${\text{5Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{ + 3Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 24}}{{\text{H}}^{\text{ + }}} \to {\text{5F}}{{\text{e}}^{{\text{3 + }}}}{\text{ + 10C}}{{\text{O}}_{\text{2}}}{\text{ + 3M}}{{\text{n}}^{{\text{2 + }}}}{\text{ + 12}}{{\text{H}}_{\text{2}}}{\text{O}}$
Thus, we can see that 5 moles of ferrous oxalate is oxidized by 3 moles of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ . Therefore, 1 mole of ferrous oxalate will be oxidized by $ = \dfrac{3}{5} = 0.6$ moles of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ .

So, the correct option is A.

Note:
The redox reactions have many important applications.
-The oxidation of fuels provides us with our daily energy needs.
-The photosynthesis process is also a redox reaction where carbon dioxide is reduced to carbohydrates and water is oxidized to oxygen.
-The redox reaction between hydrogen and oxygen in fuel cells meets the energy needs of a space capsule.