Question

Equivalent mass of ferrous oxalate \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] in the following reaction is: \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3}} - }}{\text{ + 2C}}{{\text{O}}_{\text{2}}}\] .
A) \[\dfrac{{{\text{Molecular mass}}}}{1}\]
B) \[\dfrac{{{\text{Molecular mass}}}}{2}\]
C) \[\dfrac{{{\text{Molecular mass}}}}{3}\]
D) \[\dfrac{{{\text{Molecular mass}}}}{4}\]

Answer 1

Hint: The equivalent mass of an oxidant or reductant is that mass which either supplies or consumes one mole of electrons in a redox reaction.
In a reaction, we can calculate the equivalent mass by using the following relation
\[{\text{Equivalent mass}} = \dfrac{{{\text{Molecular mass}}}}{n \text {factor}}\]
For this purpose, we need to first calculate the n factor. n factor gives the number of electrons lost or gained by an oxidant or reductant.

Complete step-by-step solution:
The reaction \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3}} - }}{\text{ + 2C}}{{\text{O}}_{\text{2}}}\] is the oxidation half reaction in which ferrous oxalate \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is oxidized to ferric ions and carbon dioxide ions. Increase in the oxidation number is oxidation and decrease in the oxidation number is reduction.
The oxidation number of iron in ferrous oxalate \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is +2. In ferric ions, the oxidation number of iron is +3. Thus, during the reaction, the oxidation number of iron increases from +2 to +3. Thus, iron is oxidized during the reaction.
\[{\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - }\]
 Ferrous ion loses a proton to form ferric ion.
 Similarly, an oxalate ion loses two electrons to form two molecules of carbon dioxide.
\[{{\text{C}}_2}{\text{O}}_4^{2 - } \to 2{\text{C}}{{\text{O}}_2} + 2{{\text{e}}^ - }\]
During this reaction, oxalate ion is oxidized to carbon dioxide. Loss of electrons is oxidation and gain of electrons is reduction.

One molecule of ferrous oxalate \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] loses three electrons to form one ferric ion and two carbon dioxide molecules.
\[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3}} - }}{\text{ + 2C}}{{\text{O}}_{\text{2}}} + 3{{\text{e}}^ - }\]
Here, the value of n factor is 3.
Equivalent mass of ferrous oxalate \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] in the reaction \[{\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to {\text{F}}{{\text{e}}^{{\text{3}} - }}{\text{ + 2C}}{{\text{O}}_{\text{2}}}\] is \[\dfrac{{{\text{Molecular mass}}}}{3}\]

Hence the correct answer is option ‘C’.

Note: In a redox reaction, n factor is the number of electrons that participate in the half reaction. We can also calculate the n factor from the change in the oxidation number.