Question

Equivalent weight of $Fe{C_2}{O_4}$ in the change $Fe{C_2}{O_4} \to F{e^{3 + }} + 2C{O_2}$, is
A. M/3
B. M/6
C. M/2
D. M/1

Answer 1

Hint: In the given reaction, both oxidation and reduction reaction takes place. When the oxidation reaction and reduction reaction takes place simultaneously then the reaction is called a redox reaction.

Complete step by step answer:
Oxidation half:
$F{e^{ + 2}} \to F{e^{ + 3}} + {e^ - }$
In this oxidation half, the iron gets oxidized and changes its oxidation state from +2 to +3 by losing 1 electron.
Reduction half:
$C_2^{ + 3}O_4^{ - 2} \to 2C{O_2}^{ + 4} + 2{e^ - }$
In the reduction half, the carbon gets reduced and changes its oxidation state from +3 to +4 by gaining 2 electrons.
Overall reaction (Redox reaction)
$Fe{C_2}{O_4} \to F{e^{3 + }} + 2C{O_2}$
In the given reaction total 3 electrons are involved in both oxidation half and reduction half of the reaction.
The equivalent weight is equal to its formula weight or molecular weight divided by the n-factor.
The formula for calculating the equivalent weight is shown below.
$Eq.Wt = \dfrac{M}{{nfactor}}$
n-factor is defined as the sum of the total electrons taking part in the oxidation reaction as well as reduction reaction.
Molecular weight is calculated by multiplying the atomic number of the atom with the total number of atoms present and then adding all the values of the atom.
Thus the equivalent weight of ferrous oxalate is given by M/3.
Therefore, the correct option is A.

Note:
Make sure that the chemical equation should be balanced, that means the number of moles of atoms present on the left side of the reaction should be equal to moles of atoms present on the right side of the reaction.