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1mol of ferric oxalate is oxidized by ‘x’ mol of $Mn{{O}_{4}}^{-}$in acidic medium. Hence, the value of x is-
[A] 1.2
[B] 1.6
[C] 1.8
[D] 1.5

Last updated date: 23rd May 2024
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Hint: The reaction given to us is a redox reaction. Here carbon will be oxidised and manganese is reduced. We can solve this question by calculating the valency factor and number of moles of each. We can equate the equivalents of the oxidised and the reduced units and solve for x to get the answer.

Complete step by step answer:
 We know that the formula of ferric oxalate is$F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$.
According to the question, ferric oxalate is oxidized by$Mn{{O}_{4}}^{-}$in acidic medium.
We can write the dissociation of ferric oxalate as-
     \[F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}\to 2F{{e}^{3+}}+3{{C}_{2}}{{O}_{4}}^{2-}\]
Here, iron is in +3 oxidation state which is its highest oxidation state therefore we cannot oxidize it further. Therefore, we have carbon left for oxidising. We can write the reaction as-
     \[{{C}_{2}}{{O}_{4}}^{2-}\to 2{{C}^{4+}}\]
Here, the oxidation state of carbon changes from +3 to +4. For one molecule of ${{C}_{2}}{{O}_{4}}^{2-}$we have two molecules of ${{C}^{4+}}$and we have three molecules${{C}_{2}}{{O}_{4}}^{2-}$. Therefore, if we want to write the oxidation reaction for ${{({{C}_{2}}{{O}_{4}})}_{3}}^{2-}$atoms considering only the carbon atom, we will get-
     \[3{{({{C}^{3+}})}_{2}}\to 6{{C}^{4+}}+6{{e}^{-}}\]
Here, carbon is oxidised from +3 to +6 by losing six electrons. Therefore, the valency factor will be 6.
Similarly, manganese in $Mn{{O}_{4}}^{-}$is reduced from +7 to +2. We can write the reaction as-
     \[M{{n}^{+7}}+5{{e}^{-}}\to M{{n}^{2+}}\]
As we can see, manganese changes its oxidation state from +7 to +2 by accepting five electrons. Therefore, the valency factor will be 5.
We can equate the equivalents of the oxidised and the reduced as it is a redox reaction.
Therefore, number of moles of $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$$\times $valency factor = number of moles of$Mn{{O}_{4}}^{-}$$\times $valency factor.
As we can see in the question, number of moles of $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$is given as 1 and that of$Mn{{O}_{4}}^{-}$is x and we have calculated that the valency factor of $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$is 6 and that of$Mn{{O}_{4}}^{-}$is 1.
Therefore, putting these values we will get-
     \[1\times 6=x\times 5\]
Or, $x=\dfrac{6}{5}=1.2$
Therefore, from the above calculations, we have found that x is 1.2.

Therefore the correct answer is option [A] 1.2.

As the above reaction is a redox reaction where ferric oxalate is oxidised by$Mn{{O}_{4}}^{-}$, it is a redox reaction as carbon is oxidised and manganese is reduced, we can also write that n-factor will be the change in oxidation numbers of carbon and manganese and it will give us the same result as change in oxidation state for manganese is +7 to +2 which gives us 5 and for carbon it is +3 to +4 which gives us +1 but we have 6 carbon atoms in the balanced reaction therefore, it will be $6(no. Of moles)\times 1$ (change in oxidation state) which will be 6.