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# The number of moles of $KMn{O_4}$ that will be needed to react completely with one mole of ferrous oxalate in acidic solution is: A: $\dfrac{2}{5}$B: $\dfrac{3}{5}$C: $\dfrac{4}{5}$D: 1

Last updated date: 12th Aug 2024
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Hint: Potassium permanganate in acidic medium will act as an oxidising agent and iron’s oxidation state will increase from ferrous to ferric. Also, the oxalate anion will get reduced to ${\text{C}}{{\text{O}}_{\text{2}}}$.

Complete step by step answer:
First thing you should know is how $KMn{O_4}$ behaves in different mediums
In acidic medium, $KMn{O_4}$ accepts ${\text{5}}{{\text{e}}^ - }$, manganese changes its oxidation. no. from $+ 7$ to $+ 2.$
In neutral medium, $KMn{O_4}$ accepts ${\text{3}}{{\text{e}}^ - }$, manganese changes its oxidation no. from $+ 7$ to $+ 4.$
In basic medium, $KMn{O_4}$ accepts ${\text{1}}{{\text{e}}^ - }$, manganese changes its oxidation no. from $+ 7$ to $+ 6.$
Secondly, ferrous ion will always get oxidized to ferrate, i.e. $+ 2$ to ${\text{ + 3}}$, that means it loses ${\text{1}}{{\text{e}}^{\text{ - }}}$. Oxalate ion gets oxidized to Carbon dioxide, i.e. carbon changes its oxidation no. from $+ 3$ to $+ 4$, but oxalate ion contains two carbons, so it looses ${\text{2}}{{\text{e}}^ - }$. that means $1$ mole ferrous oxalate is going to lose ${\text{3}}{{\text{e}}^ - }$
Now , we know from point $1$, that $1$ mole $KMn{O_4}$ in acidic medium can accept ${\text{5}}{{\text{e}}^{\text{ - }}}$-.
for ${\text{5}}{{\text{e}}^ - }$, $1$ mole $KMn{O_4}$ is produced
Therefore, ${\text{1}}{{\text{e}}^ - }$ will give $\dfrac{1}{5}$ mole $KMn{O_4}$
And ${\text{3}}{{\text{e}}^ - }$ will give $\dfrac{3}{5}$ mole $KMn{O_4}$
The skeletal equation:
$KMn{O_4} + Fe{C_2}{O_4} + {H^ + } \to F{e^{3 + }} + C{O_2} + M{n^{2 + }} + {K^ + } + {H_2}O$
Here the ${{\text{H}}^{\text{ + }}}$ can come from any strong mineral acid.
If $3$ moles of $KMn{O_4}$ react with $5$ moles of ${\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$, then with one mole of ${\text{Fe}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$, $\dfrac{3}{5}$ or $0.6$ moles of $KMn{O_4}$ will react. (This makes sense, as iron oxalate is present in more quantity per mole here

So, the correct answer is Option C .

Note:
If the reaction is not fully balanced ,you still need to balance the hydrogens by adding the suitable number of ions on the side deficient in hydrogen, and add more molecules of water on the other side to completely balance the reaction.