Q.Write the reaction of $KMnO_4$ with Fe(II) ions in acidic medium.

In acid medium, $MnO_4^-$ oxidises oxalic acid completely, breaking the C-C bond and converting both carbons to $CO_2$; this reaction is the standard method for volumetrically standardising potassium permanganate.

The redox reaction between acidified potassium permanganate and oxalic acid is a classic titrimetric reaction in analytical chemistry: $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2\uparrow + 8H_2O$ Here, $Mn$ is reduced from the $+7$ oxidation state (in $MnO_4^-$) to $+2$ (as $Mn^{2+}$), while the oxalate ion's carbon (each at $+3$ oxidation state) is oxidised to $+4$ in $CO_2$, with the C-C bond of oxalate being completely broken. The colourless-to-faint-pink endpoint of this titration (self-indicating, as excess $MnO_4^-$ imparts a pink colour) makes it the standard method for determining the exact concentration (stand …

Acidified $KMnO_4$ is a strong oxidising agent; it oxidises $Fe^{2+}$ (ferrous) to $Fe^{3+}$ (ferric) and is itself reduced to $Mn^{2+}$.

Reduction half-reaction (Mn goes from +7 in $MnO_4^-$ to +2 in $Mn^{2+}$, gaining 5 electrons):

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Oxidation half-reaction (Fe goes from +2 to +3, losing 1 electron; multiplied by 5 to balance electrons):

$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$

Adding the two half-reactions (the $5e^-$ cancel):

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In acidic medium, permanganate is reduced from $+7$ to $+2$ (gaining 5 electrons) while ferrous ion is oxidised to ferric ion (losing 1 electron each); balancing electrons needs 5 $Fe^{2+}$ per $MnO_4^-$.

Reduction half-reaction ($Mn^{+7} \to Mn^{2+}$, gain of 5 electrons):

$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$

Oxidation half-reaction ($Fe^{2+} \to Fe^{3+}$, loss of 1 electron), multiplied by 5 to balance electrons:

$5Fe^{2+} \longrightarrow 5Fe^{3+} + 5e^-$

Overall balanced ionic equation:

$MnO_4^- + 8H^+ + 5Fe^{2+} \longrightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

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