Q.Write the balanced ionic equation for the reaction of acidified potassium permanganate with oxalate ions, and state the n-factor of $\text{KMnO}_4$ in this (acidic-medium) reaction.

In acidic medium, permanganate is reduced according to the half-reaction

a change from $\text{Mn}^{7+}$ to $\text{Mn}^{2+}$, i.e. a gain of 5 electrons, so the n-factor of $\text{KMnO}_4$ in acidic medium is 5.

Oxalate ion is oxidized according to the half-reaction

a two-electron change (each carbon goes from the $+3$ state in oxalate to the $+4$ state in $\text{CO}_2$).

To balance electrons, the permanganate half-reaction (5 electrons gained) must be multiplied by 2, and the oxalate half-reaction (2 electrons lost) must be multiplied by 5, so that both supply/require 10 electrons:

Adding these gives the overall balanced ionic equation:

$2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \longrightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$ …

Write the two half-reactions (MnO4- to Mn2+, gaining 5e-; C2O4^2- to 2CO2, losing 2e-), find the LCM of electrons (10), scale e …

Using the wrong n-factor for KMnO4 by forgetting the medium matters (5 in acid, 3 in neutral, 1 in strongly alkaline -- these are NOT interchange …