Looking for classes? Ksquare Career Institute, Bengaluru →
The number of moles of $KMnO_4$ reduced by one mole of KI in alkaline medium is :-
A
One
B
Two
C
Five
D
One fifth
Detailed Solution
In alkaline (or neutral) medium, $KMnO_4$ is reduced to $MnO_2$: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ (Mn changes from +7 to +4, a gain of 3 electrons).
In alkaline medium, iodide is oxidised to iodate: $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$ (I changes from $-1$ to +5, a loss of 6 electrons).
Equating the electrons: 6 electrons lost by one $I^-$ are taken up by $\dfrac{6}{3} = 2$ moles of $MnO_4^-$.
Balanced equation: $2KMnO_4 + KI + H_2O \rightarrow 2MnO_2 + KIO_3 + 2KOH$
So one mole of KI reduces two moles of $KMnO_4$.
In alkaline medium, iodide is oxidised to iodate: $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$ (I changes from $-1$ to +5, a loss of 6 electrons).
Equating the electrons: 6 electrons lost by one $I^-$ are taken up by $\dfrac{6}{3} = 2$ moles of $MnO_4^-$.
Balanced equation: $2KMnO_4 + KI + H_2O \rightarrow 2MnO_2 + KIO_3 + 2KOH$
So one mole of KI reduces two moles of $KMnO_4$.