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The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphite ion in acidic solution is
A
$\dfrac{4}{5}$
B
$\dfrac{2}{5}$
C
1
D
$\dfrac{3}{5}$
Detailed Solution
In acidic solution: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (Mn changes from +7 to +2, gain of 5 electrons)
$SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$ (S changes from +4 to +6, loss of 2 electrons)
To balance the electrons, multiply the first by 2 and the second by 5 and add.
The balanced chemical equation is: $2MnO_4^- + 6H^+ + 5SO_3^{2-} \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O$
From the equation, the moles of $MnO_4^-$ required to oxidise 5 moles of $SO_3^{2-}$ are 2.
Moles of $MnO_4^-$ required to oxidise 1 mole of $SO_3^{2-}$ are $\dfrac{2}{5}$.
$SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$ (S changes from +4 to +6, loss of 2 electrons)
To balance the electrons, multiply the first by 2 and the second by 5 and add.
The balanced chemical equation is: $2MnO_4^- + 6H^+ + 5SO_3^{2-} \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O$
From the equation, the moles of $MnO_4^-$ required to oxidise 5 moles of $SO_3^{2-}$ are 2.
Moles of $MnO_4^-$ required to oxidise 1 mole of $SO_3^{2-}$ are $\dfrac{2}{5}$.