Question

The number of moles of $Mn{O}_4^{-}$ that will be needed to react with one mole of $S{O}_3^{2-}$ (sulphite ion) in acidic solution is :

• A. $0.4$ mol
• B. $0.6$ mol
• C. $0.8$ mol
• D. none of the above

Answer 1

The correct option is A $0.4$ mol

In acidic medium,
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$......(1)
$S{O}_3^{2-}+H_{2}O\to S{O}_4^{2-}+2H^{+}+2e^{-}$......(2)
Multiply equation (1) with 2 and equation (2) with 5 to balance the electrons.
$2Mn{O}_4^{-}+16H^{+}+10e^{-}\to 2M{n}^{2+}+8H_{2}O$......(3)
$5S{O}_3^{2-}+5H_{2}O\to 5S{O}_4^{2-}+10H^{+}+10e^{-}$......(4)
Add equations (3) and (4).
$2Mn{O}_4^{-}+5S{O}_3^{2-}+6H^{+}+10e^{-}\to 2M{n}^{2+}+5S{O}_4^{2-}+3H_{2}O$
In acidc medium 5 moles of sulphite ions react with 2 moles of $Mn{O}_4^{-}$ ions.
Hence, in acidic medium, 1 mole of sulphite ion will react with $\frac{2}{5}=0.4$ moles of $Mn{O}_4^{-}$ ions.