Question

Balance the following ionic equation for a redox reaction, using whole number coefficients. MnO−4(aq)+SO2−3(aq)+H3O+(aq)⟶Mn2+(aq)+SO2−4(aq)+H2O(l)

Answer 1

Answer : The balance ionic equation for a redox reaction will be,

$2MnO_4^-(aq)+6H_3O^+(aq)+5SO_3^{2-}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_4^{2-}(aq)+9H_2O(l)$

Explanation :

The half balanced redox reactions in acidic medium will be:

(1) $MnO_4^-(aq)+8H_3O^+(aq)+5e^-\rightarrow Mn^{2+}(aq)+12H_2O(l)$

(2) $SO_3^{2-}(aq)+3H_2O(l)\rightarrow SO_4^{2-}(aq)+2H_3O^+(aq)+2e^-$

In order of balance the electrons, we multiply the equation 1 by 2 and equation 2 by 5, we get:

(1) $2MnO_4^-(aq)+16H_3O^+(aq)+10e^-\rightarrow 2Mn^{2+}(aq)+24H_2O(l)$

(2) $5SO_3^{2-}(aq)+15H_2O(l)\rightarrow 5SO_4^{2-}(aq)+10H_3O^+(aq)+10e^-$

Now adding both the equation, we get:

The overall reaction will be,

$2MnO_4^-(aq)+6H_3O^+(aq)+5SO_3^{2-}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_4^{2-}(aq)+9H_2O(l)$