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)`