Perform Redox to the following chemical equation and balance it chemically and ionically: MnO_4^- + SO_3^-2 ---> SO_4^-2 + Mn^+2

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Here `MnO_4^-` oxidize `SO_3^(2-)` to `SO_4^(2-)` . In the process `MnO_4^-` is reduced to `Mn^(2+)` . The reaction takes place in a acidic medium.


`MnO_4^(-)+8H^+ rarr Mn^(2+) + 4H_2O`

In this reaction at right side the total charge is +2

So in the left also it should be +2. Currently we have -1+8 = +7 charge at left. So we need 5 electrons to make it +2.


`MnO_4^(-)+8H^++5e rarr Mn^(2+) + 4H_2O` ----(1)


`SO_3^(2-)+H_2O rarr SO_4^(2-)+2H^++2e` -----(2)


(1)*2+(2)*5 will cancel out the electrons.

`2MnO_4^(-)+16H^++5SO_3^(2-)+5H_2O rarr 2Mn^(2+) + 8H_2O+5SO_4^(2-)+10H^+`

So the balance reaction is;

`2MnO_4^(-)+6H^++5SO_3^(2-) rarr 2Mn^(2+) + 3H_2O+5SO_4^(2-)`





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