Question

Balance by half reaction. Mno4- + SO2. gives Mn2+ + HSO4- in acidic medium

Answer 1

First split the question up to look like this. 

SO2 ---> (SO4)2- 

MnO4- ---> (Mn)2+ 

You don't need to balance for S or for Mn so start with oxygen on each side. Then balance for hydrogen on each equation. Finally, put both together so your total charges cancel out (system of equations sort of). 

After that it's just simplification. 

SO2 + 2H2O ---> SO4(2-) + 4H+ +2e- ] Multiply by factor of 5 

MnO4- + 8H+ +5e----> Mn(2+) + 4H2O ] Multiply by factor of 2 

5SO2 + 10H2O ----> 5SO4 + 20H + 10e- 
2MnO4 + 16H + 10e- ----> 2Mn(2+) +8H2O 

electrons cancel out, now just simplify it all out. 

5SO2 + 2MnO4(-) + 2H2O ----> 5SO4(-) + 2Mn(2+) + 4H2O

Answer 2

Answer:

The correct answer of this question is 2 H2O + H+ + 5SO2 + 2MnO4- <-> 2Mn2+ + 5HSO4

Explanation:

Given -  Half-reaction. Mno4- + SO2. gives Mn2+ + HSO4- in acidic medium.

To Find - Balance by half-reaction.

5e + 8H+ + MnO4- <-> Mn2+ + 4 H2O  (x2)

2H2O + SO2 <-> HSO4- + 3H+ 2 e  (x5)

A redox reaction's half-reaction is either the oxidation or reduction reaction component. The change in oxidation states of individual chemicals participating in the redox reaction is used to calculate a half-reaction.

Multiply e by a common number, such as 10. Use it to multiply each issue separately.

10e +16H+ + 2MnO4- <-> 2Mn2+ + 8H2O

10H2O + 5SO2 + 2MnO4- <-> 5HSO4- + 15H+ + 10e

And the answer is

2 H2O + H+ + 5SO2 + 2MnO4- <-> 2Mn2+ + 5HSO4-

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