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2MnO4 + H + 5HSO3 = 2Mn + 5HSO4 + 3H2O

I was asked to write the half equations for both the oxidized and reduced reactions.

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

Reduced
• 5SO42- + 15H+ +10e- -----> 5HSO3- + 5H2O

Are these half equations correct?

2 answers

They are oxidzed and reduced equations that you wrote but you didn't write the right ones. In the equation it's MnO4^- that is reduced to Mn^2+ and the SO3^2- is oxidized to SO4^2-
You wrote the oxidized and reduced equations for the REVERSE equation.
Oxidized
• 5SO42- + 15H+ +10e- -----> 5HSO3- + 5H2O

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

I am correct now?