How to balance equations using the half cell method?

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The discussion outlines the steps for balancing a redox reaction involving sulfite (SO32-) and permanganate (MnO4-). It begins with dividing the overall reaction into half-reactions, focusing on balancing atoms other than hydrogen and oxygen first. Oxygen is balanced by adding water (H2O), while hydrogen is balanced by adding hydrogen ions (H+). The charge is balanced by adding electrons to each half-reaction. The key is to equalize the number of electrons gained and lost before combining the half-reactions. The final half-reactions are presented, with the sulfite oxidation and permanganate reduction clearly defined. The process emphasizes the importance of adjusting coefficients to achieve equal electron transfer, ultimately leading to the balanced overall equation.
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SO32- +MnO4-1+H+1 <-> Mn2+ +SO42- +H20(l)
 
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Step 1 Divide the skeleton equation into half-reactions.

Step 2 Balance atoms other than H and O.

Step 3 Balance oxygen atoms by adding H2O to the side that needs O.

Step 4 Balance hydrogen by adding H+ to the side that needs H.

Step 5 Balance the charge by adding electrons.

Step 6 Make the electrons gained equal to the electrons lost and then add the two half-reactions.

Step 7 Cancel anything that is the same on both sides.
 
this problem may seem a bit more tricky than the redox problems that you are used to since it may not be obvious which is getting reduced/oxidized based upon the superscripts alone.

The half-reactions are:

SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-
MnO_4^{-1} + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O

Make your electrons equal by multiplying the half rxns to make the number of electrons 10 for both, then add the half rxns up and you should get the correct answer.
 
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