Final answer:
The oxidation-reduction reaction in question seems to contain typographical errors; however, assuming 'SO' is an oxidized sulfur species, Mn²+ undergoes reduction while 'SO' undergoes oxidation. To properly balance the redox equation, full details of the 'SO' species are required.
Step-by-step explanation:
In this oxidation-reduction reaction, Mn²+ undergoes reduction, and SO undergoes oxidation.
Firstly, the given reaction seems to have some typographical errors, as 'SO' is not a common chemical species for redox reactions. Assuming that 'SO' represents a sulfur-containing species undergoing oxidation and forming S²-, we can say that sulfur (S) in 'SO' is oxidized as its oxidation state increases from an assumed positive state to -2. The reaction likely represents the reduction of manganese and oxidation of sulfur in another species (not specified here).
When writing half-reactions, we focus on the electron transfer involved. Oxidation involves the loss of electrons, while reduction involves the gain of electrons. Since Mn²+ is gaining electrons to become Mn (with an oxidation state of +7 to +2), it is undergoing reduction. The 'SO' molecule would be undergoing oxidation as its oxidation state increases with the formation of S²-.
To provide a balanced equation for the actual redox reaction, it would be necessary to know the full chemical formula for 'SO', since balance involves not only charges but also the number of atoms of each element on both sides of the reaction. Without complete information on the 'SO' species, we can't balance the equation fully.
Note that in a redox reaction, the total charges must balance out before and after the reaction, indicating a conservation of charge.