Balanced Half-Reactions

Can someone show me the balanced half-reactions for the conversions of:
A. Sn(OH)4 2- to Sn(OH)6 2-

and

B. Bi(OH)3 to Bi


Please and thank you!

A. Sn(OH)4 2- -> Sn(OH)6 2-

OH has a charge of -1, hence on the left OH in total contributes a total of -4, whilst on the right -6
Hence Sn must originally be +2, and end as +4. You also need to have gained the other 2 OH
groups from somewhere, most likely water. You therefore also have 2 waters going to 2 OH- and H2:

Sn2+ -> Sn4+
2H20 -> 2OH- + H2

B. Bi(OH)3 -> Bi

Again your OH has a charge of -1, so Bi must originally be +3

Bi3+ -> Bi0