What is the reducing agent in the reaction:

Balance and calculate the reducing agent in the reaction:

Sn(OH)3^-(aq) + Bi(OH)3(s) + OH^-(aq) &ραρρ; Sn(OH)6^2-(aq) + Bi(s)

I really don't understand how to balance the charges. I know that the reducing agent is the compound that gains charge in the reaction.

If someone could help me understand how to get the answer I would greatly appreciate it:)

[Sn(OH)3]^-1 has Sn at +2 -->& it gets oxidized up to --> [Sn(OH)6]-2 which has Sn at +4
[Sn(OH)3]^-1 --> [Sn(OH)6]-2 & 2 electrons lost
[Sn(OH)3]^-1 --> [Sn(OH)6]-2 & 2 e-


Bi(OH)3(s) with Bi at +3 --> gets reduced down to --> Bi at zero
Bi(OH)3(s) takes 3 electrons --> Bi(s)
Bi(OH)3(s) & 3 e- --> Bi(s)


electrons balance as:
oxidation: 3 [Sn(OH)3]^-1 --> 3 [Sn(OH)6]-2 & 6 e-
reduction: 2 Bi(OH)3(s) & 6 e- --> 2 Bi(s)

giving us
2 Bi(OH)3(s) & 3 [Sn(OH)3]^-1 --> 3 [Sn(OH)6]-2 & 2 Bi(s)

with a total -3 on the left & a total of -6 on the right,
we balance the charges with OH-'s, giving a -6 total on both sides
2 Bi(OH)3(s) & 3 [Sn(OH)3]^-1 & 3 OH- --> 3 [Sn(OH)6]-2 & 2 Bi(s)


since the oxidation number of Bismuth was reduced to a lower value,
Sn(OH)3^-(aq) is the reducing agent