J.R. S. answered 04/28/21
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
SnO32- ==> SnO22- ... unbalanced
SnO32- ==> SnO22- + H2O ... balance for Sn and O
SnO32- +2H2O ==> SnO22- + H2O + 2OH- ... balanced for Sn, O and H
SnO32- +2H2O + 2e- ==> SnO22- + H2O + 2OH- ... balanced for mass and charge
___1_SnO32- + ___2___H2O
___1__SnO22- + ___2__OH-
In the above half reaction, the oxidation state of tine changes from 4+ to 2+