Verified answer

Bi(OH)3 + K2Sn2O8 → Bi + K2SnO3

solve by half reaction,

Bi^3+ + 3e → Bi. . . . . . . .(reduction)

Sn2O8^2- + 2H2O + 6e → 2SnO3^2- + 4OH^- . . . .. . . (reduction)

i think it should be reduction and oxidation,

Bi → Bi^3+ + 3e . . . . . . . . . (multiply by 2)

Sn2O8^2- + 2H2O + 6e → 2SnO3^2- + 4OH^-

2Bi → 2Bi^3+ + 6e

Sn2O8^2- + 2H2O + 6e → 2SnO3^2- + 4OH^-

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (+)

2Bi + Sn2O8^2- + 2H2O → 2Bi^3+ + 2SnO3^2- + 4OH^-

the equation would be unable to be balanced precise side Bismuth hydrogen Potassium Sulphur Oxygen Tin is lacking Left side Bismuth Potassum Tin Oxygen Sulphur and hydrogen lacking in this side