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^-

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the equation can't be balanced

Right side

Bismuth

hydrogen

Potassium

Sulphur

Oxygen

Tin is missing

Left side

Bismuth

Potassum

Tin

Oxygen

Sulphur and hydrogen missing on this side