I2+2S2O32- xrightarrowS4O62-+2I- In the above reaction

Question

I2+2S2O32- xrightarrowS4O62-+2I- In the above reaction (A) iodine is reduced; sulphur is reduced (B) iodine is reduced; sulphur is oxidized (C) iodine is oxidised; sulphur is reduced (D) iodine is oxidised; sulphur is oxidized

Tardigrade Admin · Accepted Answer

Increase in oxidation number is oxidation while decrease in oxidation number represents reduction.

Q. I2​+2S2​O32−​​S4​O62−​+2I− In the above reaction

iodine is reduced; sulphur is reduced

iodine is reduced; sulphur is oxidized

iodine is oxidised; sulphur is reduced

iodine is oxidised; sulphur is oxidized

Increase in oxidation number is oxidation while decrease in oxidation number represents reduction.

Q. $I_{2} + 2 S_{2} O_{3}^{2 -} S_{4} O_{6}^{2 -} + 2 I^{-}$ In the above reaction