In the reaction- S-2O-2-8-2I-rightarrow 2SO-2-4-I-2-oxidation of iodide into iodine takes placereduction of iodine into iodide takes placenone of the aboveboth oxidation and reduction of iodine takes place

Question

Toppr Admin · Accepted Answer

In this reaction- the oxidation number of iodine increases from -1 -in I-x2212- to 0 -in I2- i-e- iodine undergoes loss of electrons or oxidation-Hence- Option A is the correct answer-

In the reaction: $S_{2} O_{8}^{2 -} + 2 I^{-} \to 2 S O_{4}^{2 -} + I_{2}$ .
oxidation of iodide into iodine takes place
reduction of iodine into iodide takes place
none of the above
both oxidation and reduction of iodine takes place

In this reaction, the oxidation number of iodine increases from -1 (in $I^{-}$ ) to 0 (in $I_{2}$ ) i.e. iodine undergoes loss of electrons or oxidation.
Hence, Option A is the correct answer.

In the reaction: S2O2−8+2I−→2SO2−4+I2.oxidation of iodide into iodine takes placereduction of iodine into iodide takes placenone of the aboveboth oxidation and reduction of iodine takes place

In this reaction, the oxidation number of iodine increases from -1 (in I−) to 0 (in I2) i.e. iodine undergoes loss of electrons or oxidation.Hence, Option A is the correct answer.

In the reaction: $S_{2} O_{8}^{2 -} + 2 I^{-} \to 2 S O_{4}^{2 -} + I_{2}$ .
oxidation of iodide into iodine takes place
reduction of iodine into iodide takes place
none of the above
both oxidation and reduction of iodine takes place

In this reaction, the oxidation number of iodine increases from -1 (in $I^{-}$ ) to 0 (in $I_{2}$ ) i.e. iodine undergoes loss of electrons or oxidation.
Hence, Option A is the correct answer.