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In this reaction: S(2)O(8)^(2-)+2I^(-) t...

In this reaction: `S_(2)O_(8)^(2-)+2I^(-) to 2SO_(4)^(2-)+I_(2)`

A

Oxidation of iodide into iodine takes place

B

Reduction of iodine into iodide takes place

C

Both oxidation ane reduction of iodine takes place

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
`2I^(-) rarr overset(0)I_(2)`
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The experiment rate law for the reaction S_(2)O_(8)^(2-)(aq) + 2I^(ɵ)(aq) rarr 2SO_(4)^(2-) (aq) + I_(2)(aq) is k[S_(2)O_(8)^(2-)] [I^(ɵ)] . How would the rate change if (a) Concentration of S_(2)O_(8)^(2-) is halved. (b) Concentration of S_(2)O_(8)^(2-) and I^(ɵ) are halved.

Assertion : In the reactin, 2S_(2)O_(3)^(2-)+I_(2) to S_(4)O_(6)^(2-) + 2I^(-): I_(2) is oxidised. Reason : During oxidation, loss of electron takes place.

In the reaction 2I^(-) + S_(2)O_(8)^(2-) rarr I_(2) + 2SO_(4)^(2-) the catalyst is (A) Fe^(+3) (B) Mn^(+2) (C) Cu^(+2) (D) Ni^(+2)

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