Consider the reactions :

$$2 \; S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; I_{ 2 \; (s) } \; \rightarrow \; S_{ 4 }O^{ 2- }_{ 6 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) }$$

$$S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; 2 \; Br_{ 2 \; (l) } \; + \; 5 \; H_{ 2 }O_{ (l) } \;$$

$$\rightarrow \; 2 \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; Br^{ – }_{ (aq) } \; + \; 10 \; H^{ + }_{ (aq) }​$$

Why does the same reductant, thiosulphate react differently with iodine and bromine ?

Asked by Pragya Singh | 1 year ago |  61

##### Solution :-

As $$Br_{ 2 }$$​ is a stronger oxidizing agent than $$I_{ 2 },$$ it oxidizes S of  $$S_{ 2 }O_{ 3 }^{ 2- }$$ to a higher oxidation no. of +6 in $$SO_{ 4 }^{ 2- }$$

As $$I_{ 2 }$$is a weaker oxidizing agent so it oxidizes S of $$S_{ 2 }O_{ 3 }^{ 2- }$$​ ion to a lower oxidation no. that is 2.5 in $$S_{ 4 }O_{ 6 }^{ 2- }$$​ ions.

Thus, thiosulphate react differently with$$I_{ 2 }$$​ and $$Br_{ 2 }$$​.

Answered by Abhisek | 1 year ago

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