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 | 11 months ago |  46

1 Answer

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 | 11 months ago

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