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 ?
1 Answer
Solution :-
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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 agoRelated Questions
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