Consider the reaction:
`2S_(2)O_(3)^(2-)(aq)+I_(2)(s) rarr S_(4)O_(6)^(2-)(aq) + 2I^(Θ)(aq)`
`2S_(2)O_(3)^(2-)(aq) + 2Br_(2)(l) + 5H_(2)O(l) rarr 2SO_(4)^(2-)(aq) + 4Br^(Θ)(aq)+10H^(o+)(aq)`
Why does the same reductant, thiosulphate, react differently with iodine and bromine?

Explanation

To understand why thiosulfate reacts differently with iodine and bromine, we need to analyze the oxidation states of sulfur in the reactions and the oxidizing power of iodine and bromine. Let's break down the solution step-by-step.

### Step 1: Write the reactions
We have two reactions involving thiosulfate (\(S_2O_3^{2-}\)):
1. With iodine (\(I_2\)):
\[ 2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-} \]
...