$ 2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}$
$\mathrm{S}_2 \mathrm{O}_3^{2-}+5 \mathrm{Br}_2+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-}+4 \mathrm{Br}^{-}+10 \mathrm{H}^{+}$
નીચે આપેલા માંથી કયું વિધાન થાયોસલ્ફેટ ના આ દ્રી-વર્તણૂક ને સમર્થન (Justify) કરે છે.
In the reaction of $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ with $\mathrm{Br}_2$, oxidation state of sulphur changes from +$2$ to +$6$ .
$\therefore$ Both $\mathrm{I}_2$ and $\mathrm{Br}_2$ are oxidant (oxidising agent) and $\mathrm{Br}_2$ is stronger oxidant than $\mathrm{I}_2$.
$14{H^ + } + C{r_2}O_7^{2 - } + 3Ni \to 2C{r^{3 + }} + 7{H_2}O + 3N{i^{2 + }}$
$MnO_4^ - \,\, + \,\,{C_2}O_4^{2 - }\,\, + \,\,{H^ + }\, \rightarrow \,\,M{n^{2 + }}\, + \,C{O_2} + \,{H_2}O$
${H_2}\mathop S\limits^{ } {O_4}\,\,\,\,\, \to \,\,\,\,\,\mathop S\limits^{ } {O_2}$
$xCu\,\,\, + \,\,\,\,yHN{O_3}\,\,\, \to \,\,\,\,xCu{\left( {N{O_3}} \right)_2}\,\,\, + \,\,\,NO\,\,\,\, + \,\,\,N{O_2}\,\,\, + \,\,\,3{H_2}O$ તે સર્હીુણકો $x$ અને $y$ શું હશે ?