Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.

The more positive the value of ${\mathrm{E}}^{⊝}$, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.

${\mathrm{E}}^{⊝}$values: ${\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}=+0.77; {\mathrm{I}}_2\left(\mathrm{s}\right)/{\mathrm{I}}^{-}=+0.54$

                    ${\mathrm{Cu}}^{2+}/\mathrm{Cu}=+0.34; {\mathrm{Ag}}^{+}/\mathrm{Ag}=+0.80\mathrm{V}$

${\mathrm{E}}^{⊝}$ values of some redox couples are given below. On the basis of these values choose the correct option.

${\mathrm{E}}^{⊝}$ values: ${\mathrm{Br}}_2/{\mathrm{Br}}^{-}=+1.90; {\mathrm{Ag}}^{+}/\mathrm{Ag}\left(\mathrm{s}\right)=+0.80$

                    ${\mathrm{Cu}}^{2+}/\mathrm{Cu}\left(\mathrm{s}\right)=+0.34; {\mathrm{I}}_2\left(\mathrm{s}\right)/{\mathrm{I}}^{-}=+0.54$

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

${\mathrm{E}}^{⊝}$ values: ${\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}=+0.77; {\mathrm{I}}_2/{\mathrm{I}}^{-}=+0.54$

                     ${\mathrm{Cu}}^{2+}/\mathrm{Cu}=+0.34;{\mathrm{Ag}}^{+}/\mathrm{Ag}=+0.80\mathrm{V}$

Thiosulphate reacts differently with iodine and bromine in the reactions given below:

$2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+{\mathrm{I}}_2\to {\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{I}}^{-}$

${\mathrm{S}}_2{\mathrm{O}}_3^{2-}+{\mathrm{Br}}_2+5{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{SO}}_4^{2-}+2{\mathrm{Br}}^{-}+10{\mathrm{H}}^{+}$

Which of the following statements justifies the above dual behaviour of thiosulphate?