Side$-$by$-$Side

In Reaction $2,$ sodium thiosulfate $(N{a}_2{S}_2{O}_8)$ reacts with the triiodide ion $(I_3^{-})$to reform the iodide ion $(I_3^{-}).$ The triiodide ion will only react with the starch after all the thiosulfate ion is consumed in the reaction.

Reaction $1$: $,3I^{-}(aq)+S_2{O}_8^{2-}(aq)longrightarrow{I}_3^{-}(aq)+2$

Reaction $2$: $,I_3^{-}(aq)+2S_2{O}_3^{2-}(aq)longrightarrow3I^{-}(aq)+$

Reaction $3$: $,I_3^{-}(aq)+$starch$(aq)longrightarrow[3I^{-}-$starch$]$

What is the purpose of adding sodium thiosulfate to the reaction mixture?

Sodium thiosulfate increases the iodide ion concentration in Reaction $1.$

Sodium thiosulfate acts as an indicator which forms blue$-$black solution in the presence of the triiodide ion.

Sodium thiosulfate slows the reaction so that its rate can be measured with a stopwatch or clock.

Sodium thiosulfate is a catalyst speeding up the overall rate of the reaction.