Use oxidation numbers to find the oxidant and the reductant!
The thiosulfate ion $S_2O_3^{2-}$ $(o.n.(S)=2)$ is the reductant which is oxidised to $S_4O_6^{2-}$ $(o.n.(S)=$\frac{5}{2}$)$. $I_2$ $(o.n.(I)=0)$ is the oxidant which is reduced to $I^-$ $(o.n.(I)=-1)$.
How many electrons are exchanged by the oxidant and the reductant?
Two ions $S_2O_3^{2-}$ loose 2 $e^-$. $I_2$ catches 2 $e^-$.
Write the partial oxidation and reduction reactions in an acid solution.
$2S_2O_3^{2-}$ $-$ $2e^-$ $\longrightarrow$ $S_4O_6^{2-}$ $I_2$ $+$ $2e^- $ $\longrightarrow$ $ 2I^-$
Balance by the number of exchanged electrons and write the redox reaction!
$2S_2O_3^{2-}$ $+$ $I_2$ $\longrightarrow$ $S_4O_6^{2-}$ $+$ $2I^-$
Complete for the ions which are not involved et annotate!
$2Na^+$ $+$ $2S_2O_3^{2-}$ $+$ $I_2$ $\longrightarrow$ $2Na^+$ $+$ $S_4O_6^{2-}$ $+$ $2Na^+$ $+$ $2I^-$ Sodium thiosulfate de sodium (colourless solution) and iodine (brown in solution) produce sodium tetrathionate and sodium iodide (colourless in solution)