Use oxidation numbers to find the oxidant and the reductant!
The thiosulfate ion $S_2O_3^{2-}$ (n.o.(S)=2) is the reductant which is oxidised to $SO_3^{2-}$ (n.o.(S)=4). The thiosulfate ion $S_2O_3^{2-}$ (n.o.(S)=2) is the oxidant which is reduced to $S_8$ (n.o.(S)=0).
How many electrons are exchanged by the oxidant and the reductant?
$S_2O_3^{2-}$ looses 4 $e^-$. 4 $S_2O_3^{2-}$ ions catch 16 $e^-$.
Write the partial oxidation and reduction reactions in an acid solution.
$S_2O_3^{2-}$ $-$ $4e^- $ $+$ $3H_2O$ $\longrightarrow$ $2SO_3^{2-}$ $+$ $6H^+$ $4S_2O_3^{2-}$ $+$ $16e^- $ $+$ $24H^+\longrightarrow S_8$ $+$ $$ $+$ $12H_2O$
Balance by the number of exchanged electrons and write the redox reaction!
$4S_2O_3^{2-}$ $-$ $16e^- $ $+$ $12H_2O$ $\longrightarrow$ $8SO_3^{2-}$ $+$ $24H^+$
$4S_2O_3^{2-}$ $+$ $16e^- $ $+$ $24H^+$ $\longrightarrow$ $S_8$ $+$ $12H_2O$
$8S_2O_3^{2-}$ $\longrightarrow$ $8SO_3^{2-}$ $+$ $S_8$
Complete for the ions which are not involved et annotate!
$16Na^+$ $+$ $12S_2O_3^{2-}$ $\longrightarrow$ $16Na^+$ $+$ $ 8SO_3^{2-}$ $+$ $S_8$ Sodium thiosulfate (colourless solution) produces sulfur (yellow solid) and sodium sulfite (colourless solution)
Where does the pungent odour come from?
$2H^+$ $+$ $12SO_3^{2-}$ $\longrightarrow$ $H_2O$ $+$ $SO_2$(sulfur dioxide, colourless gas with a pungent odour)