Here, Iodine is going under oxidation and Sulphur reduction.
8[tex]NaI[/tex](aq) + 5 (aq) ---> 4(aq) + (g) + 4 [tex]Na_{2} SO_{4}[/tex] (aq) + 4[tex]H_{2} O[/tex](l)
the oxidation state of I in NaI is -1 whereas, it is 0 in [tex]I_{2}[/tex] .
in the same way , the oxidation state of S in [tex]H_{2} SO_{4}[/tex] is +6 whereas it is -2 in [tex]H_{2} S[/tex].
what is oxidation?
oxidation is when the element loses its electron ( I went from -1 to 0, losing an electron ).
what is reduction ?
reduction is when the element gains its electron ( S went from +6 to -2, gaining 8 electron ).
Learn more about oxidation and reduction here:-
https://brainly.com/question/13182308
#SPJ1
