Question

Na2S2O3 + I2 gives Na2S4O6 + NaI​

Answer 1

Answer:

Na2S2O3 +I2 = Na2S4O6+NaI balance the equation by ion electron method. here we see iodine , I2 is reduced (as oxidation number changes from 0 to -1) and sulphur is oxidised. ( as oxidation number of s changes from +2 to +2.5).

Answer 2

$\huge\fcolorbox{black}{lime}{AnsweR:}$

Na2S2O3 +I2 = Na2S4O6+NaI balance the equation by ion electron method.

here we see iodine , I2 is reduced (as oxidation number changes from 0 to -1) and sulphur is oxidised. ( as oxidation number of s changes from +2 to +2.5).

half reduction reaction :$\bold{l_2}$ ⇆ $\bold{2l^++2e^- ...(1)}$

half oxidation reaction : $\bold{2S_2O_3^2^- +2e^-}$ ⇆ $\bold{S_4O_6^2^-}$

now adding $\bold{2Na^+}$ions both sides in equations (1) and (2)

then,$\bold{2Na^++l_2}$ ⇆ $\bold{2Nal+2e^- ...(1)}$

and $\bold{2Na_2S_2O_3+2e^-}$ ⇄ $\bold{Na_2S_4O_6...(2)}$

now adding equations (1) and (2) we get,

$\bold{2Na_2S_2O_3+l_2}$ ⇄ $\bold{N_2S_4O_6+2Nal}$

This is your balanced chemical equation sis ❤️