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  4. In the reaction 2 Na2S2O3 + I2 arrow Na2 S4 O6 + NaI the oxidation number of sulphur is

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Q. In the reaction $2 Na_2S_2O_3 + I_2 \rightarrow Na_2 S_4 O_6 + NaI$ the oxidation number of sulphur is

Redox Reactions

Solution:

$Na _2 S _2 O _3=+2+2 x -6=0$ or $2 x=4$ or $x=+2$.
$Na _2 S _4 O _6=+2+4 x -12=0$ or $4 x=10$ or $x=\frac{10}{4}=2.5$
Here, the O.N. of S is increased.