Thank you for reporting, we will resolve it shortly
Q.
Oxidation number of sulphur in $ H_2S_2O_8$ is
Rajasthan PMTRajasthan PMT 2007Redox Reactions
Solution:
(i) Draw the structue of oxyacid $\left( H _{2} S _{2} O _{8}\right)$ to decide the oxidation state of oxygen. Peroxide linkage has $-1$ oxidation state and normally oxygen has $-2$ oxidation state.
$\therefore $ Two oxygens form peroxide linkage.
$\therefore $ Oxidation state of two oxygens is $-1$ each and rest of the six oxygens have $-2$ oxidation state each,
$ H _{2} S _{2} O _{8} $
$+1 \times 2+2 x+(-1 \times 2)+(-2 \times 6) =0$
or $ 2+2 x-2-12 =0 $
or $ 2 x =12 $
$\therefore x =+6$
