Answer:
oxidizing element=S oxidation number is increasing i.e. +4 to +6
Reducing element=Si oxidation number is decreasing i.e. +4 to 0
Oxidizing agent: =
which oxidizes other but reduce itself.
Reducing agent =
which reduces other but oxidize itself.
formula of the oxidizing agent=![SiO_3^{2-}](https://tex.z-dn.net/?f=SiO_3%5E%7B2-%7D)
formula of the reducing agent=![SO_3^{2-}](https://tex.z-dn.net/?f=SO_3%5E%7B2-%7D)
Explanation:
first write the chemical reaction:
![SiO_3^{2-} + 2SO_3^{2-} +H_2O\rightarrow 2SO_4^{2-} +Si + 2OH^-](https://tex.z-dn.net/?f=SiO_3%5E%7B2-%7D%20%2B%202SO_3%5E%7B2-%7D%20%2BH_2O%5Crightarrow%20%20%202SO_4%5E%7B2-%7D%20%20%2BSi%20%2B%202OH%5E-)
and then write the oxidation state of all the elemet both side i.e. product and reactant:
oxidation state of element of reactant side:
Si=+4
O=-2
S=+4
H=+1
oxidation state of element of product side:
Si=0
O=-2
S=+6
H=+1
from the above it is clearly that:
oxidizing element=S oxidation number is increasing i.e. +4 to +6
Reducing element=Si oxidation number is decreasing i.e. +4 to 0
Oxidizing agent: =
which oxidizes other but reduce itself.
Reducing agent =
which reduces other but oxidize itself.
formula of the oxidizing agent=![SiO_3^{2-}](https://tex.z-dn.net/?f=SiO_3%5E%7B2-%7D)
formula of the reducing agent=![SO_3^{2-}](https://tex.z-dn.net/?f=SO_3%5E%7B2-%7D)