Respuesta :
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: = [tex]SiO_3^{2-}[/tex] which oxidizes other but reduce itself.
Reducing agent =[tex]SO_3^{2-}[/tex] which reduces other but oxidize itself.
formula of the oxidizing agent=[tex]SiO_3^{2-}[/tex]
formula of the reducing agent=[tex]SO_3^{2-}[/tex]
Explanation:
first write the chemical reaction:
[tex]SiO_3^{2-} + 2SO_3^{2-} +H_2O\rightarrow 2SO_4^{2-} +Si + 2OH^-[/tex]
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: = [tex]SiO_3^{2-}[/tex] which oxidizes other but reduce itself.
Reducing agent =[tex]SO_3^{2-}[/tex] which reduces other but oxidize itself.
formula of the oxidizing agent=[tex]SiO_3^{2-}[/tex]
formula of the reducing agent=[tex]SO_3^{2-}[/tex]
