SiO32- + 2SO32-+ H2O2SO42- + Si+ 2OH-In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent. name of the element oxidized: name of the element reduced: formula of the oxidizing agent:formula of the reducing agent:

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=

formula of the reducing agent=

Explanation:

first write the chemical reaction:

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=

formula of the reducing agent=

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