Question

Is the S₂O3^2- ion undergoing oxidation or reduction?

Answer 1

The oxidation number of sulphur atom in the given ion is calculated as shown below:

Let the oxidation number of sulphur atom is x

Oxygen is -2.

2(x)+3(-2)=-2

=>x = +2

On average the oxidation number of sulphur atom is +2

For each sulpur atom it is +6 and -2.

+6 is the highest oxidation state exhibited by sulphur atom.

It can gain electrons and hence,it acts as an oxidising agent.

-2 is the lowest oxidation state of sulphur atom and hence, it can lose electrons and can act as a reducing agent.

So the given this sulphate ion can undergo both oxidation and reduction.