Question

2I2+S2O32-+6OH- ------> 2SO32-+4I-+3H2O

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent

Answer 1

Hi,
An unbalanced equation (the reaction skeleton) is written that contains all the reactants and products of the chemical reaction. For best results, the reaction is written in the ionic form.


 I2 + S2O32- + OH- → SO32- + I- + H2O 


The oxidation numbers of each atom that appears in the reaction are determined. The oxidation number (or the degree of oxidation) is a measure of the degree of oxidation in a molecule.

I⁰₂ + S⁺²₂O^-2₃^2- + O^-2H^+1- → S⁺⁴O^-2₃^2- + I^-1- + H⁺¹₂O^-2

 

The transfer of electrons is written. The number of atoms that have been oxidized, that is reduced on both sides of the equation, must be the same.

O: S⁺²₂O^-2₃^2- → 2S⁺⁴O^-2₃^2- + 4e^-    (S)

R:     I⁰₂ + 2e^- → 2I^-1-                 (I)

 


Sulfur (S) has oxidized because its oxidation number has increased, it acts as a reducing agent. On the other hand, iodine (I) decreases its oxidation number (it is reduced), it is the oxidizing agent.



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