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The oxidation state of the most electronegative element in the products of the reaction between BaO2 and H2SO4?
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The oxidation state of the most electronegative element in the product...
Reaction Overview
The reaction between barium peroxide (BaO2) and sulfuric acid (H2SO4) can be represented as follows:
\[ \text{BaO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + \text{H}_2\text{O}_2 \]
In this reaction, barium sulfate (BaSO4) and hydrogen peroxide (H2O2) are formed.
Identifying the Most Electronegative Element
In the products, the most electronegative element is oxygen (O).
Oxidation States of Elements
- In BaSO4:
- Barium (Ba) has an oxidation state of +2.
- Sulfate (SO4) has a sulfur oxidation state of +6, and each oxygen is -2.
- In H2O2:
- Hydrogen (H) has an oxidation state of +1.
- The two oxygen atoms in hydrogen peroxide are in the -1 oxidation state each.
Oxidation State of Oxygen
- In BaSO4:
- The oxidation state of oxygen is -2.
- In H2O2:
- The oxidation state of oxygen is -1.
Hence, in the products of the reaction, the most electronegative element, oxygen, has two different oxidation states based on its compound:
Conclusion
- In BaSO4, oxygen has an oxidation state of -2.
- In H2O2, oxygen has an oxidation state of -1.
The predominant oxidation state in this reaction corresponds to its presence in the more stable product, BaSO4, where the oxidation state of oxygen is -2.
Community Answer
The oxidation state of the most electronegative element in the product...
In BaO2 is -1 and in H2SO4 is -2
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The oxidation state of the most electronegative element in the products of the reaction between BaO2 and H2SO4?
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