Question

Даны схемы окислительно-восстановительных реакций a)Cus+HNO3(разб.)-Cu(No3)2+S+NO+H2O Б)K2So3+H2So4+kmno4-k2so4+MNSO4+.... Помогите пожалуйста!

Answer 1

Извиняюсь за грязь если что

Answer 2

Oxidation-Reduction Reactions

Let's break down the given oxidation-reduction reactions:

a) Cus + HNO3 (diluted) → Cu(NO3)2 + S + NO + H2O b) K2SO3 + H2SO4 + KMnO4 → K2SO4 + MnSO4 + ...

Reaction (a)

The reaction involves the oxidation of copper(I) sulfide by nitric acid. The products are copper(II) nitrate, sulfur, nitrogen monoxide, and water.

Reaction (b)

The given reaction involves the oxidation of sulfite ions to sulfate ions by potassium permanganate in acidic medium.

Explanation

1. Reaction (a): - The oxidation state of copper changes from +1 in Cus to +2 in Cu(NO3)2, indicating its oxidation. - The oxidation state of sulfur changes from -2 in Cus to 0 in S, indicating its reduction. - Nitric acid acts as an oxidizing agent, causing the oxidation of Cus to Cu(NO3)2 and itself being reduced to nitrogen monoxide.

2. Reaction (b): - The oxidation state of sulfur changes from +4 in K2SO3 to +6 in K2SO4, indicating its oxidation. - The oxidation state of manganese changes from +7 in KMnO4 to +2 in MnSO4, indicating its reduction.

Conclusion

These reactions demonstrate the transfer of electrons from one reactant to another, resulting in changes in oxidation states and the formation of new compounds.

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