Question

подберите коэффициенты в схемах следующих окислительно-восстановительных реакций методом электронного баланса:Na2S+KMnO4+H2O=S+MnO2+NaOH+KOH

Answer 1

3Na2S+2KMnO4+4H2O=3S+2MnO2+6NaOH+2KOH

S(-2) - 2e = S(0)          |3

Mn(+7) + 3e = Mn(+4)|2

Answer 2

Oxidation-Reduction Reaction and Electron Balance

To balance the given oxidation-reduction reaction using the electron balance method, we need to determine the coefficients for each compound in the reaction.

The given reaction is: Na2S + KMnO4 + H2O = S + MnO2 + NaOH + KOH

To balance the reaction, we can follow these steps:

1. Identify the oxidation states of each element in the reaction. 2. Write down the half-reactions for oxidation and reduction. 3. Balance the atoms in each half-reaction. 4. Balance the charges in each half-reaction. 5. Multiply the half-reactions by appropriate coefficients to equalize the number of electrons transferred. 6. Add the balanced half-reactions together and simplify if necessary.

Let's proceed with balancing the given reaction using the electron balance method.

Balancing the Reaction

The oxidation state of Sulfur (S) in Na2S is -2, and the oxidation state of Manganese (Mn) in KMnO4 is +7.

The half-reactions for the oxidation and reduction are: Oxidation: S → S^6+ Reduction: Mn^7+ → Mn^4+

Balancing the atoms and charges in each half-reaction, we get: Oxidation: 2e^- + S → S^6+ Reduction: 3e^- + Mn^7+ → Mn^4+

To equalize the number of electrons transferred, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2: Oxidation: 6e^- + 3S → 3S^6+ Reduction: 6e^- + 2Mn^7+ → 2Mn^4+

Adding the balanced half-reactions together, we get: 6e^- + 3S + 2Mn^7+ → 3S^6+ + 2Mn^4+

This balanced equation represents the oxidation-reduction reaction using the electron balance method.

Conclusion

The balanced coefficients for the given oxidation-reduction reaction using the electron balance method are: 6Na2S + 2KMnO4 + 8H2O = 3S + 2MnO2 + 4NaOH + 2KOH