Question

Расставьте степени окисления элементов и уравняйте реакцию методом электронного баланса:K2S+H2O2+H2SO4 - S+K2SO4+H2OFeSO4+HNO3 - Fe(NO3)3+NO2+H2SO4+H2O

Answer 1

Fe (2+) – 1е → Fe (3+)
NO3 (–) + 2Н(+) + 1 е → NO2 + Н2О
FeSO4 + 4HNO3 → Fe(NO3)3 + NO2↑ + H2SO4 + H2O

Answer 2

Balancing the equation K2S + H2O2 + H2SO4 -> S + K2SO4 + H2O using the method of electron balance:

To balance the given equation, we need to assign oxidation numbers to each element and then balance the equation by adjusting coefficients.

1. Assigning oxidation numbers: - The oxidation number of potassium (K) is +1. - The oxidation number of sulfur (S) is -2 in K2S and 0 in S. - The oxidation number of hydrogen (H) is +1 in H2O2 and 0 in H2O. - The oxidation number of oxygen (O) is -2 in K2SO4, -1 in H2O2, and -2 in H2SO4.

2. Balancing the equation: - Start by balancing the atoms that are undergoing oxidation or reduction. - In this case, sulfur (S) is being reduced from an oxidation state of +6 in H2SO4 to 0 in S. - To balance the sulfur atoms, we need to add 3 H2SO4 on the left-hand side.

The balanced equation is: 3 K2S + 8 H2O2 + 8 H2SO4 -> 3 S + 4 K2SO4 + 16 H2O

Balancing the equation FeSO4 + HNO3 -> Fe(NO3)3 + NO2 + H2SO4 + H2O using the method of electron balance:

1. Assigning oxidation numbers: - The oxidation number of iron (Fe) is +2 in FeSO4 and +3 in Fe(NO3)3. - The oxidation number of nitrogen (N) is +5 in HNO3 and +4 in NO2. - The oxidation number of hydrogen (H) is +1 in HNO3 and 0 in H2O. - The oxidation number of sulfur (S) is +6 in H2SO4 and -2 in FeSO4.

2. Balancing the equation: - Start by balancing the atoms that are undergoing oxidation or reduction. - In this case, iron (Fe) is being oxidized from an oxidation state of +2 in FeSO4 to +3 in Fe(NO3)3. - To balance the iron atoms, we need to add 2 Fe(NO3)3 on the right-hand side.

The balanced equation is: 3 FeSO4 + 2 HNO3 -> 3 Fe(NO3)3 + NO2 + H2SO4 + H2O