Question

Для двух реакций составьте схемы электронного баланса и расставьте коэффициенты, определите окислитель, восстановитель, а также укажите процессы восстановления и окисления: H2S + HClO4 → H2SO4 + HCl; HgS + HNO3 + HCl → HgCl2 + S + NO + H2O;

Answer 1

H2S + HClO4 → H2SO4 + HCl S2- -8е=S+6|1 в-ль Cl7+ +8е =Cl-|1 о-ль 3HgS + 2HNO3 + 2HCl → HgCl2 + 3S + 2NO + 2H2O S2- -2e =S|3 в-ль N5+ +3e=N2+|2 о-ль

Answer 2

Reaction 1: H2S + HClO4 → H2SO4 + HCl

To balance the electron transfer in this reaction, we need to assign oxidation numbers to each element and then balance the atoms and charges on both sides of the equation.

1. Assign oxidation numbers: - Hydrogen (H) has an oxidation number of +1. - Sulfur (S) in H2S has an oxidation number of -2. - Chlorine (Cl) in HClO4 has an oxidation number of +7. - Oxygen (O) in HClO4 has an oxidation number of -2. - Sulfur (S) in H2SO4 has an oxidation number of +6. - Hydrogen (H) in H2SO4 has an oxidation number of +1. - Oxygen (O) in H2SO4 has an oxidation number of -2. - Chlorine (Cl) in HCl has an oxidation number of -1.

2. Balance the atoms and charges: - The sulfur (S) atoms are already balanced. - The hydrogen (H) atoms are already balanced. - The chlorine (Cl) atoms are already balanced. - The oxygen (O) atoms are not balanced. There are 4 O atoms on the left side and 6 O atoms on the right side.

3. Balance the oxygen atoms: - To balance the oxygen atoms, we need to add 2 HClO4 molecules on the left side of the equation.

The balanced equation is: 2 H2S + 4 HClO4 → H2SO4 + 4 HCl

- Oxidation: In this reaction, sulfur (S) in H2S is oxidized from an oxidation state of -2 to +6 in H2SO4. - Reduction: In this reaction, chlorine (Cl) in HClO4 is reduced from an oxidation state of +7 to -1 in HCl.

Reaction 2: HgS + HNO3 + HCl → HgCl2 + S + NO + H2O

To balance the electron transfer in this reaction, we need to assign oxidation numbers to each element and then balance the atoms and charges on both sides of the equation.

1. Assign oxidation numbers: - Hydrogen (H) has an oxidation number of +1. - Mercury (Hg) in HgS has an oxidation number of +2. - Sulfur (S) in HgS has an oxidation number of -2. - Nitrogen (N) in HNO3 has an oxidation number of +5. - Oxygen (O) in HNO3 has an oxidation number of -2. - Chlorine (Cl) in HCl has an oxidation number of -1. - Mercury (Hg) in HgCl2 has an oxidation number of +2. - Chlorine (Cl) in HgCl2 has an oxidation number of -1. - Sulfur (S) in S has an oxidation number of 0. - Nitrogen (N) in NO has an oxidation number of +2. - Hydrogen (H) in H2O has an oxidation number of +1. - Oxygen (O) in H2O has an oxidation number of -2.

2. Balance the atoms and charges: - The mercury (Hg) atoms are already balanced. - The sulfur (S) atoms are already balanced. - The nitrogen (N) atoms are not balanced. There is 1 N atom on the left side and 1 N atom on the right side. - The hydrogen (H) atoms are not balanced. There are 3 H atoms on the left side and 2 H atoms on the right side. - The chlorine (Cl) atoms are not balanced. There are 1 Cl atom on the left side and 2 Cl atoms on the right side. - The oxygen (O) atoms are not balanced. There are 6 O atoms on the left side and 3 O atoms on the right side.

3. Balance the atoms: - To balance the nitrogen (N) atoms, we need to add 2 HNO3 molecules on the left side of the equation. - To balance the hydrogen (H) atoms, we need to add 2 H2O molecules on the right side of the equation. - To balance the chlorine (Cl) atoms, we need to add 2 HCl molecules on the right side of the equation. - To balance the oxygen (O) atoms, we need to add 3 H2O molecules on the right side of the equation.

The balanced equation is: HgS + 2 HNO3 + 2 HCl → HgCl2 + S + NO + 3 H2O

- Oxidation: In this reaction, sulfur (S) in HgS is oxidized from an oxidation state of -2 to 0 in S. - Reduction: In this reaction, nitrogen (N) in HNO3 is reduced from an oxidation state of +5 to +2 in NO.