Question

Помогите пжл срочно решить. Реакции протекают по схемам: KMnO4+Cd+H2SO4-->CdSO4+MnSO4+K2SO4+H2O Cu+HNO3-->Cu(NO3)2+NO+H2O Составьте электронные уравнения.Расставьте коэффициенты в уравнениях реакций.Для каждой реакции укажите,какое вещество является окислителем,какое восстановителем,какое вещество окисляется,какое восстанавливается.

Answer 1

1) Mn(+7) + 5e=Mn(+2)    5   2
    Cd(0) -2e=Cd(+2)         2   5
2KMnO4+5Cd+8H2SO4 = 5CdSO4=2MnSO4 +K2SO4 +8H2O

2) Cu(0) -2e=Cu(+2)     2   3
    N(+5) +3e=N(+2)      3    2
3Cu+8HNO3 = 3Cu(NO3)2 + 2NO +4H2O

Answer 2

Electron equations and coefficients for the given reactions:

1. KMnO4 + Cd + H2SO4 → CdSO4 + MnSO4 + K2SO4 + H2O

- Oxidation half-reaction: Cd → Cd2+ + 2e- - Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Balancing the number of electrons in both half-reactions, we get:

- Oxidation half-reaction: 2Cd → 2Cd2+ + 4e- - Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Balancing the number of atoms on both sides, we get:

- Oxidation half-reaction: 2Cd + MnO4- + 8H+ → 2Cd2+ + Mn2+ + 4H2O - Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

To balance the charges, we add 5 electrons to the oxidation half-reaction:

- Oxidation half-reaction: 2Cd + MnO4- + 8H+ + 5e- → 2Cd2+ + Mn2+ + 4H2O - Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Finally, we balance the number of atoms on both sides:

- Oxidation half-reaction: 2Cd + MnO4- + 8H+ + 5e- → 2Cd2+ + Mn2+ + 4H2O - Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

2. Cu + HNO3 → Cu(NO3)2 + NO + H2O

- Oxidation half-reaction: Cu → Cu2+ + 2e- - Reduction half-reaction: HNO3 + 3e- + 2H+ → NO + 2H2O

Balancing the number of electrons in both half-reactions, we get:

- Oxidation half-reaction: 3Cu → 3Cu2+ + 6e- - Reduction half-reaction: HNO3 + 3e- + 2H+ → NO + 2H2O

Balancing the number of atoms on both sides, we get:

- Oxidation half-reaction: 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O - Reduction half-reaction: HNO3 + 3e- + 2H+ → NO + 2H2O

To balance the charges, we add 3 electrons to the oxidation half-reaction:

- Oxidation half-reaction: 3Cu + 8HNO3 + 6e- → 3Cu(NO3)2 + 2NO + 4H2O - Reduction half-reaction: HNO3 + 3e- + 2H+ → NO + 2H2O

Finally, we balance the number of atoms on both sides:

- Oxidation half-reaction: 3Cu + 8HNO3 + 6e- → 3Cu(NO3)2 + 2NO + 4H2O - Reduction half-reaction: 8HNO3 + 6e- + 6H+ → 2NO + 6H2O

Identification of oxidizing and reducing agents:

1. In the reaction KMnO4 + Cd + H2SO4 → CdSO4 + MnSO4 + K2SO4 + H2O: - Oxidizing agent: KMnO4 - Reducing agent: Cd - Substance being oxidized: Cd - Substance being reduced: MnO4-

2. In the reaction Cu + HNO3 → Cu(NO3)2 + NO + H2O: - Oxidizing agent: HNO3 - Reducing agent: Cu - Substance being oxidized: Cu - Substance being reduced: HNO3

Please let me know if there is anything else I can help you with!