Question

Подберите коэффициенты для окислительно восстановительной реакции,используя метод электронного баланса Sb+HNO3=HSbO3+NO+H2O

Answer 1

3Sb + 5HNO3 = 3HSbO3 + 5NO + H2O, баланс:

Sb -5e- = Sb+5      | 5 | 15 | 3 |

N+5 +3e- = N+2     | 3 |     | 5 |.

Answer 2

Balancing the Chemical Equation Sb + HNO3 = HSbO3 + NO + H2O

To balance the chemical equation Sb + HNO3 = HSbO3 + NO + H2O using the method of electron balance, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

The unbalanced equation is: Sb + HNO3 = HSbO3 + NO + H2O

First, let's identify the oxidation states of the elements involved in the reaction: - Antimony (Sb) can have oxidation states of -3, 0, +3, or +5. - Nitrogen (N) in HNO3 has an oxidation state of +5. - Oxygen (O) in HNO3 has an oxidation state of -2. - Hydrogen (H) in HNO3 has an oxidation state of +1. - Hydrogen (H) in H2O has an oxidation state of +1. - Oxygen (O) in H2O has an oxidation state of -2.

Now, let's balance the equation using the method of electron balance.

Balancing the Half-Reactions

1. Oxidation Half-Reaction: Sb → HSbO3 The oxidation state of Sb changes from 0 to +5.

The balanced half-reaction for the oxidation of Sb is: Sb → HSbO3 + 3e- 2. Reduction Half-Reaction: HNO3 → NO The oxidation state of N changes from +5 to +2.

The balanced half-reaction for the reduction of HNO3 is: 2HNO3 + 6e- → 2NO + 4H2O

Combining the Half-Reactions

To combine the half-reactions, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

Multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance the electrons, we get: 2Sb + 6HNO3 → 2HSbO3 + 3NO + 6H2O

Therefore, the balanced chemical equation using the method of electron balance is: 2Sb + 6HNO3 → 2HSbO3 + 3NO + 6H2O