Question

Расставьте коэффициенты в схемах окислительно - восстановительных реакций, использую метод электронного баланса, укажите окислитель и восстановитель. 1) KNO2 + KI + H2SO4 ---> I2 + NO + K2SO4 + H2O 2)KMnO4 + KI + H2SO4 ------>MnSO4 + I2 + K2SO4 + H2O 3) CuS + HNO3 ----> CuSO4 + NO2 + ....

Answer 1

Balancing Redox Reactions using the Electron Balance Method

To balance the given redox reactions using the electron balance method, we need to assign oxidation numbers to each element in the reactants and products. The change in oxidation numbers will help us determine the number of electrons transferred during the reaction. We can then balance the reaction by ensuring that the total number of electrons lost in the oxidation half-reaction is equal to the total number of electrons gained in the reduction half-reaction.

Let's go through each reaction and balance them step by step.

1) KNO2 + KI + H2SO4 → I2 + NO + K2SO4 + H2O

To balance this reaction, we need to assign oxidation numbers to each element:

- In KNO2, the oxidation number of K is +1, N is +3, and O is -2. - In KI, the oxidation number of K is +1 and I is -1. - In H2SO4, the oxidation number of H is +1, S is +6, and O is -2. - In I2, the oxidation number of I is 0. - In NO, the oxidation number of N is +2 and O is -2. - In K2SO4, the oxidation number of K is +1, S is +6, and O is -2. - In H2O, the oxidation number of H is +1 and O is -2.

Now, let's balance the reaction:

1. Balance the atoms other than hydrogen and oxygen: - The number of K atoms is already balanced. - The number of N atoms is already balanced. - The number of I atoms is already balanced. - The number of S atoms is already balanced.

2. Balance the oxygen atoms by adding water molecules (H2O): - There are 4 oxygen atoms on the left side (2 in KNO2 and 2 in H2SO4). - Add 2 water molecules (H2O) to the right side to balance the oxygen atoms.

3. Balance the hydrogen atoms by adding hydrogen ions (H+): - There are 4 hydrogen atoms on the left side (4 in H2SO4). - Add 4 hydrogen ions (H+) to the left side to balance the hydrogen atoms.

4. Balance the charge by adding electrons (e-): - The total charge on the left side is +2 (2K+). - The total charge on the right side is 0 (I2 has no charge). - Add 2 electrons (2e-) to the right side to balance the charge.

The balanced equation is:

2KNO2 + 2KI + 2H2SO4 → I2 + NO + K2SO4 + 2H2O

In this reaction, KNO2 is the oxidizing agent (it gets reduced) and KI is the reducing agent (it gets oxidized).

2) KMnO4 + KI + H2SO4 → MnSO4 + I2 + K2SO4 + H2O

To balance this reaction, we need to assign oxidation numbers to each element:

- In KMnO4, the oxidation number of K is +1, Mn is +7, and O is -2. - In KI, the oxidation number of K is +1 and I is -1. - In H2SO4, the oxidation number of H is +1, S is +6, and O is -2. - In MnSO4, the oxidation number of Mn is +2, S is +6, and O is -2. - In I2, the oxidation number of I is 0. - In K2SO4, the oxidation number of K is +1, S is +6, and O is -2. - In H2O, the oxidation number of H is +1 and O is -2.

Now, let's balance the reaction:

1. Balance the atoms other than hydrogen and oxygen: - The number of K atoms is already balanced. - The number of Mn atoms is already balanced. - The number of I atoms is already balanced. - The number of S atoms is already balanced.

2. Balance the oxygen atoms by adding water molecules (H2O): - There are 4 oxygen atoms on the left side (4 in KMnO4). - Add 4 water molecules (H2O) to the right side to balance the oxygen atoms.

3. Balance the hydrogen atoms by adding hydrogen ions (H+): - There are 8 hydrogen atoms on the left side (8 in H2SO4). - Add 8 hydrogen ions (H+) to the left side to balance the hydrogen atoms.

4. Balance the charge by adding electrons (e-): - The total charge on the left side is +2 (2K+). - The total charge on the right side is 0 (I2 has no charge). - Add 2 electrons (2e-) to the right side to balance the charge.

The balanced equation is:

2KMnO4 + 16KI + 8H2SO4 → 2MnSO4 + 8I2 + 8K2SO4 + 8H2O

In this reaction, KMnO4 is the oxidizing agent (it gets reduced) and KI is the reducing agent (it gets oxidized).

3) CuS + HNO3 → CuSO4 + NO2 + ...

To balance this reaction, we need to assign oxidation numbers to each element:

- In CuS, the oxidation number of Cu is +2 and S is -2. - In HNO3, the oxidation number of H is +1, N is +5, and O is -2. - In CuSO4, the oxidation number of Cu is +2, S is +6, and O is -2. - In NO2, the oxidation number of N is +4 and O is -2.

Now, let's balance the reaction:

1. Balance the atoms other than hydrogen and oxygen: - The number of Cu atoms is already balanced. - The number of S atoms is already balanced. - The number of N atoms is already balanced.

2. Balance the oxygen atoms by adding water molecules (H2O): - There are 3 oxygen atoms on the left side (3 in HNO3). - Add 3 water molecules (H2O) to the right side to balance the oxygen atoms.

3. Balance the hydrogen atoms by adding hydrogen ions (H+): - There are 3 hydrogen atoms on the left side (3 in HNO3). - Add 6 hydrogen ions (2H+) to the left side to balance the hydrogen atoms.

4. Balance the charge by adding electrons (e-): - The total charge on the left side is 0 (CuS has no charge). - The total charge on the right side is +2 (CuSO4 has a +2 charge). - Add 2 electrons (2e-) to the left side to balance the charge.

The balanced equation is:

CuS + 6HNO3 → CuSO4 + NO2 + 3H2O

In this reaction, CuS is the oxidizing agent (it gets oxidized) and HNO3 is the reducing agent (it gets reduced).

Please note that the ellipsis (...) in the reaction indicates that there may be additional products or reactants that are not provided in the given information. Without further information, it is not possible to determine the complete balanced equation.