Question

Помогите решить используя метод электронного баланса, расставьте коэффициенты в уравнении реакции , определите окислитель и восстановитель HNO2+HI->NO+I2+H2O

Answer 1

2HNO2 + 2HI -> 2NO + I2 + 2H2O

N(+3) +1e =N(+2)   2   ок-ль, в-ие

2I(-) -2e = I2(0)      1     в-ль, ок-ие

Answer 2

Balancing the Equation: HNO2 + HI -> NO + I2 + H2O

To balance the equation using the method of electronic balance, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's go step by step to balance the given equation:

1. Start by counting the number of atoms for each element on both sides of the equation: - Hydrogen (H): 2 on the left, 2 on the right - Nitrogen (N): 1 on the left, 1 on the right - Oxygen (O): 2 on the left, 1 on the right - Iodine (I): 1 on the left, 2 on the right

2. Balance the atoms that appear in the least number first. In this case, we start with iodine (I): - Add a coefficient of 2 in front of HI on the left side: 2HI - This balances the iodine atoms: 2 on both sides

The equation now becomes: HNO2 + 2HI -> NO + I2 + H2O

3. Next, balance the hydrogen atoms: - Add a coefficient of 2 in front of H2O on the right side: HNO2 + 2HI -> NO + I2 + 2H2O - This balances the hydrogen atoms: 4 on both sides

4. Now, balance the oxygen atoms: - Add a coefficient of 2 in front of HNO2 on the left side: 2HNO2 + 2HI -> NO + I2 + 2H2O - This balances the oxygen atoms: 4 on both sides

The final balanced equation is: 2HNO2 + 2HI -> NO + I2 + 2H2O

Identifying the Oxidizing and Reducing Agents

To determine the oxidizing and reducing agents in the reaction, we need to look at the changes in oxidation states of the elements involved.

In the given reaction: 2HNO2 + 2HI -> NO + I2 + 2H2O

- Nitrogen (N) in HNO2 has an oxidation state of +3, and in NO, it has an oxidation state of +2. Therefore, nitrogen is reduced from +3 to +2, indicating that HNO2 is the reducing agent. - Iodine (I) in HI has an oxidation state of -1, and in I2, it has an oxidation state of 0. Therefore, iodine is oxidized from -1 to 0, indicating that HI is the oxidizing agent.

Summary:

The balanced equation for the reaction is: 2HNO2 + 2HI -> NO + I2 + 2H2O.

In this reaction, HNO2 acts as the reducing agent as it undergoes a reduction by decreasing the oxidation state of nitrogen from +3 to +2. On the other hand, HI acts as the oxidizing agent as it undergoes oxidation by increasing the oxidation state of iodine from -1 to 0.

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