Question

Помогите пожалуйста с химией. Используя метод электронного баланса, расставьте коэффициенты в уравнении реакции, схема которой: H3PO4 + HI = H3PO3 + I2 + H2O Определите окислитель и восстановитель.

Answer 1

H₃P⁵O₄ + 2HI⁻¹ = H₃P³O₃ + I₂⁰ + H₂O

P⁵ +2e= P³   | 2  |        |   1    окислитель (восстановляется)

                    |      |   2  |

2I⁻¹ -1e= I₂⁰    | 1   |        |   2    восстановитель (окисляется)

Answer 2

Balancing the Chemical Equation

To balance the chemical 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.

The given chemical equation is: H3PO4 + HI = H3PO3 + I2 + H2O

To balance this equation, we can follow these steps:

Step 1: Count the number of atoms of each element on both sides of the equation.

| Element | Left Side | Right Side | |---------|-----------|------------| | H | 3 | 5 | | P | 1 | 1 | | O | 4 | 3 | | I | 1 | 2 |

Step 2: Start by balancing the atoms that appear in the fewest number of molecules. In this case, we can start with phosphorus (P).

P: The left side has 1 P atom, and the right side also has 1 P atom. It is already balanced.

Step 3: Next, balance the hydrogen (H) atoms.

H: The left side has 3 H atoms, and the right side has 5 H atoms. To balance the hydrogen atoms, we can add a coefficient of 2 in front of H3PO4 on the left side.

The equation becomes: 2H3PO4 + HI = H3PO3 + I2 + H2O

Step 4: Now, balance the oxygen (O) atoms.

O: The left side has 8 O atoms (4 from H3PO4 and 4 from H2O), and the right side has 3 O atoms. To balance the oxygen atoms, we can add a coefficient of 5/2 in front of H2O on the right side.

The equation becomes: 2H3PO4 + HI = H3PO3 + I2 + (5/2)H2O

Step 5: Finally, balance the iodine (I) atoms.

I: The left side has 1 I atom, and the right side has 2 I atoms. To balance the iodine atoms, we can add a coefficient of 2 in front of HI on the left side.

The balanced equation is: 2H3PO4 + 2HI = H3PO3 + I2 + (5/2)H2O

Identifying the Oxidizing and Reducing Agents

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

In the given reaction: H3PO4 + 2HI = H3PO3 + I2 + (5/2)H2O

The oxidation state of iodine (I) changes from -1 in HI to 0 in I2. This indicates that iodine is being reduced and is the reducing agent.

The oxidation state of phosphorus (P) changes from +5 in H3PO4 to +3 in H3PO3. This indicates that phosphorus is being oxidized and is the oxidizing agent.

Therefore, in the given reaction, iodine (I) is the reducing agent, and phosphorus (P) is the oxidizing agent.

Note: The coefficients in the balanced equation may not be whole numbers due to the need to balance the equation.