Question

K2cr2o7 + hclo4 + hi = cr(clo4)3 + kclo4 + i2 + h2o Расставить коэффициенты методом электронно-ионного баланса, определить эквивалент и молярную массу у окислителя и восстановителя.

Answer 1

K2Cr2O7 + 8HClO4 + 6HI = 2Cr(ClO4)3 + 2KClO4 + 3I2 + 7H2O:

(Cr2O7)2- +6e- + 14H+ = 2Cr3+ + 7H2O     | 6 | 6 | 1 | - окис-ь, ф.э. = 1/6

2HI -2e- = I2 + 2H+                                    | 2 |    | 3 | - восс-ь, ф.э. = 1/1;

(Cr2O7)2- + 6HI + 8H+ = 2Cr3+ + 3I2 + 7H2O;


М экв. окис-я = 294*(1/6) = 49 г/моль;

М экв. восс-я = 128*(1/1) = 128 г/моль.

Answer 2

Balancing the Chemical Equation

The given chemical equation is: K2Cr2O7 + HClO4 + HI = Cr(ClO4)3 + KClO4 + I2 + H2O

To balance this equation using the method of electron-ion balance, we need to follow these steps:

1. Write down the unbalanced equation. 2. Identify the oxidation states of each element. 3. Balance the atoms that change oxidation state. 4. Balance the remaining atoms. 5. Balance the charges by adding electrons. 6. Multiply each half-reaction by an integer to make the number of electrons equal in both half-reactions. 7. Add the half-reactions together and simplify if necessary.

Let's proceed with balancing the given chemical equation.

Balancing the Chemical Equation Step by Step

1. Write down the unbalanced equation: K2Cr2O7 + 6HClO4 + 6HI = 2Cr(ClO4)3 + 4KClO4 + 3I2 + 7H2O

2. Identify the oxidation states of each element: - In K2Cr2O7, the oxidation state of Cr is +6, and the oxidation state of O is -2. - In HClO4, the oxidation state of Cl is +7, and the oxidation state of O is -2. - In HI, the oxidation state of I is -1. - In Cr(ClO4)3, the oxidation state of Cr is +6, and the oxidation state of O is -2. - In KClO4, the oxidation state of Cl is +7, and the oxidation state of O is -2. - In I2, the oxidation state of I is 0. - In H2O, the oxidation state of O is -2.

3. Balance the atoms that change oxidation state: - The atoms that change oxidation state are Cr and I.

4. Balance the remaining atoms: - The atoms are already balanced.

5. Balance the charges by adding electrons: - The charges are already balanced.

6. Multiply each half-reaction by an integer to make the number of electrons equal in both half-reactions: - The number of electrons is already equal in both half-reactions.

7. Add the half-reactions together and simplify if necessary: - The balanced chemical equation is: K2Cr2O7 + 6HClO4 + 6HI = 2Cr(ClO4)3 + 4KClO4 + 3I2 + 7H2O

Determining the Equivalent and Molar Mass

To determine the equivalent and molar mass of the oxidizer and reducer, we need to first identify the oxidizer and reducer in the balanced equation. Then, we can calculate their equivalent and molar masses.

In the given balanced equation, K2Cr2O7 is the oxidizer, and HI is the reducer.

The equivalent mass of an oxidizer is calculated using the formula: Equivalent mass = Molar mass / n where n is the change in oxidation state per mole of the oxidizer.

The molar mass of K2Cr2O7 is approximately 294.18 g/mol.

The equivalent mass of K2Cr2O7 can be calculated based on the change in oxidation state of chromium.

The molar mass of HI is approximately 127.91 g/mol.

The equivalent mass of HI can be calculated based on the change in oxidation state of iodine.

Let's calculate the equivalent and molar masses for K2Cr2O7 and HI.

Calculating the Equivalent and Molar Mass

The equivalent and molar masses for K2Cr2O7 and HI can be calculated as follows:

- Equivalent mass of K2Cr2O7: - The change in oxidation state of chromium in K2Cr2O7 is 6. - Therefore, the equivalent mass of K2Cr2O7 is approximately 49.03 g/equiv.

- Molar mass of K2Cr2O7: - The molar mass of K2Cr2O7 is approximately 294.18 g/mol.

- Equivalent mass of HI: - The change in oxidation state of iodine in HI is 1. - Therefore, the equivalent mass of HI is approximately 127.91 g/equiv.

- Molar mass of HI: - The molar mass of HI is approximately 127.91 g/mol.

In summary, the equivalent mass of K2Cr2O7 is approximately 49.03 g/equiv, and its molar mass is approximately 294.18