Question

Уравняйте реакции методом электронно-ионного баланса, укажите факторы эквивалентности окислителя и восстановителя : A) KNO2+KMnO4+H2SO4=KNO3+MnSO4+K2SO4+H2O Б) HIO3+H2O2=I2+O2+H2O Помогите пожалуйста ​

Answer 1

Ответ:

Mn(+7) +5e = Mn(+2) *2  - восстановитель

N(+3)   - 2e = N(+5)   *5 - оксилитель

2KMnO4+3H2SO4+5KNO2=2MnSO4+K2SO4+5KNO3+3H2O

2O(-1) -2e= O2

2I(+5) +10e = I2

Первое домножаем на 5

5H2O2+2HIO3=I2+5O2+6H2O

Объяснение:

Answer 2

Balancing Chemical Equations using the Electron-Ion Method

To balance the given chemical equations using the electron-ion method, we need to ensure that the number of atoms of each element is the same on both sides of the equation. We will start by balancing the atoms that appear in the fewest compounds and then proceed to balance the more complex compounds.

Equation A: KNO2 + KMnO4 + H2SO4 = KNO3 + MnSO4 + K2SO4 + H2O

To balance this equation, we can follow these steps:

Step 1: Balance the atoms other than hydrogen and oxygen. - Start by balancing the potassium (K) atoms. There are two potassium atoms on the right side, so we need to add a coefficient of 2 in front of KNO2 and KNO3. - The equation now becomes: 2KNO2 + KMnO4 + H2SO4 = 2KNO3 + MnSO4 + K2SO4 + H2O.

Step 2: Balance the oxygen atoms. - Count the number of oxygen atoms on each side of the equation. There are 4 oxygen atoms on the left side and 10 oxygen atoms on the right side. - To balance the oxygen atoms, we can add water (H2O) molecules to the left side of the equation. By adding 6 H2O molecules, we will have a total of 12 oxygen atoms on both sides. - The equation now becomes: 2KNO2 + KMnO4 + H2SO4 = 2KNO3 + MnSO4 + K2SO4 + 6H2O.

Step 3: Balance the hydrogen atoms. - Count the number of hydrogen atoms on each side of the equation. There are 4 hydrogen atoms on the left side and 12 hydrogen atoms on the right side. - To balance the hydrogen atoms, we can add 8 H+ ions (protons) to the left side of the equation. By doing so, we will have a total of 12 hydrogen atoms on both sides. - The equation now becomes: 2KNO2 + KMnO4 + H2SO4 + 8H+ = 2KNO3 + MnSO4 + K2SO4 + 6H2O.

Step 4: Balance the charge. - Count the total charge on each side of the equation. On the left side, we have a charge of +8 (from the H+ ions). On the right side, we have a charge of +2 (from the K2SO4 compound). - To balance the charge, we can add 6 electrons (e-) to the left side of the equation. By doing so, we will have a total charge of +2 on both sides. - The equation now becomes: 2KNO2 + KMnO4 + H2SO4 + 8H+ + 6e- = 2KNO3 + MnSO4 + K2SO4 + 6H2O.

Step 5: Balance the remaining atoms. - Count the number of nitrogen (N) and sulfur (S) atoms on each side of the equation. There are 2 nitrogen atoms and 1 sulfur atom on both sides, so they are already balanced. - The final balanced equation is: 2KNO2 + KMnO4 + H2SO4 + 8H+ + 6e- = 2KNO3 + MnSO4 + K2SO4 + 6H2O.

Equation B: HIO3 + H2O2 = I2 + O2 + H2O

To balance this equation, we can follow these steps:

Step 1: Balance the atoms other than hydrogen and oxygen. - Start by balancing the iodine (I) atoms. There is one iodine atom on the left side, so we need to add a coefficient of 2 in front of HIO3 and I2. - The equation now becomes: 2HIO3 + H2O2 = 2I2 + O2 + H2O.

Step 2: Balance the oxygen atoms. - Count the number of oxygen atoms on each side of the equation. There are 8 oxygen atoms on the left side and 3 oxygen atoms on the right side. - To balance the oxygen atoms, we can add water (H2O) molecules to the right side of the equation. By adding 5 H2O molecules, we will have a total of 8 oxygen atoms on both sides. - The equation now becomes: 2HIO3 + H2O2 = 2I2 + O2 + 5H2O.

Step 3: Balance the hydrogen atoms. - Count the number of hydrogen atoms on each side of the equation. There are 6 hydrogen atoms on the left side and 10 hydrogen atoms on the right side. - To balance the hydrogen atoms, we can add 4 H+ ions (protons) to the left side of the equation. By doing so, we will have a total of 10 hydrogen atoms on both sides. - The equation now becomes: 2HIO3 + H2O2 + 4H+ = 2I2 + O2 + 5H2O.

Step 4: Balance the charge. - Count the total charge on each side of the equation. On the left side, we have a charge of +4 (from the H+ ions). On the right side, we have a charge of 0. - To balance the charge, we can add 4 electrons (e-) to the left side of the equation. By doing so, we will have a total charge of 0 on both sides. - The equation now becomes: 2HIO3 + H2O2 + 4H+ + 4e- = 2I2 + O2 + 5H2O.

Step 5: Balance the remaining atoms. - Count the number of iodine (I) atoms on each side of the equation. There are 2 iodine atoms on the right side, so we need to add a coefficient of 2 in front of I2. - The final balanced equation is: 2HIO3 + H2O2 + 4H+ + 4e- = 2I2 + O2 + 5H2O.

Factors of Equivalence for the Oxidizing and Reducing Agents

In Equation A, the oxidizing agent is KMnO4, which is reduced to MnSO4. The factor of equivalence for the oxidizing agent is the coefficient in front of KMnO4, which is 1.

In Equation A, the reducing agent is KNO2, which is oxidized to KNO3. The factor of equivalence for the reducing agent is the coefficient in front of KNO2, which is 2.

In Equation B, the oxidizing agent is HIO3, which is reduced to I2. The factor of equivalence for the oxidizing agent is the coefficient in front of HIO3, which is 2.

In Equation B, the reducing agent is H2O2, which is oxidized to O2. The factor of equivalence for the reducing agent is the coefficient in front of H2O2, which is 1.

Please note that the factors of equivalence represent the stoichiometric coefficients of the substances being oxidized or reduced in the balanced chemical equations.