Question

Виставте ступені окиснення та розставте коефіцієнти в рівняннях реакцій методом електронного балансу, вкажіть окисники й відновники, а також процес окиснення та процес відновлення. Cu + H₂SO CuSO4 + SO,+ H₂O​

Answer 1

Ответ:

Объяснение:

Cu + H₂SO₄  → CuSO₄ + SO₂ + 2 H₂O

відновник  Cu⁰ - 2 e- → Cu²⁺ процес окислення

окисник      S⁶⁺ + 2 e- → S⁴⁺   процес відновлення

Answer 2

Stupeni okysnennya (oxidation states) and coefficients in the reaction equations using the method of electronic balance, indicating the oxidizers and reducers, as well as the oxidation process and the reduction process.

To balance the given chemical equation using the method of electronic balance, we need to determine the oxidation states (stupeni okysnennya) of the elements involved in the reaction. The oxidation state of an element indicates the number of electrons it has gained or lost in a compound or ion.

The given chemical equation is: Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O

Let's determine the oxidation states of the elements in the equation:

1. Copper (Cu): - Copper is a transition metal and can have multiple oxidation states. - In this equation, copper is being oxidized from an oxidation state of 0 to +2 in CuSO₄. - Therefore, the oxidation process for copper is: Cu → Cu²⁺

2. Hydrogen (H): - Hydrogen typically has an oxidation state of +1, except when it is bonded to metals, where it has an oxidation state of -1. - In this equation, hydrogen is being reduced from an oxidation state of +1 in H₂SO₄ to 0 in H₂O. - Therefore, the reduction process for hydrogen is: H⁺ → H₂

3. Sulfur (S): - Sulfur can have multiple oxidation states. - In this equation, sulfur is being oxidized from an oxidation state of +6 in H₂SO₄ to +4 in SO₂. - Therefore, the oxidation process for sulfur is: S⁶⁺ → S⁴⁺

Now, let's balance the equation using the method of electronic balance:

Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. Balancing copper (Cu): - There is 1 Cu atom on both sides of the equation, so copper is already balanced.

2. Balancing hydrogen (H): - There are 2 H atoms on the left side and 2 H atoms on the right side of the equation, so hydrogen is already balanced.

3. Balancing sulfur (S): - There is 1 S atom on the left side and 1 S atom on the right side of the equation, so sulfur is already balanced.

Now, let's balance the oxygen (O) atoms:

1. Balancing oxygen (O): - There are 4 O atoms on the right side of the equation (2 in CuSO₄ and 2 in H₂O). - To balance the O atoms on the left side, we need to add a coefficient of 2 in front of H₂SO₄: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + H₂O

The balanced equation is: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + H₂O

In this balanced equation: - Copper (Cu) is being oxidized from an oxidation state of 0 to +2 in CuSO₄. It is the reducing agent (vіdnovnyk). - Hydrogen (H) is being reduced from an oxidation state of +1 in H₂SO₄ to 0 in H₂O. It is the oxidizing agent (okysnyk). - The oxidation process is Cu → Cu²⁺, and the reduction process is H⁺ → H₂.

Please note that the coefficients in front of the compounds in the balanced equation may vary depending on the specific conditions of the reaction. The coefficients provided here are based on balancing the equation using the method of electronic balance.

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