Question

4. Расставить коэффеценты в схемах реакций методом электронного баланса, указать окислитель и восстановитель, процессы окисления и восстановления: 1)HNO3 + Cu → Cu(NO3)2 + NO2 +H20 2) Zn + H2SO4 → ZnS04 + S + H20

Answer 1

1)N(+5) + e = N(+4) 2 окислитель восст-ся
Cu(0) - 2e = Cu(+2) 1 восст-ль окисл-ся
4HNO3 + Cu = Cu(NO3)2 + 2NO2 + 2H2O
2)Zn(0) - 2e = Zn(+2) 3 восс-ль оксил-ся
S(+6) + 6e = S(0) 1 ок-ль восст-ся
3Zn + 4H2SO4 = 3ZnSO4 + S + 4H2O

Answer 2

Balancing Chemical Equations and Identifying Redox Reactions

To balance the given chemical equations and identify the oxidizing and reducing agents, we will use the method of electron balance. This method involves balancing the number of electrons gained and lost during the reaction to ensure that the equation is balanced both in terms of mass and charge.

Let's analyze each equation separately:

1) HNO3 + Cu → Cu(NO3)2 + NO2 + H2O

To balance this equation, 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.

First, let's balance the nitrogen (N) atoms: - There is one nitrogen atom in HNO3 and two nitrogen atoms in Cu(NO3)2. - To balance the nitrogen atoms, we need to multiply HNO3 by 2: 2HNO3 + Cu → Cu(NO3)2 + NO2 + H2O

Next, let's balance the hydrogen (H) atoms: - There are six hydrogen atoms in 2HNO3 and two hydrogen atoms in H2O. - To balance the hydrogen atoms, we need to multiply H2O by 3: 2HNO3 + Cu → Cu(NO3)2 + NO2 + 3H2O

Now, let's balance the oxygen (O) atoms: - There are six oxygen atoms in 2HNO3 and ten oxygen atoms in Cu(NO3)2. - To balance the oxygen atoms, we need to multiply Cu(NO3)2 by 5: 2HNO3 + 5Cu → 5Cu(NO3)2 + NO2 + 3H2O

The equation is now balanced.

To identify the oxidizing and reducing agents, we need to determine which elements undergo changes in oxidation state. In this reaction: - Copper (Cu) is oxidized from an oxidation state of 0 to +2 in Cu(NO3)2. It is the reducing agent. - Nitrogen (N) is reduced from an oxidation state of +5 in HNO3 to +4 in NO2. It is the oxidizing agent.

The processes of oxidation and reduction are as follows: - Oxidation: Cu → Cu^2+ - Reduction: 2N^5+ → 2N^4+

2) Zn + H2SO4 → ZnSO4 + S + H2O

To balance this equation, we will follow the same steps as before.

First, let's balance the zinc (Zn) atoms: - There is one zinc atom on both sides of the equation.

Next, let's balance the sulfur (S) atoms: - There is one sulfur atom in H2SO4 and one sulfur atom in ZnSO4. - The sulfur atoms are already balanced.

Now, let's balance the hydrogen (H) atoms: - There are two hydrogen atoms in H2SO4 and two hydrogen atoms in H2O. - The hydrogen atoms are already balanced.

Finally, let's balance the oxygen (O) atoms: - There are four oxygen atoms in H2SO4 and four oxygen atoms in ZnSO4. - The oxygen atoms are already balanced.

The equation is already balanced.

To identify the oxidizing and reducing agents, we need to determine which elements undergo changes in oxidation state. In this reaction: - Zinc (Zn) is oxidized from an oxidation state of 0 to +2 in ZnSO4. It is the reducing agent. - Sulfur (S) is reduced from an oxidation state of +6 in H2SO4 to 0 in S. It is the oxidizing agent.

The processes of oxidation and reduction are as follows: - Oxidation: Zn → Zn^2+ - Reduction: S^6+ → S^0

I hope this helps! Let me know if you have any further questions.