Question

Составить электронный баланс, определить окислитель и восстановитель, расставить коэффициенты в уравнениях a)O2+ P-> P2 O3 b)NaOH + H2O + Si->Na2SiO3+ H2 c)NO2+ N2O+O2 ->HNO3 d)SO2 +Br2 + H2O->H2SO4 +HBr e)PbS+HNO3 -> S + Pb(NO3)2 +NO + H2O

Answer 1

Ответ на фотографии

Answer 2

Electron Balancing and Identifying Redox Reactions

To balance the given chemical equations and identify the oxidizing and reducing agents, we need to follow a step-by-step process. Let's go through each equation one by one.

a) O2 + P -> P2O3

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. Let's start by balancing the phosphorus (P) atoms:

Step 1: Count the number of P atoms on each side of the equation. There is 1 P atom on the left side and 2 P atoms on the right side.

Step 2: To balance the P atoms, we need to put a coefficient of 2 in front of the P on the left side:

2 O2 + 4 P -> P2O3

Now, let's check if the equation is balanced for oxygen (O) atoms:

Step 3: Count the number of O atoms on each side of the equation. There are 4 O atoms on the left side and 3 O atoms on the right side.

Step 4: To balance the O atoms, we need to put a coefficient of 2 in front of the O2 on the left side:

2 O2 + 4 P -> 2 P2O3

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the change in oxidation states for each element. In this case, the oxidation state of P changes from 0 to +3, indicating that it is being oxidized. The oxidation state of O changes from 0 to -2, indicating that it is being reduced.

Therefore, in this reaction, P is the reducing agent (it is being oxidized) and O is the oxidizing agent (it is being reduced).

b) NaOH + H2O + Si -> Na2SiO3 + H2

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. Let's start by balancing the sodium (Na) atoms:

Step 1: Count the number of Na atoms on each side of the equation. There is 1 Na atom on the left side and 2 Na atoms on the right side.

Step 2: To balance the Na atoms, we need to put a coefficient of 2 in front of the NaOH on the left side:

2 NaOH + H2O + Si -> Na2SiO3 + H2

Now, let's check if the equation is balanced for hydrogen (H) atoms:

Step 3: Count the number of H atoms on each side of the equation. There are 4 H atoms on the left side and 4 H atoms on the right side.

Step 4: The equation is already balanced for hydrogen.

Now, let's check if the equation is balanced for oxygen (O) atoms:

Step 5: Count the number of O atoms on each side of the equation. There are 3 O atoms on the left side and 3 O atoms on the right side.

Step 6: The equation is already balanced for oxygen.

Therefore, the balanced equation is:

2 NaOH + H2O + Si -> Na2SiO3 + H2

To identify the oxidizing and reducing agents, we need to determine the change in oxidation states for each element. In this case, the oxidation state of Si changes from 0 to +4, indicating that it is being oxidized. The oxidation state of Na changes from +1 to +2, indicating that it is being reduced.

Therefore, in this reaction, Si is the reducing agent (it is being oxidized) and NaOH is the oxidizing agent (it is being reduced).

c) NO2 + N2O + O2 -> HNO3

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. Let's start by balancing the nitrogen (N) atoms:

Step 1: Count the number of N atoms on each side of the equation. There are 3 N atoms on the left side and 1 N atom on the right side.

Step 2: To balance the N atoms, we need to put a coefficient of 2 in front of the N2O on the left side:

NO2 + 2 N2O + O2 -> HNO3

Now, let's check if the equation is balanced for oxygen (O) atoms:

Step 3: Count the number of O atoms on each side of the equation. There are 5 O atoms on the left side and 3 O atoms on the right side.

Step 4: To balance the O atoms, we need to put a coefficient of 5 in front of the O2 on the right side:

NO2 + 2 N2O + 5 O2 -> HNO3

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the change in oxidation states for each element. In this case, the oxidation state of N changes from +4 to +5, indicating that it is being oxidized. The oxidation state of O changes from 0 to -2, indicating that it is being reduced.

Therefore, in this reaction, N2O is the reducing agent (it is being oxidized) and O2 is the oxidizing agent (it is being reduced).

d) SO2 + Br2 + H2O -> H2SO4 + HBr

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. Let's start by balancing the sulfur (S) atoms:

Step 1: Count the number of S atoms on each side of the equation. There is 1 S atom on the left side and 1 S atom on the right side.

Step 2: The equation is already balanced for sulfur.

Now, let's check if the equation is balanced for oxygen (O) atoms:

Step 3: Count the number of O atoms on each side of the equation. There are 2 O atoms on the left side and 4 O atoms on the right side.

Step 4: To balance the O atoms, we need to put a coefficient of 2 in front of the H2O on the left side:

SO2 + Br2 + 2 H2O -> H2SO4 + HBr

Now, let's check if the equation is balanced for hydrogen (H) atoms:

Step 5: Count the number of H atoms on each side of the equation. There are 4 H atoms on the left side and 6 H atoms on the right side.

Step 6: To balance the H atoms, we need to put a coefficient of 4 in front of the HBr on the right side:

SO2 + Br2 + 2 H2O -> H2SO4 + 4 HBr

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the change in oxidation states for each element. In this case, the oxidation state of S changes from +4 to +6, indicating that it is being oxidized. The oxidation state of Br changes from 0 to -1, indicating that it is being reduced.

Therefore, in this reaction, SO2 is the reducing agent (it is being oxidized) and Br2 is the oxidizing agent (it is being reduced).

e) PbS + HNO3 -> S + Pb(NO3)2 + NO + 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. Let's start by balancing the sulfur (S) atoms:

Step 1: Count the number of S atoms on each side of the equation. There is 1 S atom on the left side and 1 S atom on the right side.

Step 2: The equation is already balanced for sulfur.

Now, let's check if the equation is balanced for nitrogen (N) atoms:

Step 3: Count the number of N atoms on each side of the equation. There is 1 N atom on the left side and 2 N atoms on the right side.

Step 4: To balance the N atoms, we need to put a coefficient of 2 in front of the HNO3 on the left side:

PbS + 2 HNO3 -> S + Pb(NO3)2 + NO + H2O

Now, let's check if the equation is balanced for oxygen (O) atoms:

Step 5: Count the number of O atoms on each side of the equation. There are 6 O atoms on the left side and 6 O atoms on the right side.

Step 6: The equation is already balanced for oxygen.

Now, let's check if the equation is balanced for hydrogen (H) atoms:

Step 7: Count the number of H atoms on each side of the equation. There are 6 H atoms on the left side and 8 H atoms on the right side.

Step 8: To balance the H atoms, we need to put a coefficient of 4 in front of the H2O on the right side:

PbS + 2 HNO3 -> S + Pb(NO3)2 + NO + 4 H2O

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the change in oxidation states for each element. In this case, the oxidation state of S changes from -2 to 0, indicating that