Question

РЕБЯТА, ОЧЕНЬ НУЖНА ВАША ПОМОЩЬ!!! ПОМОГИТЕ ПОЖАЛУЙСТА! Методом электронного баланса расставить коэффициенты в уравнении. Указать вещество-окислитель и вещество-восстановитель. 1)NaI+NaIO3+H2SO4=I2+Na2SO4+H2O 2)HNO3+Mg=Mg(NO3)2+N2O↑+H2O 3)MnSO4+PbO2+HNO3→HMnO4+Pb(NO3)2+PbSO4+H2O

Answer 1

в первом уравнении окислителем и восстановителем является йод I

во втором магнийMg восстановитель а азотN окислитель

в третьем азотN окислитель а марганецMn восстановитель

1 10NaI+2NaIO3+6H2SO4=6I2+6Na2SO4+6H2O

2I(-) -2e I2(0)          окислился  восстановитель  10

2I(+5) +10e 2I(0)     восстановился окислитель   2

Answer 2

Balancing Chemical Equations and Identifying Redox Reactions

Balancing chemical equations involves ensuring that the number of atoms of each element is the same on both sides of the equation. Additionally, identifying the substances that undergo oxidation (lose electrons) and reduction (gain electrons) is important in redox reactions.

Let's go through each equation and balance them while identifying the oxidizing and reducing agents.

Equation 1: NaI + NaIO3 + H2SO4 = I2 + Na2SO4 + H2O

To balance this equation, we can follow these steps:

1. Start by balancing the atoms that appear in the fewest compounds. In this case, we have iodine (I) and hydrogen (H). - There are 2 iodine atoms on the right side, so we need 2 iodine atoms on the left side. We can achieve this by placing a coefficient of 2 in front of NaI: 2NaI + NaIO3 + H2SO4 = I2 + Na2SO4 + H2O. - There are 4 hydrogen atoms on the right side, so we need 4 hydrogen atoms on the left side. We can achieve this by placing a coefficient of 4 in front of H2O: 2NaI + NaIO3 + H2SO4 = I2 + Na2SO4 + 4H2O.

2. Next, balance the remaining atoms. In this case, we have sodium (Na) and sulfur (S). - There are 2 sodium atoms on the left side, so we need 2 sodium atoms on the right side. We can achieve this by placing a coefficient of 2 in front of Na2SO4: 2NaI + NaIO3 + H2SO4 = I2 + 2Na2SO4 + 4H2O. - There is 1 sulfur atom on the left side, so we need 1 sulfur atom on the right side. We can achieve this by placing a coefficient of 2 in front of H2SO4: 2NaI + NaIO3 + 2H2SO4 = I2 + 2Na2SO4 + 4H2O.

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the changes in oxidation states of the elements involved.

In this equation, iodine (I) changes from an oxidation state of -1 in NaI to 0 in I2. It gains electrons and is reduced, so it is the reducing agent.

On the other hand, iodine in NaIO3 changes from an oxidation state of +5 to 0 in I2. It loses electrons and is oxidized, so it is the oxidizing agent.

Equation 2: HNO3 + Mg = Mg(NO3)2 + N2O↑ + H2O

To balance this equation, we can follow these steps:

1. Start by balancing the atoms that appear in the fewest compounds. In this case, we have nitrogen (N) and oxygen (O). - There is 1 nitrogen atom on the right side, so we need 1 nitrogen atom on the left side. We can achieve this by placing a coefficient of 2 in front of HNO3: 2HNO3 + Mg = Mg(NO3)2 + N2O↑ + H2O. - There are 6 oxygen atoms on the right side, so we need 6 oxygen atoms on the left side. We can achieve this by placing a coefficient of 6 in front of HNO3: 6HNO3 + Mg = Mg(NO3)2 + N2O↑ + H2O.

2. Next, balance the remaining atoms. In this case, we have magnesium (Mg) and hydrogen (H). - There is 1 magnesium atom on the right side, so we need 1 magnesium atom on the left side. We can achieve this by placing a coefficient of 1 in front of Mg: 6HNO3 + Mg = Mg(NO3)2 + N2O↑ + H2O. - There are 6 hydrogen atoms on the right side, so we need 6 hydrogen atoms on the left side. We can achieve this by placing a coefficient of 6 in front of H2O: 6HNO3 + Mg = Mg(NO3)2 + N2O↑ + 6H2O.

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the changes in oxidation states of the elements involved.

In this equation, nitrogen (N) changes from an oxidation state of +5 in HNO3 to 0 in N2O. It gains electrons and is reduced, so it is the reducing agent.

On the other hand, magnesium (Mg) changes from an oxidation state of 0 to +2 in Mg(NO3)2. It loses electrons and is oxidized, so it is the oxidizing agent.

Equation 3: MnSO4 + PbO2 + HNO3 → HMnO4 + Pb(NO3)2 + PbSO4 + H2O

To balance this equation, we can follow these steps:

1. Start by balancing the atoms that appear in the fewest compounds. In this case, we have manganese (Mn) and hydrogen (H). - There is 1 manganese atom on the right side, so we need 1 manganese atom on the left side. We can achieve this by placing a coefficient of 1 in front of MnSO4: MnSO4 + PbO2 + HNO3 → HMnO4 + Pb(NO3)2 + PbSO4 + H2O. - There are 2 hydrogen atoms on the right side, so we need 2 hydrogen atoms on the left side. We can achieve this by placing a coefficient of 8 in front of HNO3: MnSO4 + PbO2 + 8HNO3 → HMnO4 + Pb(NO3)2 + PbSO4 + H2O.

2. Next, balance the remaining atoms. In this case, we have lead (Pb), oxygen (O), and sulfur (S). - There is 1 lead atom on the right side, so we need 1 lead atom on the left side. We can achieve this by placing a coefficient of 3 in front of Pb(NO3)2: MnSO4 + PbO2 + 8HNO3 → HMnO4 + 3Pb(NO3)2 + PbSO4 + H2O. - There are 4 oxygen atoms on the right side, so we need 4 oxygen atoms on the left side. We can achieve this by placing a coefficient of 4 in front of HMnO4: MnSO4 + PbO2 + 8HNO3 → 4HMnO4 + 3Pb(NO3)2 + PbSO4 + H2O. - There is 1 sulfur atom on the left side, so we need 1 sulfur atom on the right side. We can achieve this by placing a coefficient of 1 in front of PbSO4: MnSO4 + PbO2 + 8HNO3 → 4HMnO4 + 3Pb(NO3)2 + PbSO4 + H2O.

Now, the equation is balanced.

To identify the oxidizing and reducing agents, we need to determine the changes in oxidation states of the elements involved.

In this equation, manganese (Mn) changes from an oxidation state of +2 in MnSO4 to +7 in HMnO4. It gains electrons and is reduced, so it is the reducing agent.

On the other hand, lead (Pb) changes from an oxidation state of +4 in PbO2 to +2 in Pb(NO3)2. It loses electrons and is