Question

Расставьте коэффициенты методом электронного баланса, укажите процессы окисления и восстановления, укажите окислитель и восстановитель. H2O2+PbS=H2O+PbSO4

Answer 1

Степени окисления буду писать в скобках.
S(-2) - 8e →S(+6)  | 8 |2| 1, восстановитель, процесс окисления
O(-1) +2e → O(-2) | 2 | 8| 4 ,окислитель, процесс восстановления
 1 и 4  - это коэффициенты в уравнении реакции
4 H2O2 + PbS= 4 H2O + PbSO4

Answer 2

Balancing the Equation using the Method of Electron Balance

To balance the equation H2O2 + PbS = H2O + PbSO4 using the method of electron balance, we need to ensure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction.

Let's break down the equation into the oxidation and reduction half-reactions:

Oxidation Half-Reaction: PbS -> PbSO4

Reduction Half-Reaction: H2O2 -> H2O

To balance the oxidation half-reaction, we need to balance the number of atoms on both sides of the equation. In this case, we have one Pb atom on the left side and one Pb atom on the right side, so the Pb atoms are already balanced. However, we have one S atom on the left side and four O atoms on the right side, so we need to balance the S atoms by adding a coefficient of 4 in front of PbS:

Oxidation Half-Reaction: 4PbS -> PbSO4

Next, let's balance the reduction half-reaction. We have two H atoms on the left side and two H atoms on the right side, so the H atoms are already balanced. However, we have two O atoms on the left side and one O atom on the right side, so we need to balance the O atoms by adding a coefficient of 2 in front of H2O:

Reduction Half-Reaction: H2O2 -> 2H2O

Now, let's balance the electrons in both half-reactions. In the oxidation half-reaction, we have four S atoms on the left side and four O atoms on the right side. Each S atom gains two electrons, so a total of eight electrons are gained in the oxidation half-reaction. To balance the electrons, we need to add an eight in front of the reduction half-reaction:

Oxidation Half-Reaction: 8H2O2 -> 16H2O

Finally, let's balance the number of H atoms in the reduction half-reaction. We have four H atoms on the left side and 16 H atoms on the right side. To balance the H atoms, we need to add a coefficient of 4 in front of H2O2:

Reduction Half-Reaction: 4H2O2 -> 8H2O

Now, the balanced equation is:

4H2O2 + 4PbS -> 16H2O + PbSO4

In this balanced equation, the oxidizing agent is H2O2, which is reduced, and the reducing agent is PbS, which is oxidized.

Please note that the sources provided did not directly address the specific equation you provided. However, the method used to balance the equation is a standard approach based on the principles of electron balance in redox reactions.