Question

Уравняйте методом электронного баланса , укажите окислитель и восстановитель для следующих окислительно-восстановительных процессов : KMnO4+ Na2SO3 + H2SO4 --> MnSO4 + Na2SO +K2SO4 + H2O

Answer 1

2KMnO4+ 5Na2SO3 + 3H2SO4 --> 2MnSO4 + 5Na2SO4 + K2SO4 + 3H2O
Mn(+7) --(+5e)--> Mn(+2)    | 2     восстановление
S(+4) --(-2e)--> S(+6)         | 5     окисление
KMnO4 - окислитель, за счёт Mn(+7)
Na2SO3 - восстановитель, за счёт S(+4)

Answer 2

2KMnO4 + 5Na2SO3 + 3H2SO4 = 5Na2SO4 + 2MnSO4 + K2SO4 + 3H2О
Mn(+7) +5e = Mn(+2)
S(+4) - 2e  =S(+6)

Answer 3

Balancing the Equation using the Electron Balance Method

To balance the given equation using the electron balance method, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

The given equation is:

KMnO4 + Na2SO3 + H2SO4 → MnSO4 + Na2SO4 + K2SO4 + H2O

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

Oxidation Half-Reaction:

In the oxidation half-reaction, the element that gets oxidized loses electrons. In this case, the element that gets oxidized is Sulfur (S), which changes its oxidation state from +4 in Na2SO3 to +6 in Na2SO4.

The oxidation half-reaction can be written as:

S in Na2SO3 → S in Na2SO4

To balance the number of sulfur atoms, we need to add 2 water molecules (H2O) to the right side of the equation:

S in Na2SO3 → S in Na2SO4 + 2H2O

Now, let's balance the oxygen atoms. There are 3 oxygen atoms on the left side and 4 oxygen atoms on the right side. To balance the oxygen atoms, we need to add 1 oxygen molecule (O2) to the left side of the equation:

S in Na2SO3 + O2 → S in Na2SO4 + 2H2O

Next, let's balance the hydrogen atoms. There are 6 hydrogen atoms on the right side and none on the left side. To balance the hydrogen atoms, we need to add 6 hydrogen ions (H+) to the left side of the equation:

S in Na2SO3 + O2 + 6H+ → S in Na2SO4 + 2H2O

Finally, let's balance the charge. There is a charge of -2 on the left side and a charge of 0 on the right side. To balance the charge, we need to add 2 electrons (e-) to the left side of the equation:

S in Na2SO3 + O2 + 6H+ + 2e- → S in Na2SO4 + 2H2O

Reduction Half-Reaction:

In the reduction half-reaction, the element that gets reduced gains electrons. In this case, the element that gets reduced is Manganese (Mn), which changes its oxidation state from +7 in KMnO4 to +2 in MnSO4.

The reduction half-reaction can be written as:

Mn in KMnO4 → Mn in MnSO4

To balance the number of manganese atoms, we need to add 1 manganese ion (Mn2+) to the left side of the equation:

MnO4- + Mn2+ → MnSO4

Now, let's balance the oxygen atoms. There are 4 oxygen atoms on the left side and 4 oxygen atoms on the right side. The oxygen atoms are already balanced.

Next, let's balance the hydrogen atoms. There are no hydrogen atoms on either side, so we don't need to balance them.

Finally, let's balance the charge. There is a charge of -1 on the left side and a charge of 0 on the right side. To balance the charge, we need to add 1 electron (e-) to the left side of the equation:

MnO4- + Mn2+ + e- → MnSO4

Balancing the Overall Equation:

To balance the overall equation, we need to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

The oxidation half-reaction involves the transfer of 2 electrons, while the reduction half-reaction involves the transfer of 1 electron. To balance the number of electrons, we need to multiply the reduction half-reaction by 2:

2MnO4- + 2Mn2+ + 2e- → 2MnSO4

Now, let's combine the oxidation and reduction half-reactions to get the balanced equation:

2MnO4- + 2Mn2+ + 2e- + S in Na2SO3 + O2 + 6H+ + 2e- → 2MnSO4 + S in Na2SO4 + 2H2O

Simplifying the equation by canceling out the electrons:

2MnO4- + 2Mn2+ + S in Na2SO3 + O2 + 6H+ → 2MnSO4 + S in Na2SO4 + 2H2O

Identifying the Oxidizing and Reducing Agents:

In the given reaction, the oxidizing agent is the species that gets reduced. In this case, the oxidizing agent is KMnO4 because it undergoes reduction from +7 to +2 oxidation state.

The reducing agent is the species that gets oxidized. In this case, the reducing agent is Na2SO3 because it undergoes oxidation from +4 to +6 oxidation state.

To summarize: - Oxidizing agent: KMnO4 - Reducing agent: Na2SO3

Please note that the above explanation is based on the information provided and the electron balance method.