Question

Методом электронного баланса расставьте коэффициенты в приведенных схемах окислительно-восстановительных реакций, укажите окислитель и восстановитель: 1) P + KClO3 → P2O5 + KCl; 2) PH3 + AgNO3 + H2O → H3PO4 + Ag + HNO3.

Answer 1

сокращения:

ок-ль окислитель

в-ль восстановитель

пр. ок-ния процесс окисления

пр. в-ния процесс восстановления

Answer 2

Balancing Oxidation-Reduction Reactions

To balance the coefficients in the given oxidation-reduction reactions using the method of electronic balance, we need to identify the oxidation and reduction half-reactions and balance them individually. Then, we can combine the balanced half-reactions to obtain the balanced overall reaction.

1) P + KClO3 → P2O5 + KCl

To balance this reaction, we first need to identify the oxidation and reduction half-reactions.

The oxidation half-reaction involves the element that is being oxidized, which is P (phosphorus) in this case. Phosphorus is going from an oxidation state of 0 to +5 in P2O5. Therefore, the oxidation half-reaction is:

P → P2O5

The reduction half-reaction involves the element that is being reduced, which is Cl (chlorine) in this case. Chlorine is going from an oxidation state of +5 to -1 in KCl. Therefore, the reduction half-reaction is:

KClO3 → KCl

Now, let's balance the half-reactions individually:

Oxidation half-reaction:

P → P2O5

To balance the phosphorus atoms, we need to put a coefficient of 4 in front of P:

4P → P2O5

Reduction half-reaction:

KClO3 → KCl

To balance the chlorine atoms, we need to put a coefficient of 3 in front of KCl:

KClO3 → 3KCl

Now, let's balance the oxygen atoms by adding water (H2O) molecules:

KClO3 → 3KCl + 3H2O

Finally, let's balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs them:

KClO3 + 6H+ → 3KCl + 3H2O

The balanced overall reaction is:

P + KClO3 + 6H+ → P2O5 + 3KCl + 3H2O

In this reaction, P is the reducing agent (it undergoes oxidation) and KClO3 is the oxidizing agent (it undergoes reduction).

2) PH3 + AgNO3 + H2O → H3PO4 + Ag + HNO3

To balance this reaction, we first need to identify the oxidation and reduction half-reactions.

The oxidation half-reaction involves the element that is being oxidized, which is P (phosphorus) in this case. Phosphorus is going from an oxidation state of -3 to +5 in H3PO4. Therefore, the oxidation half-reaction is:

PH3 → H3PO4

The reduction half-reaction involves the element that is being reduced, which is Ag (silver) in this case. Silver is going from an oxidation state of +1 to 0 in Ag. Therefore, the reduction half-reaction is:

AgNO3 → Ag

Now, let's balance the half-reactions individually:

Oxidation half-reaction:

PH3 → H3PO4

To balance the phosphorus atoms, we need to put a coefficient of 4 in front of PH3:

4PH3 → H3PO4

Reduction half-reaction:

AgNO3 → Ag

The silver atoms are already balanced, so we don't need to change anything.

Now, let's balance the nitrogen atoms by adding nitrogen dioxide (NO2) molecules:

AgNO3 → Ag + NO2

Finally, let's balance the oxygen atoms by adding water (H2O) molecules:

AgNO3 + H2O → Ag + NO2 + H2O

The balanced overall reaction is:

4PH3 + AgNO3 + 4H2O → H3PO4 + Ag + NO2 + 4H2O

In this reaction, PH3 is the reducing agent (it undergoes oxidation) and AgNO3 is the oxidizing agent (it undergoes reduction).

Please note that the coefficients provided above are the result of balancing the reactions using the method of electronic balance.