Question

Расставьте коэффициенты в схемах реакций методом электронного баланса. Укажите окислитель и восстановитель, процессы окисления и восстановления. 1) Ca + N2 - Ca3N22) Al + O2 - AL2O33) P + O2 - P2O5Пожалуйста! Срочно!

Answer 1

решение вложила............

Answer 2

Balancing Redox Reactions using the Electron Balance Method

To balance the given redox reactions using the electron balance method, we need to assign oxidation numbers to each element in the reactants and products, determine the changes in oxidation numbers, and balance the electrons transferred.

1) Ca + N2 -> Ca3N2

Let's start by assigning oxidation numbers to the elements in the reaction: - The oxidation number of Ca is +2 in Ca and +3 in Ca3N2. - The oxidation number of N is 0 in N2 and -3 in Ca3N2.

Based on the changes in oxidation numbers, we can see that Ca is oxidized from +2 to +3, while N is reduced from 0 to -3.

To balance the electrons transferred, we need to multiply the oxidation half-reaction by the appropriate coefficient so that the number of electrons gained equals the number of electrons lost.

The oxidation half-reaction is: Ca -> Ca3+ + 3e-

Now, let's balance the reduction half-reaction: N2 + 6e- -> 2N3-

By multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3, we can balance the electrons transferred: 2Ca -> 2Ca3+ + 6e- 3N2 + 6e- -> 6N3-

Now, we can combine the two half-reactions and cancel out the electrons: 2Ca + 3N2 -> 2Ca3N2

In this reaction, Ca is the reducing agent (it is oxidized) and N2 is the oxidizing agent (it is reduced). The process of oxidation occurs to Ca, while the process of reduction occurs to N2.

2) Al + O2 -> Al2O3

Let's assign oxidation numbers to the elements: - The oxidation number of Al is 0 in Al and +3 in Al2O3. - The oxidation number of O is 0 in O2 and -2 in Al2O3.

Based on the changes in oxidation numbers, we can see that Al is oxidized from 0 to +3, while O is reduced from 0 to -2.

The oxidation half-reaction is: Al -> Al3+ + 3e-

The reduction half-reaction is: O2 + 4e- -> 2O2-

By multiplying the oxidation half-reaction by 4 and the reduction half-reaction by 3, we can balance the electrons transferred: 4Al -> 4Al3+ + 12e- 3O2 + 12e- -> 6O2-

Now, we can combine the two half-reactions and cancel out the electrons: 4Al + 3O2 -> 2Al2O3

In this reaction, Al is the reducing agent (it is oxidized) and O2 is the oxidizing agent (it is reduced). The process of oxidation occurs to Al, while the process of reduction occurs to O2.

3) P + O2 -> P2O5

Let's assign oxidation numbers to the elements: - The oxidation number of P is 0 in P and +5 in P2O5. - The oxidation number of O is 0 in O2 and -2 in P2O5.

Based on the changes in oxidation numbers, we can see that P is oxidized from 0 to +5, while O is reduced from 0 to -2.

The oxidation half-reaction is: P -> P5+ + 5e-

The reduction half-reaction is: O2 + 10e- -> 5O2-

By multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 5, we can balance the electrons transferred: 2P -> 2P5+ + 10e- 5O2 + 10e- -> 25O2-

Now, we can combine the two half-reactions and cancel out the electrons: 2P + 5O2 -> P2O5

In this reaction, P is the reducing agent (it is oxidized) and O2 is the oxidizing agent (it is reduced). The process of oxidation occurs to P, while the process of reduction occurs to O2.

I hope this helps! Let me know if you have any further questions.