Question

For the complete redox reactions given below, do each of the following. (Use the lowest possible coefficients. Omit states-of-matter in your answer.) (i) Break down each reaction into its half-reactions. (ii) Identify the oxidizing agent. (iii) Identify the reducing agent. (a) 2 Sr O2 → 2 SrO

Answer 1

(i) The half oxidation-reduction reactions are:

Oxidation reaction :
`Sr\rightarrow Sr^(2+)+2e^-`

Reduction reaction :
`O_2+4e^-\rightarrow 2O^(2-)`

(ii)
`O_2` is oxidized species.

(iii)
`Sr` is reducing species.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

`2Sr+O_2\rightarrow 2SrO`

The half oxidation-reduction reactions are:

Oxidation reaction :
`Sr\rightarrow Sr^(2+)+2e^-`

Reduction reaction :
`O_2+4e^-\rightarrow 2O^(2-)`

From this we conclude that the
`'Sr'` is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and
`'O_2'` is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

In this reaction, 'Sr' is oxidized from oxidation (0) to (+2) and 'O' is reduced from oxidation state (0) to (-2). Hence, 'Sr' act as a reducing agent and 'O' act as a oxidizing agent.

Thus,
`Sr` is reduced species and
`O_2` is oxidized species.

Answer 2

See the explanation.

Step-by-step explanation:

Hello,

In this case, I found that the reactions are: (a) 2Sr+O₂ → 2 SrO, (b) 2Li+H₂→ 2LiH, (c) 2Cs+Br₂ → 2CsBr and (d) 3Mg+N₂ → Mg₃N₂

Thus, for each reaction we obtain:

(a) 2Sr+O₂ → 2 SrO:

(i) Half-reactions:

oxidation:
`Sr^0-->Sr^(+2)+2e^-`

reduction:
`O_2^0+4e^--->2O^(-2)`

(ii) Oxidizing agent: oxygen as it is reduced.

(iii) Reducing agent: strontium as it is oxidized.

(b) 2Li+H₂→ 2LiH:

(i) Half-reactions:

oxidation:
`Li^0-->Li^(+1)+1e^-`

reduction:
`H_2^0+2e^--->2H^(-1)`

(ii) Oxidizing agent: hydrogen as it is reduced.

(iii) Reducing agent: lithium as it is oxidized.

(c) 2Cs+Br₂ → 2CsBr

(i) Half-reactions:

oxidation:
`Cs^0-->Cs^(+1)+1e^-`

reduction:
`Br_2^0+2e^--->2Br^(-1)`

(ii) Oxidizing agent: bromine as it is reduced.

(iii) Reducing agent: cesium as it is oxidized.

(d) 3Mg+N₂ → Mg₃N₂

(i) Half-reactions:

oxidation:
`3Mg^0-->Mg_3^(+2)+6e^-`

reduction:
`N_2^0+6e^--->N_2^(-3)`

(ii) Oxidizing agent: nitrogen as it is reduced.

(iii) Reducing agent: magnesium as it is oxidized.

Best regards.