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

Answer :

(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

Answer:

See the explanation.

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.