Respuesta :
Answer :
(i) The half oxidation-reduction reactions are:
Oxidation reaction : [tex]Sr\rightarrow Sr^{2+}+2e^-[/tex]
Reduction reaction : [tex]O_2+4e^-\rightarrow 2O^{2-}[/tex]
(ii) [tex]O_2[/tex] is oxidized species.
(iii) [tex]Sr[/tex] 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 :
[tex]2Sr+O_2\rightarrow 2SrO[/tex]
The half oxidation-reduction reactions are:
Oxidation reaction : [tex]Sr\rightarrow Sr^{2+}+2e^-[/tex]
Reduction reaction : [tex]O_2+4e^-\rightarrow 2O^{2-}[/tex]
From this we conclude that the [tex]'Sr'[/tex] is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and [tex]'O_2'[/tex] 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, [tex]Sr[/tex] is reduced species and [tex]O_2[/tex] is oxidized species.
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: [tex]Sr^0-->Sr^{+2}+2e^-[/tex]
reduction: [tex]O_2^0+4e^--->2O^{-2}[/tex]
(ii) Oxidizing agent: oxygen as it is reduced.
(iii) Reducing agent: strontium as it is oxidized.
(b) 2Li+H₂→ 2LiH:
(i) Half-reactions:
oxidation: [tex]Li^0-->Li^{+1}+1e^-[/tex]
reduction: [tex]H_2^0+2e^--->2H^{-1}[/tex]
(ii) Oxidizing agent: hydrogen as it is reduced.
(iii) Reducing agent: lithium as it is oxidized.
(c) 2Cs+Br₂ → 2CsBr
(i) Half-reactions:
oxidation: [tex]Cs^0-->Cs^{+1}+1e^-[/tex]
reduction: [tex]Br_2^0+2e^--->2Br^{-1}[/tex]
(ii) Oxidizing agent: bromine as it is reduced.
(iii) Reducing agent: cesium as it is oxidized.
(d) 3Mg+N₂ → Mg₃N₂
(i) Half-reactions:
oxidation: [tex]3Mg^0-->Mg_3^{+2}+6e^-[/tex]
reduction: [tex]N_2^0+6e^--->N_2^{-3}[/tex]
(ii) Oxidizing agent: nitrogen as it is reduced.
(iii) Reducing agent: magnesium as it is oxidized.
Best regards.
