Question

Use electron transfer or electron shift to identify what is oxidized and what is reduced in each reaction :

Answer 1

A) Na is oxidised Br is reduced
b) H is oxidised Cl is reduced
c) Li is oxidised F is reduced
d)S is oxidised Cl is reduced
e) N is oxidised O is reduced
f) Mg is oxidised and N is reduced

Remember: Oxidation= loss and Reduction= gains

Answer 2

Answer :

Oxidation-reduction reaction : It is a type of reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction : It is the reaction in which a substance looses its electrons. In the oxidation reaction, the oxidation state of an element increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In the reduction reaction, the oxidation state of an element decreases.

(a) The balanced chemical reactions is,

$2Na(s)+Br_2(l)\rightarrow 2NaBr(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Na\rightarrow Na^{1+}+1e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^{1-}$

From this we conclude that, 'Na' is oxidized and $'Br_2'$ is reduced in this reaction. The reducing agent is, 'Na' and oxidizing agent is, $'Br_2'$.

(b) The balanced chemical reactions is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

Half reactions of oxidation and reduction are :

Oxidation : $H_2\rightarrow H^{1+}+1e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^{1-}$

From this we conclude that, $'H_2'$ is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, $'H_2'$ and oxidizing agent is, $'Cl_2'$.

(c) The balanced chemical reactions is,

$2Li(s)+F_2(g)\rightarrow 2LiF(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Li\rightarrow Li^{1+}+1e^-$

Reduction : $F_2+2e^-\rightarrow 2F^{1-}$

From this we conclude that, 'Li' is oxidized and $'F_2'$ is reduced in this reaction. The reducing agent is, 'Li' and oxidizing agent is, $'F_2'$.

(d) The balanced chemical reactions is,

$S(s)+Cl_2(g)\rightarrow SCl_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $S\rightarrow S^{2+}+2e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^{1-}$

From this we conclude that, 'S' is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, 'S' and oxidizing agent is, $'Cl_2'$.

(e) The balanced chemical reactions is,

$N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $N_2\rightarrow N^{4+}+4e^-$

Reduction : $O_2+4e^-\rightarrow 2O^{2-}$

From this we conclude that, $'N_2'$ is oxidized and $'O_2'$ is reduced in this reaction. The reducing agent is, $'N_2'$ and oxidizing agent is, $'O_2'$.

(f) The balanced chemical reactions is,

$Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Mg\rightarrow Mg^{2+}+2e^-$

Reduction : $Cu^{2+}+2e^-\rightarrow Cu$

From this we conclude that, $'Mg'$ is oxidized and $'Cu'$ is reduced in this reaction. The reducing agent is, 'Mg' and oxidizing agent is, 'Cu'.