Question

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

a) 2Na(s) + Br2(l) ----> 2NaBr(s)
b) H2(g) + Cl2(g) ----> 2HCl(g)
c) 2Li(s) + F2(g) ----> 2LiF(s)
d) S(s) + Cl2(g) ----> SCl2(g)
e)N2(g) + 2O2(g) ----> 2NO2(g)
f) Mg(s) +Cu(NO3)2(aq) = Mg(NO3)2(aq) + Cu(s)

For each reaction above, identify the reducing agent and the oxidizing agent

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

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'.