Chemistry, 23.08.2019 20:40 hbkakabryce0p3fkoq

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

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Answer from: wrolle

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 from: Vanesssajohn6294

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' .

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

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'.

Answer from: Quest

the earth would become to overheated and everything alive on it will eventually die from overheating. plants will eventually die as well, water will evaporate, and it will be nothing but a hot planet without anything living on it.

Answer from: Quest

It has to be b i hope this