Question

Write the following redox reactions in the oxidation and reduction half reaction

(i)
2K(s) + Cl2(g) →2KCl(s)

(ii)
2Al(s) + 3Cu2+(aq) →2A1+(aq) + 3Cu(s)​

Answer 1

Answer:

oxidation half reaction

2K -> 2K+ + 2e-

reduction reaction

Cl2 + 2e- ->2Cl

Answer 2

Part (i)

The oxidation-reduction half reaction will be :

Oxidation : $K\rightarrow K^++e^-$

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

Part (ii)

The oxidation-reduction half reaction will be :

Oxidation : $Al\rightarrow Al^{3+}+3e^-$

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

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.

(i) The given chemical reaction is,

$2K(s)+Cl_2(g)\rightarrow 2KCl(s)$

In this reaction, the oxidation state of potassium changes from (0) to (+1) which shows oxidation and oxidation state of chlorine changes from (0) to (-1) which shows reduction.

The oxidation-reduction half reaction will be :

Oxidation : $K\rightarrow K^++e^-$

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

(ii) The given chemical reaction is,

$2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)$

In this reaction, the oxidation state of aluminum changes from (0) to (+1) which shows oxidation and oxidation state of copper changes from (+2) to (0) which shows reduction.

The oxidation-reduction half reaction will be :

Oxidation : $Al\rightarrow Al^{3+}+3e^-$

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

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