Question

Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the

Answer 1

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.

(a) The balanced chemical reaction will be:

$K+H_2O\rightarrow KOH+H_2$

The oxidation-reduction half reaction will be :

Oxidation : $K\rightarrow K^++1e^-$

Reduction : $2H^++2e^-\rightarrow H_2$

In this reaction, the oxidation state of 'K' changes from (0) to (+1) that means 'K' lost 1 electron and it shows oxidation and the oxidation state of 'H' changes from (+1) to (0) that means 'H' gains 1 electron and it shows reduction.

In this reaction, potassium is the oxidizing atom that has the highest oxidation state is (+1).

(b) The balanced chemical reaction will be:

$Ba+2HBr\rightarrow BaBr_2+H_2$

The oxidation-reduction half reaction will be :

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

Reduction : $2H^++2e^-\rightarrow H_2$

In this reaction, the oxidation state of 'Ba' changes from (0) to (+2) that means 'Ba' lost 2 electrons and it shows oxidation and the oxidation state of 'H' changes from (+1) to (0) that means 'H' gains 1 electron and it shows reduction.

In this reaction, barium is the oxidizing atom that has the highest oxidation state is (+2).

(c) The balanced chemical reaction will be:

$Sn+I_2\rightarrow SnI_2$

The oxidation-reduction half reaction will be :

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

Reduction : $I_2+2e^-\rightarrow 2I^-$

In this reaction, the oxidation state of 'Sn' changes from (0) to (+2) that means 'Sn' lost 2 electron and it shows oxidation and the oxidation state of 'I' changes from (0) to (-1) that means 'I' gains 1 electron and it shows reduction.

In this reaction, tin is the oxidizing atom that has the highest oxidation state is (+2).