Question

QUESTION 8

Answer 1

answer to Question will be 2 electron as it forms 2H+

Answer 2

Answer :

(8) The number of electrons added to balance the reaction should be, Two electrons.

(9) The correct option is, $2I^-\rightarrow I_2+2e^-$ (oxidized) and $2IO_3^-+2e^-\rightarrow I_2$ (reduced)

Explanation :

(8) The given oxidation half-reaction is,

$H_2O_2\rightarrow 2H^++O_2$

The given oxidation half-reaction is an unbalanced reaction because in this reaction the charges are not balanced. So, in order to balance the half reaction we are adding two electrons on product side of the reaction and the oxidation means loss of electrons.

Thus, the balanced oxidation half-reaction will be,

$H_2O_2\rightarrow 2H^++O_2_2e^-$

Hence, the number of electrons added to balance the reaction should be, Two electrons.

(9) 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.

The given oxidation-reduction reaction is :

$I^-+IO_3^-\rightarrow I_2$

The oxidation-reduction half reaction will be :

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

Reduction : $2IO_3^-+2e^-\rightarrow I_2$

In this reaction, the oxidation state of 'I' changes from (-1) to (0) that means 'I' lost 2 electrons and it shows oxidation and the oxidation state of 'I' changes from (+5) to (0) that means 'I' gains 2 electrons and it shows reduction.

Hence, the correct option is, $2I^-\rightarrow I_2+2e^-$ (oxidized) and $2IO_3^-+2e^-\rightarrow I_2$ (reduced)