Question

in which direction do electrons move during a Redox reaction: from oxidizing agent to reducing agent or vice versa?​

Answer 1

Answer:

Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing ent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. Following illustrations justify this.

(i) Oxidizing agent is F

2

and reducing agent is P

4

. When excess P

4

reacts with F

2

, PF

3

is produced in which P has +3 oxidation number.

P

4

( excess ) +F

2

→PF

3

But if fluorine is in excess, PF

5

is formed in which P has oxidation number of +5.

P

4

+F

2

( excess ) →PF

5

(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen, K

2

O is formed in which oxygen has oxidation number of -2.

4K( excess ) +O

2

→2K

2

O

But if oxygen is in excess, then K

2

O

2

is formed in which O has oxidation number of -1.

2K+O

2

( excess ) →K

2

O

2

(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.

C( excess ) +O

2

→CO

When excess of oxygen is used, CO

2

is formed in which C has +4 oxidation number.

C+O

2

( excess ) →CO

2