Help with chemistry?

Question

help how and what is being oxidized /reduced clearly label the oxidizing agent

Pb (s) +2Ag+ (aq)>>>>> Pb2+ (aq) +2 Ag (s)

Answer 1

Pb(s) +  2Ag+(aq) -->  Pb2+(aq) + 2Ag(s)

OILRIG - oxidation is loss, reduction is gain

Pb  -->  Pb2+ + 2e-  (oxidation half reaction - Pb loses electrons)

2(Ag+ + e- --> Ag)  (reduction half reaction - Ag gains electron

------------------------------

Pb + 2Ag+ --> Pb2+ + 2Ag

The oxidizing agent is the one that contains the element reduced.  Since silver was reduced, silver ion must be the oxidizing agent, and lead metal the reducing agent.

Answer 2

Pb is being oxidized and Ag is being reduced. The oxidizing agent is what is being reduced which is Ag

Answer 3

The oxidation number of element in its pure elemental form is zero.

The oxidation number of a monoatomic ion(ion with just one atom) is the charge on the ion itself.

Based on the above two rules, the Oxidation number of the species in the reaction are:

Pb : 0

Ag+  :  +1

Pb2+  :  +2

Ag :  0

Now, oxidation can be defined as the increase in oxidation number, and reduction is the decrease in oxidation number.

A substance that gets oxidized is a good reducing agent, and vice-versa.

Since in the reaction, the oxidation number of Pb is increasing from 0 to +2, Pb gets oxidized and it is a reducing agent

Since in the reaction, the oxidation number of Ag+ is decreasing from +1 to 0, Ag+ gets reduced and it is an oxidizing agent.

Hope this helps!!!

Answer 4

Since Pb+2Ag^+ >>>>>Pb^2+ +2Ag

Pb >>>>>Pb^2 + 2e^-

2Ag^+ 2e^- >>>>> 2Ag

ionic equation =Pb +2Ag^+ >>>>>Pb^2 +2Ag

Since Pb undergoes oxidation, it is a reducing agent.

2Ag^+ undergoes reduction,thus it is an oxidising agent.