Question

For a concentration cell:

Zn/Zn2+ (0.05M)//Zn2+(1M)/Zn

E= 0.035 volt

a. What can you say about the measured emf of this cell and the concentration cell of Cu/Cu2+ (0.05M)//Cu2+(1M)/Cu ?

b. Why is the standard emf (E) not a factor in these two cells?

c. Use the Nernst equation to calculate the emf of this cell.

Answer 1

b)

it is a concentration cell

the cathode and anode is made up of same material

we know that

Eo cell = EO cathode - EO anode

as cathode and anode are same

Eo cell= EO cathode - EO cathode

Eo cell = 0

so

the standard EMF is not a factor in these two cells

c)

according to nernst equation

E= EO - ( 0.0591/n) log Q

consider the cell reactio

Zn(s) + Zn+2 (cathode) ----> Zn+2 (anode) + Zn

so the reaction quotient Q is given by

Q= [Zn+2]anode / [Zn+2] cathode

using given values

Q= 0.05 /1

Q = 0.05

also

consider the oxidation reaction : anode

Zn ----> Zn+2 + 2e-

as two electrons are transferred

n=2

now

E = EO - ( 0.0591/n) log Q

E= 0 - ( 0.0591/2) log 0.05

E= 0.038

so

the emf of this cell is 0.038

a)

now the measured emf of both the cell are same because

both are concentration cell , i.e Eo = 0

in both the cells two electrons are transferred i.e n=2

both the cells have the same conc of anode and cathode

so

the emf of both the cells is same