Question

Compare the potentials between the Zn/Zn2+ and Cu/Cu2+ half cells before and after adding excess NH3 to the Cu/Cu2+ half cell. Did the potential increase or decrease? Explain the result in terms of the Nernst equation.

Answer 1

the reduction potential values are

Eo Zn+2/Zn = -0.7618 V

Eo Cu+2/Cu = + 0.337 V

the electrode with more postive reduction potential act as cathode

so

Cu as cathode

we know that

At cathode reduction takes place

So

Cathode reaction :

Cu+2 + 2e- ----> Cu

now

Zn acts as anode

we know that

At anode oxidation takes place

So

Anode reaction :

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

th final reaction is

Zn + Cu+2 ---> Zn+2 + Cu

now

according to nernst equation

E= Eo - ( 0.05916/n) log Q

consider the final reaction

Zn (S) + Cu+2 (aq) ---> Zn+2 (aq)   + Cu (S)

reaction quotient is given by

Q = [Zn+2] /[Cu+2]

also

two electrons are transferred in the reactio

so n=2

E= Eo - (0.05916/2) log [Zn+2/Cu+2]

E= Eo - 0.02958 log [Zn+2/Cu+2]

E= Eo + 0.02958 log [Cu+2/Zn+2]

now

given that

NH3 is added to Cu/Cu+2 half cell

we know that

Cu+2 reacts with Nh4 to given [Cu(NH3)4]+2

Cu+2 + 4NH3 ----> [Cu(NH3)4]+2

so on addition of NH3 , the conc of Cu+2 decreases

now consider the nernst equation

E= Eo + 0.02958 log [Cu+2/Zn+2]

the conc of Cu+2 decreases

so by looking at the above equation we can say that

As [Cu+2] decreases the potential also decreases

SO

finally

on addition of NH3 , the potential will decrease