A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4.

Question

A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4. The Zn wire is attached to the positive terminal and the Ag/AgCl electrode is attached to the negative terminal of the potentiometer. Calculate the [Zn2+] in the solution if the cell potential, Ecell, is -1.061 V. The standard reduction potential of the Zn2+/Zn half-reaction is –0.762 V.

 

[Zn2+]= ? M

Answer 1

 Step1: Write the two half reactions  and their standard potentials

 

2AgCl(s) +2e^- → 2Ag(s) + 2Cl^- (aq)                Eº=0.197 V

Zn²⁺(aq) +2e^- → Zn(s)                                   Eº=-0.762 V

 

Step2: Write a Nernst equation for the net reaction and put in all known quantities.

 

Know quantities:

E =-1.061

E^º(Ag/AgCl (sat. KCl)=0.197 V

Eº(Zn²⁺/Zn=-0.762 V

 

E=(E₊ - E_-)

Where, E₊ is the potential of the electrode attached to the positive terminal of the potentiometer and E_- is the potential of the negative terminal of the potentiometer.

 

E=(E₊ - E_-)=[Eº(Zn²⁺/Zn) - (0.05916/2)log (1/Zn²⁺)]-[Eº(Ag/AgCl) - (0.05916/2) log(1/Cl^-)]

 

Put in known quantities:

 

-1.061 V=[-0.762 V - (0.05916/2)log (1/[Zn²⁺])]-[0.197 V-0)

-1.061 V=-0.762 V-0.197 V- (0.05916/2)log (1/[Zn²⁺])

-1.061V + 0.959V = -0.05916/2 log (1/[Zn²⁺])

(-0.102 x2)/-0.05916 = log (1/[Zn²⁺])

3.448 V= log (1/[Zn²⁺])

10^ (3.448)= (1/[Zn²⁺])=2805 M^-1

[Zn²⁺]=1/2805 M^-1=3.56 x10^-4 M