Question

# A Zn wire and Ag/AgCl reference electrode saturated with KCl (E = 0.197 V) are placed...

A Zn wire and Ag/AgCl reference electrode saturated with KCl (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 -1.067 V. The standard reduction potential of the Zn2 /Zn half-reaction is –0.762 V. Please show all the steps.

Step1

Zn2+(aq) +2e- → Zn(s) Eº=-0.762 V

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

Step2: Nernst equation for the net reaction and put the values.

Know quantities:

E =-1.061

Eº (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 and E- is the potential of the negative terminal.

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

Putting the values:

-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+])

103.448= (1/[Zn2+])=2805 M-1

[Zn2+]=1/2805 M-1=3.56 x10-4 M

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