Answer to Question #71738 in General Chemistry for Whitney

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.

[Zn^2+]= ? M

Answer 1

Ecell = -1.061 V
E0Zn2+ /Zn = –0.762 V
E0 Ag/AgCl = +0.197 V
Ecell = {E0Zn2+ /Zn – 0.05916/2* log[Zn2+]/[Zn]} – { E0 Ag/AgCl - 0.05916* log[Сl-]} =
-1.061 = –0.762 - 0.05916/2* log[Zn2+]/[Zn] – 0.197
-0.102 = - 0.05916/2* log[Zn2+]/[Zn]
log[Zn2+]= 3.448
[Zn2+]= 2805.43