Question

Use the table of Standdard Reduction Potentials and the Nernst Equation to calculate the concentration of Cu2+ present in the Cu/Cu2+ half-cell after the addition NH3 (aq) while it was coupled with

a) Zn/Zn2+

b) Ag/Ag+

Table 2 has your measured voltages for these cells. SHOW YOUR SETUP OF THE NERST EQUATION WITH ALL VARIABLES FILLED IN. SHOW THE CALCULATED [Cu2+] CONCENTRATION.

Table 2:

METAL	Mg	Ag	Ni	Zn	Pb
Cu	1.230 V	0.642 V	0.030 V	0.648 V	0.256 V

Answer 1

Yoou are missing the concentrations of Zn and Ag, to easily calculate that. However, let me show you how you would do it so you can get an idea of how to solve it, once you use the concentrations of those ions. You can also assume a concentration of Ag and Zn, maybe 1 M (standard condition) or 0.1 M to do it:

a) E^0 Zn /Zn^+ 2 = -0.76 v E^0 Cu/Cu^+ 2 = 0.34 v Zn (s) rightarrow Zn^+ 2 + 2 e^- Cu^2 + + 2 e^- rightarrow Cu 0.76 0.34 E^0 = 1.10 v E = E^0 = 0.059/n log [Zn^2 +] /[Cu^2 +] (E - E^0) times n/0.059 = -log [Zn^2 +]/[Cu^2 +] 10^[- (E - E^0) times n/0.059] [Zn^2 +]/[Cu^2 +] [Cu^2 +] = [2 n^2 +]/10^- [(E - E^0) n/0.059] Assuming [Zn^2 +] = 1M [Cu^2 +] = 1/10^[Z(0.648 - 1.10) times 2/0.059] [Cu^2 +] = 4.76 times 10^- 16 M b) E^0 Ag/Ag^+ = 0.80 v E^0 = 0.80 - 0.34 = 0.46 v [Cu^2 +] = 1 times 10^- [0.642 - 0.46/0.0295] [Cu^2 +] = 1 times 6.77 10^- 7 [Cu^2 +] = 6.77 times 10^- 7 M