Question

Print out this page and solve the two problems below (attach extra pages as needed). Write the following quantities from your solutions below. (1) Final [Cu]- (2) Equilibrium [Fe]- 1. A test for the completeness of the plating of Cu metal from a solution of Cu"(aq) is to add NHs(aq). A blue color signifies the formation of the complex ion [Cu(NH)'广with K.-1.1 x 10', Let 250.0 mL oro. 1000 M CuSO4(aq) be electrolyzed with a 3.512 A current for 1368 s. At the end of this time, add enough NIH,(aq) to complex any remaining Cu2+ and to maintain a free [N1b] = 0-10 M. If [Cu(NII山]2 is detectable at concentrations as low as 1 x 10 M, should the blue color appear? 2. A Galvanic cell is constructed based on the following reaction and initial concentrations: Fe2. (0.0050 M) Ag, (2.0 M) Fel. (0.0050 M) + Ag(s) + Calculate [FeJ when the cell reaction reaches equilibrium

Answer 1

58547 . Amount of Coppes lt behide 88 Ts- 1s854 2 2 0:003193 9m moles32930.oo0 SIT4 mle Com centeoct emaing G en -L S9-yemaining_. CnContente ahon..is-geher han

Final [Cu²⁺] = 0.0002M

For Ag/Ag+ half reaction

E = 0.8 + 0.0591log [Ag+]

For Fe2+/Fe3+ half reaction

E = 0.771 - 0.0591log([Fe2+]/[Fe3+])

don't need to use n since n=1

The overall reaction is Fe²⁺ + Ag⁺ ===> Fe³⁺ + Ag

and K = [Fe³⁺]/([Fe²⁺] [Ag⁺])

Therefore:
0.80 - 0.771 = 0.0591log[Fe3+]/([Fe2+][Ag+]) or log K = 0.491. so K = 3.10

Using the I.C.E table then will get at equilibrium

[Fe2+] = 0.005 - X
[Ag+] = 2 - X

[Fe3+] = 0.005 + X * adding cause its producing
[Ag(s)] = nothing its a solid it doesn't apply to equilibrium

therefore use the quadratic equation

3.10 = [0.005-x] / [2 -X][0.005-X]
Solving for X we have
[Fe2+] = 0.00304M