[Sis290025]

Please tell me if my final answers are correct.

1) Determine the number of grams of silver deposited when 19.000A are passed through an aqueous silver nitrate solution for 21.000 minutes. Report your answer to 2 decimal places.


A = 19.000
t = 21.000 min(60 s/1 min) = 1260 s

19.000 A(1260 s) = 23940 C(1 mol e-/96500 C)(1 mol Ag/1 mol e-)(107.868 g Ag/1 mol Ag) = 26.7602 g Ag
= 26.76 g Ag


2) Calculate the cell potential for the following reaction:
Cu+2 (0.1014M) + H2 (1atm) <--> Cu (s) + 2H+ (pH=3.73)
You are not at standard state.

Cu (2+) --> Cu reduced
H2 --> 2H+ + 2e- oxidized

pH = -log[H+]
antilog(-3.73) = 1.862087137E-4 M

E* = E_red + E_ox
= E_Cu + E_H2
= 0.34 V + 0V = 0.34 V

Use E = E* - (0.0592 V/n)*log Q

where Q = [H+]^2/[Cu(2+)](PH2)
Q = [1.862087137E-4 M]^2/[0.104 M](1 atm) = (3.467368505E-8]/[0.1014 M] = 3.419495567E-7

E = 0.34 V - (0.0592 V/2 mol e-)* log(3.419495567E-7)
= 0.34 - 0.0296(-6.466037955)
= 0.5313947
= 0.53 V


Once again, thank you very much.