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27 Nov 2019
Write the cell reaction for the following voltaic cell andcalculate the concentration of copper (II) in the cell at 25degrees Celsius given that Ecell= 0.25 volts. Be sure to show allwork(ie: E^o cell, n and Q). Fe |Fe^+3(1.25M) || Cu^+2 (? M) | Cu I just need you guys to look at my work and tell me what I'mdoing wrong so that I can get the correct answer of x = 5.07 x10^-5 which is the answer that our prof gave us for thisproblem. E=E^o - 0.059/n (log Q) Q = [Fe+2]^2/[Cu+2]^3
My half reactions are: Fe ---> Fe3+ +3e- 0.04volts
Cu+2 + 2e- ---> Cu 0.34 volts So the E^0 of this cell is 0.38 volts.
Now I have to make the electrons balance out in both equationsbecause the Fe half reaction has 3 and the Cu has 2. So I multiplythe Fe one by 3 and the Cu one by 2 and get these newhalf-reactions:
2Fe ---> 2Fe3+ + 6e-
3Cu+2 + 6e- ---> Cu
Combining these reactions gives me the cell reaction for thevoltaic cell: 2Fe + 3Cu2+ ---> 2Fe3+ + 3Cu I know I have everything above correct, but something goeswrong when I try to plug my numbers into the Nerstequation.
So I plug my voltage into the Nerst equation of E=E^o - 0.059/n X (log Q) 0.25 volts = 0.38 volts - (.059/6) (log Q) -0.13 volts = -(.059/6) (log Q) -.013 volts = (-.0098333333) (log Q) 1.322033903 = - (log Q)
-1.322033903 = log Q 10^-1.322033903 = .0476393796 Q= .0476393796 .0476393796 = [Fe+3]^2/[Cu+2]^3 .0476393796 = [1.25]^2/[x]^3 .0476393796[x]^3 = 1.25^2 .0476393796[x]^3 = 1.5625
[x]^3 = 32.79849783 [x] = 3.20 [Cu2+] = 3.20 M My teacher gave us the answer as [Cu2+] = 5.07 X 10^-5 M, so Iam way off.
Write the cell reaction for the following voltaic cell andcalculate the concentration of copper (II) in the cell at 25degrees Celsius given that Ecell= 0.25 volts. Be sure to show allwork(ie: E^o cell, n and Q).
Fe |Fe^+3(1.25M) || Cu^+2 (? M) | Cu
I just need you guys to look at my work and tell me what I'mdoing wrong so that I can get the correct answer of x = 5.07 x10^-5 which is the answer that our prof gave us for thisproblem.
E=E^o - 0.059/n (log Q) Q = [Fe+2]^2/[Cu+2]^3
My half reactions are:
Fe ---> Fe3+ +3e- 0.04volts
Cu+2 + 2e- ---> Cu 0.34 volts
So the E^0 of this cell is 0.38 volts.
Now I have to make the electrons balance out in both equationsbecause the Fe half reaction has 3 and the Cu has 2. So I multiplythe Fe one by 3 and the Cu one by 2 and get these newhalf-reactions:
2Fe ---> 2Fe3+ + 6e-
3Cu+2 + 6e- ---> Cu
Combining these reactions gives me the cell reaction for thevoltaic cell:
2Fe + 3Cu2+ ---> 2Fe3+ + 3Cu
I know I have everything above correct, but something goeswrong when I try to plug my numbers into the Nerstequation.
So I plug my voltage into the Nerst equation of
E=E^o - 0.059/n X (log Q)
0.25 volts = 0.38 volts - (.059/6) (log Q)
-0.13 volts = -(.059/6) (log Q)
-.013 volts = (-.0098333333) (log Q)
1.322033903 = - (log Q)
-1.322033903 = log Q
10^-1.322033903 = .0476393796
Q= .0476393796
.0476393796 = [Fe+3]^2/[Cu+2]^3
.0476393796 = [1.25]^2/[x]^3
.0476393796[x]^3 = 1.25^2
.0476393796[x]^3 = 1.5625
[x]^3 = 32.79849783
[x] = 3.20
[Cu2+] = 3.20 M
My teacher gave us the answer as [Cu2+] = 5.07 X 10^-5 M, so Iam way off.
Beverley SmithLv2
24 Feb 2019