Question

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.