I can't do this completely but your answer for A is correct. However, I don't like the way it is done.
If you write the Pb reaction as you have it,
Anode (Oxidation): Pb(s) -> Pb ^+2 (aq)+ 2e^- E^0 = -0.126V
it is written AS AN OXIDATION and Eo AS WRITTEN is +1.26 and
Eo cell = Eas ab oxidation + Eas a reduction = 0.126 + 1.61 = 1.736 v which I would round to 1.74 v. I see on the web a number of sites that are doing it as you have calculated but I think that is confusing (at least confusing to me). I think oxidations must be treated as oxidations and reductions as reductions.
I didn't go through the E from log Q because I couldn't tell what the concentrations were. I don't think Pb(s) can be 0.015 and you have only one number listed for the Ce^3+/Ce^+4 couple.
I'm not sure if I'm doing the problem correctly and would appreciate if someone could take a look. Thank you.
Question: For the following cell reaction at 25 degree C
Ce^+3 | Ce ^+4 0.80M || Pb ^+2 | Pb^0 0.015M |
A) Calculate E^0 for the cell.
Anode (Oxidation): Pb(s) -> Pb ^+2 (aq)+ 2e^- E^0 = -0.126V
Cathode (Reduction): Ce^+4 (aq) + e^- -> Ce^+3(aq) E^0 = +1.61V
Pb(s) + 2 Ce^+4(aq) -> Pb^+2 (aq) + 2 Ce^+3 (aq)
E^0 = E_cathode- E_anode
E^0 = 1.61 V - (-0.126V)
E^0 = 1.736V
B) Calculate the EMF (E) for the cell under the given conditions
E = E^0 - 0.0592/n *logQ
E = 1.736V - 0.0592/2 * log [0.80M] ^2/ [0.015M]
E = 1.69V
C) Calculate K_eq cell reaction at 25 degree C under standard conditions
E_cell = 0.0592/n * logK
1.69V = 0.0592/2 * logK
2(1.69V)/0.0592 = logK
K_eq = 1.24 x 10^27
D) Calculate delta G
delta G_rxn = -nFE^0
delta G_rxn = -2 mol (9.6485 x 10^4 J/V*mol) (1.69V)
delta G_rxn = -3.26 x 10^5 J
J) Spontaneous?
Yes
2 answers
I'm sorry it was confusing. The question was written just like I've typed there. I'm not sure if it was a typo or if the question was written poorly. Thank you for your help though.
1 answer