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
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 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.
To calculate the standard cell potential (E^0) for the given cell reaction, you need to subtract the standard reduction potential of the anode from the standard reduction potential of the cathode. The standard reduction potentials can be found in a table of standard reduction potentials.
In this case, the anode (oxidation) is Pb(s) -> Pb^+2(aq) + 2e^-, which has a standard reduction potential of -0.126V. The cathode (reduction) is Ce^+4(aq) + e^- -> Ce^+3(aq), which has a standard reduction potential of +1.61V.
E^0 = E_cathode - E_anode
E^0 = 1.61V - (-0.126V)
E^0 = 1.736V
To calculate the cell EMF (E) under the given conditions, you can use the Nernst equation. The Nernst equation relates the cell potential (E) to the standard cell potential (E^0), the number of electrons involved in the balanced cell reaction (n), the temperature (T) in Kelvin, and the reaction quotient (Q).
E = E^0 - (0.0592/n) * log(Q)
For the given cell:
E = 1.736V - (0.0592/2) * log([Ce^3+]/[Pb^2+])
E = 1.736V - (0.0296) * log([0.80M]^2/[0.015M])
E = 1.69V
To calculate the equilibrium constant (K_eq) for the cell reaction, you can use the relation between the cell potential (E) and the equilibrium constant. The equation is:
E_cell = (0.0592/n) * log(K_eq)
For the given cell:
1.69V = (0.0592/2) * log(K_eq)
2(1.69V)/0.0592 = log(K_eq)
K_eq = 1.24 x 10^27
To calculate the change in Gibbs free energy (delta G) for the cell reaction, you can use the equation:
delta G_rxn = -nFE^0
Where delta G_rxn represents the change in Gibbs free energy, n is the number of moles of electrons transferred in the balanced cell reaction, F is the Faraday constant (9.6485 x 10^4 J/V*mol), and E^0 is the standard cell potential.
For the given cell:
delta G_rxn = -2 mol (9.6485 x 10^4 J/V*mol) (1.69V)
delta G_rxn = -3.26 x 10^5 J
Finally, to determine if the cell reaction is spontaneous, you can check the sign of delta G_rxn. If delta G_rxn is negative, the reaction is spontaneous. In this case, delta G_rxn is -3.26 x 10^5 J, which is negative, so the reaction is spontaneous.