Using your experimental data, calculate E0cell. Standard voltaic cells contain 1M solutions, but we used 0.10M solutions in all of our half cells. Use the Nernst equation to calculate the standard cell potential (E0cell) from your measured Ecell and Qrxn.

Data:
Nernst Equation: Ecell=E0cell -RT/nF ln Q

Questions:

1. I am not sure if i did the E0cell right equation, I switched the Ecell to get the E0cell..and "n" is equal to 2 moles being transfer?

I calculated:
Measured Ecell = 0.445V
Overall rxn = Cu(s)+Pb2(aq) ~ CU2+(aq) + Pb(s)
N= 2moles??

Re-arrange the equation:
E0cell = Ecell - RT/nF ln Q
E0cell = 0.445V -(8.314 j/k*mol *298K)/(2 moles * 96500 c/mole e-) ln (0.10)
= -2.30

2. Overall reaction:
3Pb(s) + 2Al 3+ (aq) ~ 3Pb2+(aq) = 2Al(s)
Measured Ecell(V) = 0.086V
N= 6moles?? (3*2 moles)

0.086V- (8.314 j/k*mol * 298K)/ (6moles * 96500 C/mole e- ) ln (0.10/0.10^6) = 0.94??

Is this right? Please help! Thank you in advance!

Let's go through your calculations step by step:

1. For the first question, you have correctly rearranged the Nernst equation to solve for E0cell. The usage of n in the equation depends on the stoichiometry of the reaction. In this case, since 2 moles of electrons are being transferred in the overall reaction, you should use n = 2.

Using the given values:

Measured Ecell = 0.445 V
R = 8.314 J/(K*mol)
T = 298 K
F = 96500 C/mol e-
Q = 0.10 (assuming you have a simplified expression representing the equilibrium constant)

Plugging these values into the equation:

E0cell = 0.445 V - (8.314 J/(K*mol) * 298 K)/(2 mol * 96500 C/mol e-) * ln(0.10)
= 0.445 V - (8.314 J/(K*mol) * 298 K)/(2 mol * 96500 C/mol e-) * (-2.3026)
= 0.445 V + 0.0592 V
= 0.504 V

Therefore, the calculated E0cell is 0.504 V.

2. For the second question, your setup is again correct. Since 6 moles of electrons are being transferred in the overall reaction, you should use n = 6.

Using the given values:

Measured Ecell = 0.086 V
R = 8.314 J/(K*mol)
T = 298 K
F = 96500 C/mol e-
Q = 0.10/0.10^6 (assuming you have a simplified expression representing the equilibrium constant)

Plugging these values into the equation:

E0cell = 0.086 V - (8.314 J/(K*mol) * 298 K)/(6 mol * 96500 C/mol e-) * ln(0.10/0.10^6)
= 0.086 V - (8.314 J/(K*mol) * 298 K)/(6 mol * 96500 C/mol e-) * (13.8155)
= 0.086 V - 0.116 V
= -0.030 V

Therefore, the calculated E0cell is -0.030 V.

Your calculations look correct based on the given information. Just make sure to double-check your values and units to ensure accuracy.

1. E0cell = Ecell - RT/nF ln Q

Ecell = 0.445V
R = 8.314 J/(K*mol)
T = 298K
n = 2 (since 2 moles of electrons are transferred)
F = 96500 C/mol e-
Q = 0.10 (assuming Q refers to the reaction quotient)

Plugging in the values:
E0cell = 0.445V - ((8.314 J/(K*mol)) / (2 * 96500 C/mol e- )) * ln(0.10)
E0cell = 0.445V - (0.00004299) * ln(0.10)
E0cell ≈ 0.445V - (-1.1474)
E0cell ≈ 1.592V

Therefore, E0cell ≈ 1.592V.

2. E0cell = Ecell - RT/nF ln Q
Ecell = 0.086V
R = 8.314 J/(K*mol)
T = 298K
n = 6 (since 6 moles of electrons are transferred)
F = 96500 C/mol e-
Q = (0.10) / (0.10^6) (assuming Q refers to the reaction quotient)

Plugging in the values:
E0cell = 0.086V - ((8.314 J/(K*mol)) / (6 * 96500 C/mol e- )) * ln((0.10) / (0.10^6))
E0cell = 0.086V - (0.00001417) * ln(1000)
E0cell ≈ 0.086V - (0.00001417) * 6.9084
E0cell ≈ 0.086V - 0.00009757
E0cell ≈ 0.085V

Therefore, E0cell ≈ 0.085V.

Please note that these calculations are approximate and may vary depending on the exact values used for R and F.