Please check a Nernst equation for me?
Calculate the cell potential for the following reaction as written at 61 °C, given that [Zn2 ] = 0.800 M and [Fe2 ] = 0.0170 M.
Half reactions:
Zn-->Zn^2+ + 2e^- (potential= -0.76)(anode)
Fe^2+ + 2e^- -->Fe (potential= -0.44)
(cathode)
E^nought=cathode-anode
E^nought=(-0.44)-(-0.76)
E^nought=0.32
Q=products/reactants
Q=.8M/.0170
Q=47.0588
(lnQ=3.851)
E=E^nought-(RT/nF)lnQ
E=[.32-(8.3145x334.15)/(2x96485)]3.851
E=1.177
Thank you for your help!
It looks ok until the final step. Isn't that 0.32 - some number. So how can it be larger than 0.32?
Doesn't make sense to me either but I can't figure it out :( The way I arrived at that answer was:
E=[.32-(8.3145x334.15)/(2x96285)]3.851
E=[.32-(2778.29)/192570)]3.851
E=(.32-.014427)3.851
E=(.305573)3.851
E=1.177
It's not right but for the life of me I don't know why. Maybe I'm doing something out of order?
You are. You can't subtract terms like that.
You have
0.32 - [(8.3145*334.15)/(2*96,485)]3.851 =
0.32 - (2778.29)/192,570)*3.851
0.32 - (0.01443)3.851
Next you complete the multiplication you started 3 steps above.
0.32 - (0.05556)
Now you subtract.
0.32 - 0.05556 = ? about 0.26444 which is 0.26 to 2 s.f. (You're allowed only 2 since this is subtraction; we can have hundredths.)
That was it!
Apparently it has been WAY too long since I did order of operations but I have refreshed myself now. I should have known better! Thank you for your help.
To check the Nernst equation calculation, we need to determine if the calculated cell potential of 1.177 V is correct for the given conditions.
The Nernst equation is given by:
E = E^0 - (RT/nF) * ln(Q)
where:
E = cell potential
E^0 = standard cell potential (the difference in potential between the cathode and the anode)
R = gas constant (8.3145 J/(mol⋅K))
T = temperature in Kelvin (61 °C = 334.15 K)
n = number of moles of electrons transferred in the balanced equation (2 in this case)
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient (concentrations of products divided by concentrations of reactants)
So, let's plug in the values into the equation:
E = 0.32 - ((8.3145 * 334.15) / (2 * 96485)) * 3.851
Calculating this:
E ≈ 1.177 V
Therefore, the calculation is correct, and the cell potential for the given reaction at 61 °C is approximately 1.177 V.