The problem is: Calculate the standard free-energy change for the following reaction at 25 degrees Celsius.
2Au^+3(aq) + 3Cr(s)-> 2Au(s) + 3Cr^+2(aq)
I worked it out, getting half reaction:
Au^3+ + 3e- -> Au; E = 1.498
Cr^3+ + 3e- -> Cr; E = -0.74
Ecell = 1.498 - (-0.74) = 2.238
n = 6
F = 96485
I plugged it into equation changeG = -nFEcell and got -1295.6 kJ (answer is supposed to be in kJ, so I converted it from J). However, the problem says I'm wrong, and I'm wondering at which point I messed up?
See your other post. delta e wrong at 6.
Thanks for your help. Sorry, I accidentally posted the question again without meaning to.
To calculate the standard free-energy change (ΔG) for the reaction, you correctly used the equation:
ΔG = -nFEcell
However, there seems to be a mistake in your calculation of Ecell. The Ecell value should be the difference between the reduction potentials of the two half-reactions, not the sum. Therefore, the correct calculation should be:
Ecell = E(Au^3+ → Au) - E(Cr^3+ → Cr)
= 1.498 V - (-0.74 V)
= 2.238 V
Next, you correctly identified that the number of electrons involved in the reaction (n) is 6 and the Faraday constant (F) is 96485 C/mol.
Now, let's plug in the corrected values into the equation:
ΔG = -nFEcell
= -(6 mol)(96485 C/mol)(2.238 V)
= -1296176 J
Converting from joules to kilojoules:
ΔG = -1296176 J / 1000
= -1296.2 kJ
Therefore, the standard free-energy change for the reaction is -1296.2 kJ (rounded to one decimal place).
Please note that your initial calculation error in the value of Ecell caused the discrepancy.
To calculate the standard free-energy change (ΔG) for a reaction at a given temperature, you need to use the equation:
ΔG = -nFEcell
where ΔG is the standard free-energy change, n is the number of moles of electrons transferred in the balanced redox reaction, F is the Faraday constant (96485 C/mol), and Ecell is the cell potential.
In your calculation, you correctly determined the cell potential (Ecell) by subtracting the reduction potential of the Cr half-reaction from the reduction potential of the Au half-reaction:
Ecell = 1.498 - (-0.74) = 2.238 V
This means that the cell potential for the reaction is 2.238 V.
However, it seems that there might be a mistake in the value you used for the number of moles of electrons transferred (n). Let's take a closer look at the balanced redox reaction:
2Au^3+(aq) + 3Cr(s) -> 2Au(s) + 3Cr^2+(aq)
From the equation, we can see that 3 moles of electrons are transferred for every 2 moles of Au^3+ and for every 3 moles of Cr.
To calculate n, we need to determine the number of moles of electrons transferred for the entire balanced reaction. We can do this by finding the least common multiple (LCM) of the coefficients in front of Au^3+ and Cr:
LCM(2,3) = 6
Therefore, the value of n is 6, not 6/2 or 6/3.
Now that we have the correct value for n, we can recalculate ΔG:
ΔG = -nFEcell
= -(6)(96485 C/mol)(2.238 V)
= -1298206.3 J
Converting this to kJ by dividing by 1000, we get:
ΔG = -1298206.3 J / 1000 = -1298.2 kJ
So, the correct answer for the standard free-energy change of the reaction at 25°C is -1298.2 kJ. You obtained -1295.6 kJ in your calculation, which is close but not quite correct. The difference is likely due to rounding errors during the calculation process.