# 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?

10 years ago

10 years ago

## Thanks for your help! I got it now. I messed up with second half reaction, it was supposed to be Cr^2+ + 2e- -> Cr; E = -0.91.

8 years ago

## It actually is 6 electrons when you plug it in Bob. You need 3 for each Au that is fully oxidized.

4 years ago

## If anyone needs this, the correct answer is -1394 kJ/mol. n=6 and Ecell=2.408

1 year ago

## Well, it seems like you're in a bit of a "gold" rush to find the mistake in your calculations. Don't worry, I'm here to help you "dig" it out!

From your half-reactions, it looks like you've got the correct reduction potentials for Au^3+ and Cr^3+. Great job on that!

The error lies in your calculation of Ecell. Remember, in order to calculate Ecell, you need to subtract the reduction potential of the anode (Cr^3+ in this case) from the reduction potential of the cathode (Au^3+ in this case). So the correct equation should be:

Ecell = E(Au^3+/Au) - E(Cr^3+/Cr)

= 1.498 V - (-0.74 V)

= 2.238 V

So your value for Ecell is correct. However, you made a small mistake when plugging it into the equation for standard free-energy change:

Î”G = -nFEcell

You correctly identified n (the number of moles of electrons transferred) as 6, and F (Faraday's constant) as 96485 C/mol.

But here's the "crucial" part: when you plug in the values, make sure to make the units consistent. Since Ecell was given in volts (V), you should use the same unit for Faraday's constant. So, the correct equation should be:

Î”G = -nFEcell

= -(6 mol)(96485 C/mol) (2.238 V)

Now, let's do the math:

Î”G = -1295.675 kJ/mol

So it looks like your previous conversion to kJ was correct, and the standard free-energy change for the reaction is indeed -1295.675 kJ/mol.

I hope this helps! Keep up the good work and never "Au-t" of humor!

11 months ago

## To calculate the standard free-energy change (Î”GÂ°) for a reaction using the Nernst equation, you need to make sure you follow the correct steps.

Here are the steps to calculate Î”GÂ°:

1. Write the two half-reactions involved in the overall reaction:

a. Au^3+(aq) + 3e^- â†’ Au(s) (redox potential: EÂ°(Au^3+/Au))

b. Cr(s) â†’ Cr^2+(aq) + 2e^- (redox potential: EÂ°(Cr^2+/Cr))

2. Determine the overall reaction by summing the half-reactions:

2Au^3+(aq) + 3Cr(s) â†’ 2Au(s) + 3Cr^2+(aq)

3. Find the standard cell potential (EÂ°cell) by subtracting the EÂ° value of the reduction (or oxidation) half-reaction from the EÂ° value of the oxidation (or reduction) half-reaction:

EÂ°cell = EÂ°(reduction) - EÂ°(oxidation)

EÂ°cell = EÂ°(Au^3+/Au) - EÂ°(Cr^2+/Cr)

In your calculation, you correctly evaluated EÂ°cell as 2.238 V.

4. Calculate the standard free-energy change using the equation:

Î”GÂ° = -nF EÂ°cell

In your calculation, you substituted n = 6 (number of electrons transferred), F = 96485 C/mol (Faraday's constant), and EÂ°cell = 2.238 V.

However, there is an error in your calculation. You mentioned that you converted the answer from J to kJ, but you forgot to divide the value by 1000.

So, if we redo the calculation with the correct conversion:

Î”GÂ° = -(6)(96485)(2.238) / 1000 = -1293.87 kJ

Therefore, the correct value for the standard free-energy change is -1293.87 kJ.

It's possible that the discrepancy between your answer and the problem's expected answer might be due to rounding during intermediate steps or using different standard reduction potentials. Make sure to check if the given standard reduction potentials you used are correct.