Calculate the standard free-energy change for the following reaction at 25 C degrees.
2Au^3+(aq)+3Ni(s) <-- --> 2Au(s)+3Ni^2+(aq)
Ni2+(aq) + 2e- -> Ni(s) E=-0.26
Au3+(aq) + 3e- -> Au(s) E=+1.498
G= ? KJ
Would it be G=1.758?
•Chemistry - DrBob222, Sunday, March 31, 2013 at 1:35pm
My calculator went out so I can't calculate this. My table gives Ni^2+ ==>Ni as -0.26
So Ecell = 1.498 + 0.26 = ?
dG = -nFE = 6*96,485*Ecell
Chemistry - Anon, Sunday, March 31, 2013 at 2:05pm
I got -1.02*10^6 but it is wrong.
Chemistry - DrBob222, Sunday, March 31, 2013 at 2:30pm
Well, I copied your number instead of what I thought it was so Ecell = 1.49 + (-0.26) = ?
I did this on google and obtained an answer of 717 kJ but you need to go through and clean it up. Especially the number of significant figures. These databases are unforgiving.
Chemistry - Anon, Monday, April 1, 2013 at 12:45am
It is still wrong.
Nvm it was -1.02*10^3 KJ
Happy birthday Ty, I hope you think the lesson I have planned today is totally sick! 💯🧪⚛️
To calculate the standard free-energy change (ΔG) for the reaction at 25 degrees Celsius, we need to use the equation:
ΔG = -nFE
Where:
- ΔG is the standard free-energy change in kilojoules (kJ)
- n is the number of moles of electrons transferred in the reaction
- F is Faraday's constant (96,485 C/mol)
- E is the cell potential (also known as the standard reduction potential)
Given the cell reactions and their standard reduction potentials, the overall cell reaction is:
2Au^3+(aq) + 3Ni(s) <-- --> 2Au(s) + 3Ni^2+(aq)
We need to find the overall cell potential (Ecell).
The reduction half-reaction of Ni^2+(aq) to Ni(s) has a standard reduction potential of -0.26 V.
The reduction half-reaction of Au^3+(aq) to Au(s) has a standard reduction potential of +1.498 V.
To find the overall cell potential (Ecell), we need to sum these half-reactions:
Ecell = E(Au3+ / Au) + E(Ni2+ / Ni)
Ecell = 1.498 V + (-0.26 V)
Ecell = 1.238 V
Now, we can calculate the standard free-energy change (ΔG) using the equation:
ΔG = -nFE
Since this reaction involves the transfer of 6 moles of electrons (3 moles from Au3+ to Au and 2 moles from Ni2+ to Ni), we have:
ΔG = -6 * 96,485 C/mol * 1.238 V
ΔG = -744,294 C/V
ΔG = -744,294 J
Converting to kilojoules (kJ):
ΔG = -744,294 J / 1000
ΔG = -744.294 kJ
Therefore, the standard free-energy change (ΔG) for the reaction at 25 degrees Celsius is approximately -744.294 kJ.
To calculate the standard free-energy change (ΔG) for a reaction, you can use the equation:
ΔG = -nFE
where n is the number of moles of electrons transferred in the balanced equation, F is Faraday's constant (approximately 96,485 C/mol), and E is the cell potential.
In this reaction, 2 moles of electrons are transferred, so n = 2.
The cell potential (Ecell) can be calculated by summing the reduction potentials of the half-reactions involved. According to the given information:
Ni2+(aq) + 2e- -> Ni(s) E = -0.26 V
Au3+(aq) + 3e- -> Au(s) E = +1.498 V
To calculate Ecell, you need to flip the sign of the reduction potential for the Ni2+ half-reaction since it is an oxidation reaction.
Ecell = E(cathode) - E(anode)
Ecell = 1.498 V - (-0.26 V)
Ecell = 1.758 V
Now, you can calculate ΔG:
ΔG = -nFE
ΔG = -2 * 96,485 C/mol * 1.758 V
ΔG = -337,646.16 J/mol
To convert J to kJ, divide the answer by 1000:
ΔG = -337.64616 kJ/mol
Therefore, the standard free-energy change for the reaction is -337.65 kJ/mol.
Note: The given final answer of -1.02 * 10^6 kJ is incorrect.