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¨C ¡ú Ni(s) E=¨C0.26
Au3+(aq) + 3e¨C ¡ú Au(s) E=+1.498

G= ? KJ

Would it be G=1.758?

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

I got -1.02*10^6 but it is wrong.

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.

It is still wrong.

To calculate the standard free-energy change (ΔG) for the given reaction at 25°C, you need to use the Nernst equation and the standard reduction potentials (E°) of the half-reactions involved.

First, write the balanced equation for the overall reaction:

2Au^3+(aq) + 3Ni(s) ↔ 2Au(s) + 3Ni^2+(aq)

Next, find the standard cell potential (E°cell) by subtracting the reduction potential of the anode reaction from the reduction potential of the cathode reaction:

E°cell = E°cathode - E°anode

In this case, the reduction potential for the Au3+ + 3e- → Au(s) half-reaction is +1.498 V, and the reduction potential for the Ni2+ + 2e- → Ni(s) half-reaction is -0.26 V.

E°cell = (+1.498 V) - (-0.26 V)
E°cell = +1.758 V

Now, use the formula for the relationship between ΔG and E°cell:

ΔG° = -nFE°cell

Where:
- ΔG° is the standard free-energy change
- n is the number of electrons transferred in the balanced equation
- F is Faraday's constant (F = 96485 C/mol)

In this case, the balanced equation shows that 3 electrons are transferred, so n = 3.

ΔG° = -(3)(96485 C/mol)(1.758 V)
ΔG° = -516051.63 J/mol

Finally, convert the value to kilojoules (kJ):

ΔG° = -516.05 kJ/mol

Therefore, the standard free-energy change for the given reaction at 25°C is approximately -516.05 kJ/mol.