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1. Assuming these half reactions make a galvanic cell, what is the cell potential in volts.

𝐹2(𝑔) +2π‘’βˆ’β†’πΉ^βˆ’(π‘Žπ‘ž)..........𝐸^oπ‘Ÿπ‘’π‘‘ = 2.87
𝐴𝑙^3+ +3π‘’βˆ’β†’π΄π‘™(𝑠)..............𝐸^π‘œπ‘Ÿπ‘’π‘‘ = βˆ’1.66

2. What is the Gibbs Free energy for this galvanic cell in kJ/mol (hint: make sure your reaction is balanced correctly)

𝐹2(𝑔)+2π‘’βˆ’β†’πΉβˆ’(π‘Žπ‘ž)........𝐸^oπ‘Ÿπ‘’π‘‘ = 2.87
𝐴𝑙3++3π‘’βˆ’β†’π΄π‘™(𝑠)............𝐸^oπ‘Ÿπ‘’π‘‘ = βˆ’1.66
3. Fill in the blank for the equilibrium constant (remember standard temperature is 25^oC):

ln(K) = _________________________

𝐹2(𝑔) +2π‘’βˆ’β†’πΉβˆ’(π‘Žπ‘ž)...........𝐸^oπ‘Ÿπ‘’π‘‘ = 2.87
𝐴𝑙^3+ +3π‘’βˆ’β†’π΄π‘™(𝑠).............𝐸^oπ‘Ÿπ‘’π‘‘ = βˆ’1.66

I think I worked this earlier. See

https://www.jiskha.com/questions/1864203/assuming-these-half-reactions-make-a-galvanic-cell-what-is-the-cell-potential-in-volts

I tried doing the first one

1. Ecell = 1.66 + 2.87 = 4.53
2. deltaG = -nFEcell = - 2mol e-/1mol * 96485 * 4.53 = -874 KJ/mol but it was wrong. What did I do wrong here?
3. I don't know how to step that one up.

n is 6.

The balanced equation is
3*(F2 + 2e ==> 2F^-)
2*(Al ==> Al^3+ + 3e)

3F2 + 2Al --> Al^+3 + 6F^-

1. To find the cell potential in volts (Ecell) for a galvanic cell, you need to add the reduction potentials (EΒ°red) for the half-reactions together. The reduction potential is a measure of the tendency of a species to gain electrons and undergo reduction. The half-reaction with the more positive reduction potential will act as the reduction reaction, and the half-reaction with the more negative reduction potential will act as the oxidation reaction.

In this case, the reduction half-reaction with the more positive EΒ°red is:
F2(g) + 2e- β†’ 2F-(aq) EΒ°red = 2.87 V

And the reduction half-reaction with the more negative EΒ°red is:
Al3+(aq) + 3e- β†’ Al(s) EΒ°red = -1.66 V

Since reduction is happening at the cathode and oxidation at the anode, the cell potential Ecell is given by:

Ecell = Ereduction - Eoxidation

Ecell = 2.87 V - (-1.66 V) = 4.53 V

Therefore, the cell potential in volts is 4.53 V.

2. To find the Gibbs Free energy (Ξ”GΒ°) for the galvanic cell, you need to use the equation:

Ξ”GΒ° = -nFEΒ°cell

Where n is the number of moles of electrons transferred in the balanced equation and F is the Faraday's constant (96,485 C/mol).

Looking at the balanced reactions for the half-reactions, we can see that each reaction involves the transfer of 2 moles of electrons. Hence, n = 2.

Therefore, Ξ”GΒ° = -2 * 96,485 C/mol * 4.53 V = -874,754 J/mol

To convert from Joules (J) to kilojoules (kJ), divide by 1000:

Ξ”GΒ° = -874,754 J/mol / 1000 = -874.754 kJ/mol

Therefore, the Gibbs Free energy for this galvanic cell is -874.754 kJ/mol.

3. To calculate the equilibrium constant (K) from the standard cell potential (EΒ°cell), you can use the equation:

ln(K) = -nF(EΒ°cell)/(RT)

In this equation, n is the number of moles of electrons transferred in the balanced equation, F is the Faraday's constant (96,485 C/mol), R is the ideal gas constant (8.314 J/(molΒ·K)), and T is the temperature in Kelvin.

Since standard temperature is 25Β°C, which is equivalent to 298 K, we can substitute the values:

ln(K) = -2 * 96,485 C/mol * 4.53 V / (8.314 J/(molΒ·K) * 298 K)

Calculating this expression will give us the value for ln(K).

Please note that the value for ln(K) is specific to the given cell potential and temperature.