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

๐น2(๐‘”) +2๐‘’โˆ’โ†’๐น^โˆ’(๐‘Ž๐‘ž)..........๐ธ^o๐‘Ÿ๐‘’๐‘‘ = 2.87
๐ด๐‘™^3+ +3๐‘’โˆ’โ†’๐ด๐‘™(๐‘ )..............๐ธ^๐‘œ๐‘Ÿ๐‘’๐‘‘ = โˆ’1.66

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

๐น2(๐‘”)+2๐‘’โˆ’โ†’๐นโˆ’(๐‘Ž๐‘ž)........๐ธ^o๐‘Ÿ๐‘’๐‘‘ = 2.87
๐ด๐‘™3++3๐‘’โˆ’โ†’๐ด๐‘™(๐‘ )............๐ธ^o๐‘Ÿ๐‘’๐‘‘ = โˆ’1.66
3. Fill in the blank for the equilibrium constant (remember standard temperature is 25^oC):

ln(K) = _________________________

๐น2(๐‘”) +2๐‘’โˆ’โ†’๐นโˆ’(๐‘Ž๐‘ž)...........๐ธ^o๐‘Ÿ๐‘’๐‘‘ = 2.87
๐ด๐‘™^3+ +3๐‘’โˆ’โ†’๐ด๐‘™(๐‘ ).............๐ธ^o๐‘Ÿ๐‘’๐‘‘ = โˆ’1.66

3 answers

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^-