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
1. Assuming these half reactions make a galvanic cell, what is the cell potential in volts.
๐น2(๐) +2๐โโ๐น^โ(๐๐)..........๐ธ^o๐๐๐ = 2.87
๐ด๐^3+ +3๐โโ๐ด๐(๐ )..............๐ธ^๐๐๐๐ = โ1.66
2. What is the Gibbs Free energy for this galvanic cell in kJ/mol (hint: make sure your reaction is balanced correctly)
๐น2(๐)+2๐โโ๐นโ(๐๐)........๐ธ^o๐๐๐ = 2.87
๐ด๐3++3๐โโ๐ด๐(๐ )............๐ธ^o๐๐๐ = โ1.66
3. Fill in the blank for the equilibrium constant (remember standard temperature is 25^oC):
ln(K) = _________________________
๐น2(๐) +2๐โโ๐นโ(๐๐)...........๐ธ^o๐๐๐ = 2.87
๐ด๐^3+ +3๐โโ๐ด๐(๐ ).............๐ธ^o๐๐๐ = โ1.66
3 answers
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.
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^-
The balanced equation is
3*(F2 + 2e ==> 2F^-)
2*(Al ==> Al^3+ + 3e)
3F2 + 2Al --> Al^+3 + 6F^-