Use the Nernst equation and data from Appendix D in the textbook to calculate E cell for each of the following cells.
Mg(s)|Mg+2 (0.012 M) || [Al(OH)4]- (0.25M), OH- (0.048M)|Al(s)
Ecell = Ecell(standard) - 0.0591/n * log Q
The Attempt at a Solution
Half reaction of Mg: -2.356
Half reaction of [Al(OH)4]-: -2.310
Ecell (standard) = -2.310-(-2.356) = .046 V
Equation written out in spontaneous form would be...
3Mg + 2[Al(OH)4]- -> 3Mg+2 + 8OH- + 2Al
Then solving for Q... which is [product]/[reactant] would be...
Q = ([.048]^8 * [.012]^3)/[.25]^2 = 7.79E-16
Number of moles of electrons transferred is 2.
Plug everything in...
Ecell = .046 - (.0591/2)*log (7.79E-16)
Which gives me
Ecell = 0.492 V
However, the program I'm entering this into says it's incorrect.
Any pointers on where I went wrong?
Thanks in advance.