posted by Mike on .
Calculate Delta G for the spontaneous cell produced from the following:
Al^3+ + 3e^- -> Al (s)
Mg^2+ + 2e^- -> Mg (s)
In order to solve this, you need to determine the half reactions..ie which is reduced and which is oxidized. How do I know which one is being reduced and which is being oxidized? :s
You look up the Eo values for Al and Mg. Some quick numbers I remember, but you need to look them up and do the work more accurately,is
Eo for Al^+3 + 3e = -1.66 volt
Eo for Mg^+2 + 2e = -2.37
Turn the most negative value around to make it an oxidation and add to the other (in order to obtain a positive value for the cell potential).
3(Mg ==> Mg^+2e) E = +2.37
2(Al^+3 + 3e ==> Al) E = -1.66
3Mg + 2Al^+3 + 6e ==>3Mg^+2 + 2Al + 6e
Then, remember oxidation is the loss of electrons; reduction is the gain of electrons.
n= number of electrons transferred in the reaction
3Mg= 3Mg+2 + 6e-
2Al+3 + 6e- = 2Al
F= faradays constant= 96500 J/V mol e-
Cathode: Al+3 + 3e- = Al E= -1.66 V
Anode: Mg+2 + 2e- = Mg E= -2.37
Ecell= -1.66 - -2.37
deltaG= - (6) (.71) (96500)
deltaG= -410000 J/mol or -410 kJ/mol