Using the activity series, either from the textbook (page 80) or your data sheets, determine whether the reactions given below would be spontaneous or not. JUSTIFY your answer in each case.
a. Cu2+ (aq) + Au (s) → Au+ (aq) + Cu (s)
b. Fe2+ (aq) + Al (s) → Al3+(aq) + Fe (s)
c. Pb2+ (aq) + Sn (s) → Sn2+ (aq) + Pb (s)
d. Mg2+ (aq) + Ag (s) → Ag+ (aq) + Mg (s)
I will do one for you. Here is what you do for the first one.
Cu2+ (aq) + Au (s) → Au+ (aq) + Cu (s)
Look up the EMF (Eo values) for Cu^2+(aq) + 2e ==> Cu(s)
You text or tables will have something close to + 0.337 volts
Next you look up Au(s) to Au^+(aq) + e. That will NOT be in the table for it is an oxidation and your table/text is one of reduction potentials. So you will see Au^+ + e ==> Au(s) and will show a voltage of 1.692 v. You want the reverse of that reaction so that is -1.692 volts. So you have these two half equations which you add together like this.
Cu^2+(aq) + 2e ==> Cu(s) ...........Eo reduction = +0.337 v or close to that.
Au(s) ==> Au^+(aq) + 1e...............Eo oxidation = -1.692 v
Now add them together to get a voltage of -1.356 v. Now you have the information to tell you if the reaction is spontaneous or not. If the voltage is negative the reaction is NOT spontaneous (it will not occur as written but WILL occur spontaneously in the opposite direction). Reactions with a positive Eo value WILL occur spontaneously.
If you need the equation for that reaction it is as follows and note that I've multiplied the second half equation by 2 to make the electrons equal so they cancel in the final equation. NOTE ALSO THAT I DID NOT MULTIPLY THE Eo value for that half equation.
Cu^2+(aq) + 2e ==> Cu(s) ...........Eo reduction = +0.337 v or close to that.
2Au(s) ==> 2Au^+(aq) + 2e...........Eo oxidation = -1.692 v
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Cu^2+(aq) +2Au(s) ==> Cu(s) + 2Au^+(aq).....Eo reaction = -1.356 v
I'll leave the others for you. Post your work if you run into problems.