Question

IS THIS RIGHT?
Experiments were performed with four metals and solutions of their ions: A-A^+, B-B^2+, C-C^2+, D-D^2+. The results of the experiments were:

D^+2 + C ---> No reaction
B^+2 + C---> B + C^+2
A^+ + B ---> A + B^+2

a.) Write the equation for each reduction half reaction.
I got..
D^+2 + 2e^- -->2A^-
C^2+ + 2e^- -->2B^-
B^2+ + 2e^- -->2C^-
A^2+ + 2e^- -->2D^-

B.) Arrange them in a relative reduction potential table with the easiest to reduce at the top and the hardest to reduce at the bottom.

I got...
D^2+ + 2A^- --> 2e^-
C^2+ + 2B^- --> 2e^-
B^2+ + 2C^- --> 2e^-
A^2+ + 2C^- --> 2e^-

C.) write the equation for each oxidation half reaction. I cant get this one

D.) Arrange them in a relative oxidation potential table with the easiest to oxidize at the top and the hardest to oxidize at the bottom.
I cant get this one either

Answer 1

a)What you have written is not correct. There is no way D^2+ can go to A anything or that C^2+ can go to B anything.
Here is what they want.
D^2+ + 2e ==> D
C^2+ + 2e ==> C
B^2+ + 2e ==> B
etc.

b) I have
A
B
C
D
Here is how you do it. Think of the activity series. Here is a link.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
This activity series (actually written as oxidations) tells us that a metal will displace any ION BELOW it in the series. With that information you can place the metals of A, B, C, and D according to their activity series.
C will not displace D^2+ ion; therefore, C is below D.
B displaces C^2+; therefore, C is above B. etc. If you follow through with this they are arranged according to the activity series as D, C, B, A. and those are as oxidations. So if you want them arranges as reductions you reverse that.
c) The oxidation half reaction is just the reverse of the reduction half equation written as part a.
d) Is the reverse of B.