The answer is A) #V^(2+)#, #U^(3+)#, #Y^(3+)#.
I'll give you the quick answer first, then focus more on explaining what's going on.
Right from the get-go, you can sahy that #Y^(3+)# wil be the weakest oxidizing agent. If you only use the information given to judge each cation's oxidizing strength, the fact that #Y^(3+)# cannot oxidize #U# placed it at the bottom of the stack.
By comparison, #U^(3+)# can oxidize one species, #Y#, and #V^(2+)# can oxidize two species, #U# and #Y#.
So there you have it
#V^(2+) > U^(3+) > Y^(3+)#
Now, here's what's going on. Since we've been working with standard electrode potentials, you can write the reduction half-reactions for #U#, #V#, and #Y#
#U^(3+) + 3e^(-) rightleftharpoons U#, #E_U^@#
#V^(2+) + 2e^(-) rightleftharpoons V#, #E_V^@#
#Y^(3+) + 3e^(-) rightleftharpoons Y#, #E_Y^@#
So, you start by placing #Y# in a solution that contains the #U^(3+)# cation and you observe that a reaction takes place. This means that #U^(3+)# oxidizes #Y#.
Automatically, #E_U^@# must be more positive than #E_Y^@#, since #U^(3+)# can take electrons from #Y#. Keep this in mind
#E_U^@ > E_Y^@#
Now you place #U# in a solution that contains #V^(2+)# cations and you observe that a reaction takes place. The same thing that happensed before happens again.
#V^(2+)# oxidizes #U#, which must mean that
#E_V^@ > E_U^@#
Moving on. You place #Y# in a solution that contains #V^(2+)# cations, and the same thing happens again. #V^(2+)# oxidizes #Y#, which implies that
#E_V^@ > E_Y^@#
Finally, you place #V# in a solution that contains #Y^(3+)# and observe that the two don't react. This means that #Y^(3+)# cannot oxidize #V#, so
#E_V^@ > E_Y^@#
As you can see, it all comes down to solving the famous "if A is greater than B, and B is greater than C" inequalities. If you have
#{ (E_U^@ > E_Y^@), (E_V^@ > E_U^@) :} => color(green)(E_V^@ > E_U^@ > E_Y^@)#
So, if you have a standard electrode potential table listed from most positive to most negative, you'll get
#V^(2+) + 2e^(-) rightleftharpoons V#, #E_V^@#
#U^(3+) + 3e^(-) rightleftharpoons U#, #E_U^@#
#Y^(3+) + 3e^(-) rightleftharpoons Y#, #E_Y^@#
So, to summarize, you now know that
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#V^(2+)# can oxidize anything lower and to the right of the equilibrium, in your case #U# and #Y#.;
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#U^(3+)# can oxidize anything lower and to the right of the equilibrium, in your case #Y#;
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#Y^(3+)# cannot oxidize anything higher and to the right of the equilibrium, in your case #V#.
