Question #0cfba

The answer is D) #Br_(2(l))#.

Think of what you're dealing with here. You need to find a species that will oxidize the iron (II) cation, meaning that it will take away electrons from #Fe^(2+)#.

In order for this to happen, you need to find a species that gains electrons, or gets reduced, more readily than #Fe^(2+)#,

Once again, the order in which the #E^@# values are listed in your table does not matter all that much. You need to find a standard electrode potential that is more positive than the standard electrode potential for this reaction

#Fe_((aq))^(3+) + e^(-1) rightleftharpoons Fe_((aq))^(2+)#, #E^@ = "+0.77 V"#

Why this reaction and not the one that forms iron metal? Because you need the iron (II) cation to be oxidized, not reduced, like it would happen if you had

#Fe_((aq))^(2+) + 2e^(-) rightleftharpoons Fe_((s))#

Remember that you're dealing with equilibrium reactions, so the reaction in which #Fe^(2+)# is "being produced by #Fe^(3+)#" can actually go both ways.

Now, find the electrode potentials for your species

#I_(2(s)) + 2e^(-) rightleftharpoons 2I_((aq))^(-)#, #E^@ = "+0.53 V"#

#Ni_((aq))^(2+) + 2e^(-) rightleftharpoons Ni_((s))#, #E^@ = "-0.25 V"#

#Zn_((aq))^(2+) + 2e^(-) rightleftharpoons Zn_((s))#, #E^@ = "-0.76 V"#

#Br_(2(l)) + 2e^(-) rightleftharpoons 2Br_((aq))^(-)#, #E^@ = "+1.07 V"#

Right from the get-go, two species are eliminated, more specifically #Ni_((s))# and #Zn_((s))#, because they are on the same side of the equilibrium as #Fe^(2+)#.

This leaves you with iodine and bromine. Remember that a more positive electrode potential means that a species can gain electrons more readily than another species.

In your case, you need #Fe^(2+)# to lose electrons in order to get oxidized to #Fe^(3+)#. This means that you need a species that will gain electrons more readily than #Fe^(2+)#.

In order for that to happen, you need a more positive electrode potential, which implies that the answer can only be bromine, #Br_(2(l))#, which has #E^@ = "+1.07 V"#.

When you put these two species together, the equilibrium will lie to the right for bromine, which will be reduced to #Br^(-)#, and to the left for iron (II), which will be oxidized to #Fe^(3+)#.