Question

Question #0cfba

Answer 1

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+)`.