Question #0cfba

May 18, 2015

The answer is D) $B {r}_{2 \left(l\right)}$.

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 $F {e}^{2 +}$.

In order for this to happen, you need to find a species that gains electrons, or gets reduced, more readily than $F {e}^{2 +}$,

Once again, the order in which the ${E}^{\circ}$ 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

$F {e}_{\left(a q\right)}^{3 +} + {e}^{- 1} r i g h t \le f t h a r p \infty n s F {e}_{\left(a q\right)}^{2 +}$, ${E}^{\circ} = \text{+0.77 V}$

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

$F {e}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s F {e}_{\left(s\right)}$

Remember that you're dealing with equilibrium reactions, so the reaction in which $F {e}^{2 +}$ is "being produced by $F {e}^{3 +}$" can actually go both ways.

Now, find the electrode potentials for your species

${I}_{2 \left(s\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 {I}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+0.53 V}$

$N {i}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s N {i}_{\left(s\right)}$, ${E}^{\circ} = \text{-0.25 V}$

$Z {n}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s Z {n}_{\left(s\right)}$, ${E}^{\circ} = \text{-0.76 V}$

$B {r}_{2 \left(l\right)} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s 2 B {r}_{\left(a q\right)}^{-}$, ${E}^{\circ} = \text{+1.07 V}$

Right from the get-go, two species are eliminated, more specifically $N {i}_{\left(s\right)}$ and $Z {n}_{\left(s\right)}$, because they are on the same side of the equilibrium as $F {e}^{2 +}$.

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 $F {e}^{2 +}$ to lose electrons in order to get oxidized to $F {e}^{3 +}$. This means that you need a species that will gain electrons more readily than $F {e}^{2 +}$.

In order for that to happen, you need a more positive electrode potential, which implies that the answer can only be bromine, $B {r}_{2 \left(l\right)}$, which has ${E}^{\circ} = \text{+1.07 V}$.

When you put these two species together, the equilibrium will lie to the right for bromine, which will be reduced to $B {r}^{-}$, and to the left for iron (II), which will be oxidized to $F {e}^{3 +}$.