# Question #485d5

May 18, 2015

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 $Y$

${U}^{3 +} + 3 {e}^{-} r i g h t \le f t h a r p \infty n s U$, ${E}_{U}^{\circ}$

${V}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s V$, ${E}_{V}^{\circ}$

${Y}^{3 +} + 3 {e}^{-} r i g h t \le f t h a r p \infty n s Y$, ${E}_{Y}^{\circ}$

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, ${E}_{U}^{\circ}$ must be more positive than ${E}_{Y}^{\circ}$, since ${U}^{3 +}$ can take electrons from $Y$. Keep this in mind

${E}_{U}^{\circ} > {E}_{Y}^{\circ}$

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

${E}_{V}^{\circ} > {E}_{U}^{\circ}$

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

${E}_{V}^{\circ} > {E}_{Y}^{\circ}$

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

${E}_{V}^{\circ} > {E}_{Y}^{\circ}$

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

$\left\{\begin{matrix}{E}_{U}^{\circ} > {E}_{Y}^{\circ} \\ {E}_{V}^{\circ} > {E}_{U}^{\circ}\end{matrix}\right. \implies \textcolor{g r e e n}{{E}_{V}^{\circ} > {E}_{U}^{\circ} > {E}_{Y}^{\circ}}$

So, if you have a standard electrode potential table listed from most positive to most negative, you'll get

${V}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s V$, ${E}_{V}^{\circ}$

${U}^{3 +} + 3 {e}^{-} r i g h t \le f t h a r p \infty n s U$, ${E}_{U}^{\circ}$

${Y}^{3 +} + 3 {e}^{-} r i g h t \le f t h a r p \infty n s Y$, ${E}_{Y}^{\circ}$

So, to summarize, you now know that

• ${V}^{2 +}$ can oxidize anything lower and to the right of the equilibrium, in your case $U$ and $Y$.;

• ${U}^{3 +}$ can oxidize anything lower and to the right of the equilibrium, in your case $Y$;

• ${Y}^{3 +}$ cannot oxidize anything higher and to the right of the equilibrium, in your case $V$.