# Question 0aadc

Aug 10, 2016

Here's what I got.

#### Explanation:

The first thing to do when dealing with similar problems is to write out the reduction equilibria given to you

$\text{Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) rightleftharpoons "Ce"_ ((aq))^(3+)" "E^@ = +"1.72 V}$

$\text{Fe"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Fe"_ ((s))" "E^@ = -"0.44 V}$

For a given reduction equilibrium, the value of the standard reduction potential, ${E}^{\circ}$, tells you the position of the equilibrium in relation to a reference hydrogen electrode.

When ${E}^{\circ}$ is positive, the equilibrium lies to the right, meaning that the chemical species loses electrons less readily than hydrogen.

When ${E}^{\circ}$ is negative, the equilibrium lies to the left, meaning that the chemical species loses electrons more readily than hydrogen.

Now, when you compare two ${E}^{\circ}$ values, you should know that

• the reduction equilibrium that has the less negative / more positive ${E}^{\circ}$ value will shift to the right
• the reduction equilibrium that the more negative / less positive ${E}^{\circ}$ value will shift to the left

In your case, the ${E}^{\circ}$ value for the first equilibrium is negative and the ${E}^{\circ}$ value for the second equilibrium is positive. This tells you that when you connect these two half-cells, you will get

$\text{Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) rightleftharpoons "Ce"_ ((aq))^(3+)" "E_"red"^@ = +"1.72 V}$

$\stackrel{\textcolor{red}{\rightarrow}}{\textcolor{w h i t e}{a a \textcolor{\mathrm{da} r k g r e e n}{\text{shift to the right}} a a a a}}$

Since the forward reaction proceeds here, you can say that this is your reduction half-reaction

$\text{Ce"_ ((aq))^(3+) + color(white)(1)"e"^(-) -> "Ce"_ ((aq))^(3+)" "E_"red"^@ = +"1.72 V}$

Likewise, you will get

$\text{Fe"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Fe"_ ((s))" "E_"oxi"^@ = -(-"0.44 V") = +"0.44 V}$

$\stackrel{\textcolor{red}{\leftarrow}}{\textcolor{w h i t e}{a a \textcolor{\mathrm{da} r k g r e e n}{\text{shift to the left}} a a a a}}$

The reverse reaction proceeds here, so reverse the sign of ${E}^{\circ}$ and write the oxidation half-reaction

$\text{Fe"_ ((s)) -> "Fe"_ ((aq))^(2+) + 2"e"^(-)" "E_"oxi"^@ = +"0.44 V}$

You can now balance and add the two half-reactions to get

$\left\{\begin{matrix}{\text{Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) -> "Ce"_ ((aq))^(3+)" " " " " "| xx 2 \\ color(white)(aaaaaa)"Fe"_ ((s)) -> "Fe"_ ((aq))^(2+) + 2"e}}^{-}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

2"Ce"_ ((aq))^(4+) + "Fe"_ ((s)) + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"Ce"_ ((aq))^(3+) + "Fe"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-))))

You thus have

$2 {\text{Ce"_ ((aq))^(4+) + "Fe"_ ((s)) -> 2"Ce"_ ((aq))^(3+) + "Fe}}_{\left(a q\right)}^{2 +}$

The standard cell potential will be equal to

${E}_{\text{cell"^@ = E_"red"^@ + E_"oxi}}^{\circ}$

${E}_{\text{cell"^@ = +"1.72 V" + "0.44 V" = }} 2.16 V$

• ${\text{Ce}}^{4 +}$ is a weaker oxidizing agent than ${\text{Fe}}^{2 +}$

This one is not true. The fact that the reaction proceeds in this direction, i.e. ${E}_{\text{cell}}^{\circ} > 0$, which implies that this redox reaction is spontaneuos, tells you that ${\text{Ce}}^{4 +}$ is a stronger oxidizing agent than ${\text{Fe}}^{2 +}$.

Similarly, iron metal, $\text{Fe}$, is a stronger reducing agent than ${\text{Ce}}^{3 +}$. Kepp in mind that for a spontaneous redox reaction, you have

overbrace(2"Ce"_ ((aq))^(4+))^(color(blue)("stronger oxi agent")) + underbrace("Fe"_ ((s)))_ (color(purple)("stronger red agent")) -> underbrace(2"Ce"_ ((aq))^(3+))_ (color(purple)("weaker red agent")) + overbrace("Fe"_ ((aq))^(2+))^(color(blue)("weaker oxi agent"))#

• ${\text{Ce}}^{4 +}$ will reduce ${\text{Fe}}^{2 +}$

This one is also not true. ${\text{Ce}}^{4 +}$ cannot reduce ${\text{Fe}}^{2 +}$ because it oxidizes $\text{Fe}$ to ${\text{Fe}}^{2 +}$.

• ${\text{Ce}}^{4 +}$ is a stronger oxidizng agent than ${\text{Fe}}^{2 +}$

This one is true, as shown above.

• ${\text{Ce}}^{4 +}$ will oxidize $\text{Fe}$

This one is also true, as shown above.