Chapter 16 Source Answers and Examples

Step 1: Identify the changed Oxidation States

This reaction has 4 involved atoms, Na, Cl, H, and O.

Sodium remains in its +1 oxidation state before and after the reaction, so there is no redox there.

Chlorine changes from a -1 to a 0 oxidation state, so that is involved in redox.

Hydrogen starts out at +1, and ends up in two separate molecules, one of which also has a +1 state, but the other which has a 0 state. Therefore, the hydrogen is also involved in redox, but only the hydrogen used to make $H_2$.

Oxygen goes from a -2 to a -2 state. Therefore, there is no electron transfer involved there.

Our Redox atoms are Chlorine and Hydrogen.

Step 2: Determine which atom gets reduced and which gets oxidized.

The reduced atom gains electrons, which drops its oxidation state. Our choices are Chlorine, which goes from a -1 to a 0 state, or Hydrogen, which goes from a +1 to a 0 state.

Hydrogen decreases in oxidation state, meaning it is the reduced atom. The oxidized atom increases in oxidation state, as we see in Chlorine, increasing from -1 to 0. Therefore, Hydrogen is the reduced atom, and Chlorine is the oxidized atom.

Step 3: Determine the oxidizing and reducing agents.

The oxidizing agent is the atom which gets reduced, allowing its partner to be oxidized. This means that Hydrogen is the oxidizing agent, as it causes Chlorine to be oxidized.

Similarly, Chlorine is the reducing agent, causing Hydrogen to get reduced.

Our answers are as follows:

Reduced Atom: Hydrogen
Oxidized Atom: Chlorine
Reducing Agent: Chlorine
Oxidizing Agent: Hydrogen

Step 1: Draw out the Lewis Diagram.

For $H_2CO_3$, the Lewis Diagram looks like:

Step 2: Determine the number of electrons on each atom.

We have 6 atoms to catalog. Let’s start on the outside with each hydrogen. Both hydrogen are in the same situation, singly bonded to an oxygen atom. Oxygen is much more electronegative, meaning that the electrons in the oxygen-hydrogen bond are both associated with the oxygen, leaving hydrogen without any electrons:

Hydrogen: 0

Next we look at the Carbon atom. The carbon atom is bonded to 3 oxygen atoms with 4 bonds. Oxygen is still the more electronegative atom, and as carbon doesn’t have any lone pairs, it doesn’t retain any electrons.

Carbon: 0

Finally, we have Oxygen to look at. Let’s start with the doubly bonded oxygen first. That oxygen has two bonds with carbon, and two lone pairs. It is more electronegative than carbon, so it claims all 4 electrons from the bonds. Coupling that with the 2 lone pairs it has, the oxygen atom has 8 electrons.

As for the other two oxygen atoms, they are each singly bonded to a hydrogen and singly bonded to a carbon. Neither atom is more electronegative than oxygen, so oxygen again has all the electrons in the two bonds. Each oxygen also has its two lone pairs, so these two oxygen are also full with 8 electrons claimed.

Oxygen: 8

Step 3: Determine the valence electrons on each atom.

If we were to draw the Lewis structure for each individual atom, we would see the following:

By counting the number of electrons around each atom, we see the valence electrons for that atom:

Oxygen: 6
Carbon: 4
Hydrogen: 1

Step 4: Determine the Oxidation state of each atom

Oxidation state is Valence - Associated, so for each atom we have:

Oxygen: $OS=6-8=-2$

Carbon: $OS = 4-0=+4$

Hydrogen: $OS=1-0=+1$

Notice that we added a + to the positive oxidation states, instead of leaving it implied. This is similar to labeling charge on an ion.

So our answer for this problem is:

$\begin{aligned} \phantom{Align}&\text{Both hydrogen have an oxidation state of +1} \\ \phantom{Align}&\text{Carbon has an oxidation state of +4}\\ \phantom{Align}&\text{All three oxygen have an oxidation state of -2} \end{aligned}$

Step 1: Draw out the Lewis Diagram.

For $S_2F_{10}$ , the Lewis Diagram looks like:

Step 2: Determine the number of electrons on each atom.

There are 12 atoms in this molecule, that’s a huge number! However, all the Fluorine and both Sulfur atoms are in the exact same setup. That makes it quite a bit easier.

So starting with the easiest one, fluorine:

Each fluorine exists with a single bond between itself and a sulfur atom. Fluorine is the most electronegative atom on the periodic table, so we can safely say it is more electronegative than sulfur. Therefore, the fluorine carries both electrons in that bond. As there are also three lone pairs on each fluorine, bringing the total electrons up to 8 on each fluorine.

Fluorine: 8

Now for sulfur. Each sulfur is connected to 5 fluorine atoms and the other sulfur atom. All the electrons involved in each fluorine bond are held by the fluorine, and as there

are no lone pairs, the only electrons available are in the sulfur-sulfur bond. As this bond is between two of the same atom, their electronegativities are equal. This means that they share the electrons in the bond, each getting one. With no other electrons available, this means that each sulfur atom has one electron.

Sulfur: 1

Step 3: Determine the valence electrons on each atom.

There are only two unique atoms in this molecule, sulfur and fluorine. Their Lewis Diagrams are:

As can be seen, fluorine has 7 valence electrons, and Sulfur has 6.

Step 4: Determine the oxidation states.

For sulfur:
$OS=6-1=+5$

For fluorine:
$OS = 7-8=-1$

So our answer for this problem is:

$\begin{aligned} \phantom{align}&\text{Sulfur has an oxidation state of +5}\\ \phantom{align}&\text{Fluorine has and oxidation state of -1} \end{aligned}$

Step 1: Determine priority of the involved atoms.

Priority for determining oxidation state is Alkali Metals, Alkali Earth Metals, Halogens, Hydrogen, Oxygen, other.
We only have two atoms, carbon, which fits into other, and chlorine, a halogen. Therefore, chlorine is a higher priority in this molecule.

Step 2: Assign states in order of priority.

As chlorine is a higher priority, we will set it first. As a halogen, its oxidation state is -1.

Chlorine: -1.

Now we need to set carbon’s state. To do so, we will use the fact that the molecule has a net charge of 0.
With 4 atoms of chlorine each at an oxidation state of -1, we set up the total oxidation state for the molecule:

$OS_{C}+4\cdot(-1)=0$
$OS_C-4=0$
$OS_C=4$

Therefore, the oxidation state of carbon is +4.

Carbon:+4

Step 1: Determine priority of the involved atoms.

We have three types of atoms in our molecule, Carbon, Hydrogen, and Oxygen. The priority puts Hydrogen as highest priority, Oxygen below that, and then Carbon as lowest priority.

Step 2: Assign states in order of priority.

Hydrogen in this molecule is of the highest priority. Therefore, we start by setting the oxidation state of hydrogen to +1.

Hydrogen: +1

Next on priority list is oxygen, which we can set to -2.

Oxygen: -2

Finally, we have carbon, which will have to be set by utilizing the overall charge of 0. Let’s set up the equation:

$OS_C+3\cdot(+1)+OS_C+(-2)+(-2)+(+1)=0$
$2\cdot OS_C+4\cdot(+1)+2\cdot (-2)=0$
$2\cdot OS_C+4-4=0$
$2\cdot OS_C=0$
$OS_C=0$

Carbon: 0.

Step 1: Draw out the Lewis Diagram.

For $CH_3COOH$, the Lewis Diagram looks like:

Step 2: Determine the number of electrons on each atom.

The Hydrogens are easy, as all four are singly bonded to an atom more electronegative than itself (carbon or oxygen). This means none of the Hydrogens have any associated electrons.

Hydrogens: 0

At first glance, the two oxygens appear to be in a different set-up, but both simply have two bonds to less electronegative atom(s) and two lone pairs, meaning they each have 8 electrons.

Oxygen: 8

Now, the two carbon atoms are completely different from one another. The carbon in the $CH_3$ complex is bonded to three hydrogen atoms, each of which it is more electronegative then, and it is also bonded with another carbon, meaning it gets 6 electrons from its bonds with the three hydrogen, and 1 more from the carbon-carbon bond.

Carbon ($CH_3$ complex): 7

The other carbon has three bonds between two oxygen atoms, which is more electronegative than itself, meaning it doesn’t get any electrons from those bonds. It also is bonded to another carbon atom, though, netting it a single electron.

Carbon ($COOH$ complex): 1

Step 3: Determine the valence electrons on each atom.

Our three atoms are Carbon, Hydrogen, and Oxygen. Their Lewis Diagrams are:

This means that carbon has 4 valence electrons, hydrogen has 1, and oxygen has 6.

Step 4: Determine the oxidation states.

Oxygen:
$OS_O=6-8=-2$

Hydrogen:
$OS_H=1-0=+1$

Carbon ($CH_3$):
$OS_C=4-7=-3$

Carbon ($COOH$):
$OS_C=4-1=+3$

Therefore our answer is:

$\begin{aligned} \phantom{align}&\text{Oxygen has an oxidation state of -2}\\ \phantom{align}&\text{Hydrogen has an oxidation state of +1}\\ \phantom{align}&\text{Carbon in }CH_3\text{ has an oxidation state of -3}\\ \phantom{align}&\text{Carbon in }COOH\text{ has an oxidation state of +3}\\ \end{aligned}$

Step 1: Determine priority of the involved atoms.

We have Nitrogen, Hydrogen, Carbon, and Oxygen. This puts the priority as Hydrogen, then Oxygen, then Carbon or Nitrogen.

Step 2: Assign states in order of priority.

Hydrogen starts us off with an oxidation state of +1. Then oxygen lets us set a state of -2. But once we get to nitrogen and carbon, we need to use the overall charge. As we can’t do that for more than one atom, we need to split this one.

Step 3: Split up the molecule.

We have a nice ionic molecule here, ammonium carbonate. Therefore, we can easily split the molecule into constituent ions.

For ammonium carbonate, we split it into ammonium and carbonate. Who would’ve guessed that?

$(NH_4)_2CO_3$
$(NH_4)_2|CO_3$
$2NH_4$ and $CO_3$

But wait! These are ions, so we need to add in the charges associated with each.

$2NH_4^+$ and $CO_3^{2-}$

Perfect, now we can solve for the oxidation states.

Step 4: Assign states in order of priority

We can start with the ammonium ion, $NH_4^+$. We have the high priority hydrogen and the low priority nitrogen. So therefore, we set hydrogen’s priority first.

Hydrogen: 1

Now all that’s left is one nitrogen atom. This means we can use overall charge to determine the oxidation state of the nitrogen.

$OS_N+4\cdot(+1)=+1$

$OS_N+4=+1$

$OS_N=(+1)-(+4)$

$OS_N=-3$

We have successfully determine the oxidation state of nitrogen:

Nitrogen: -3

Now we can move onto the other ion, Carbonate. Here, we have high priority oxygen and lower priority carbon. We assign oxygen’s value first:

Oxygen: -2

Now we use the overall charge to solve for carbon’s state.

$OS_C+3\cdot (-2)=-2$

$OS_C-6=-2$

$OS_C=(-2)-(-6)$

$OS_C=+4$

That is all the atoms determined, now that carbon has been found.

Carbon: +4

Step 5: Reassemble molecule and check the state.

We need to make sure that our logic worked. Let’s look at the whole molecule.

$(NH_4)_2CO_3$

We need to check the overall charge is going to be 0 with our calculated states for each atom.

$(OS_N+4\cdot OS_H)\cdot2+OS_C+3\cdot OS_O\overset ?=0$

$[(-3)+4\cdot (+1)]\cdot2+(+4)+3\cdot (-2)\overset ?=0$

$[-3+4]\cdot2+4-6\overset ?=0$

$[1]\cdot2-2\overset ?=0$

$2-2\overset ?=0$

$0\overset\checkmark=0$

Hooray! Our oxidation states gave us the correct charge on the molecule. This means that we correctly split the molecule.

Step 1: Identify the $\boldsymbol{E^\circ}$ of each half-cell

Mercury has $E^\circ=0.85V$.

Aluminum has $E^\circ=-1.66V$.

Step 2: Identify which Electrode is the anode.

The anode of a galvanic cell is the electrode where oxidation occurs. This means it is the reducing agent. A lower the $E^\circ$ relates to a higher reducing potential, meaning the half-cell with the lower $E^\circ$ will be the anode, as it will reduce the other half-cell. Therefore, our anode is Aluminum.

The aluminum electrode is the anode.

Step 1: Write out equation and find $\boldsymbol n$.

Galvanic cell implies a positive cell potential, so we know that the electrode with a lower $E^\circ$ will be getting oxidized. In our cell, the half-reactions are:

$Cu^{2+}+2e^-\rightarrow Cu_{(s)}$ $E^\circ=0.34V$
$Zn^{2+}+2e^-\rightarrow Zn_{(s)}$ $E^\circ = -0.76V$

This means that zinc will be getting oxidized, and we already have equal numbers of electrons involved in each reaction.

Therefore our reaction is:
$Cu^{2+}+Zn_{(s)}\rightarrow Cu_{(s)}+Zn^{2+}$
From our half-reactions we see that $n=2$.

Step 2: Determine $\boldsymbol Q$.

For this reaction, our reaction coefficient is found by:
$Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$
$Q=\frac{0.5}{4}$

Step 3: Determine $\boldsymbol{E_{cell}}$

From our reaction, we can find $E^\circ_{cell}$.

$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$

$E^\circ_{cell}=0.34-(-0.76)=1.1V$

Now we know everything we need for $E$. As we are not at $25^\circ C$, we have to use the full equation:
$E = E^\circ-\frac{RT}{nF}\ln{Q}$

$E=1.1-\frac{8.314\cdot283.15}{2\cdot96\,485}\ln0.125$

$E=1.125V$

Step 1: Determine Cell set-up.

This battery is our familiar zinc-copper cell, with the anode zinc and cathode copper:

$Cu_{(aq)}^{2+}+Zn_{(s)}\rightarrow Cu_{(s)}+Zn^{2+}_{(aq)}$

This reaction shows that we have a 2 electron reaction between copper and zinc.

Step 2: Determine limiting electrode.

To figure out our limiting electrode, we need to know the electrode with the highest equivalent weight.

Copper: $EqM_{Cu^{2+}}=\frac{M_{Cu^{2+}}}2=\frac {63.55}2=31.775$
Zinc: $EqM_{Zn}=\frac{M_{Zn}}{2}=\frac{65.38}2=32.69$

Just the equivalent mass isn’t enough, though, as we have unequal starting values for the electrode mass.

In order to determine which electrode is limiting, we need the one with the shortest lifetime on a charge-basis. Equivalent mass gives us the value of $\frac{g_A}{mol\ e^-}$. We need to use the initial mass of each electrode to figure out which one will pass the fewest electrons.

$n_{e^-}=\frac{m_A}{EqM_A}$

So for copper:
$n_{e^-,\ Cu^{2+}}=\frac{m_{Cu^{2+}}}{EqM_{Cu^{2+}}}=\frac{6g}{31.775\frac{g_{Cu^{2+}}}{mol\ e^-}}=0.1888mol\ e^-$

and zinc:
$n_{e^-,\ Zn}=\frac{m_{Zn}}{EqM_{Zn}}=\frac{4g}{32.69\frac{g_{Zn}}{mol\ e^-}}=0.1224mol\ e^-$

This means that the Zinc electrode can pass the fewest electrons, causing it to be the limiting electrode.

Step 3: Determine the rating of the battery

We have determined that the zinc electrode will only provide $0.1224mol\ e^-$. We can convert that to coulombs with Faraday’s constant:
$0.1224mol\ e^-\cdot 96\,485\frac C{mol\ e^-}=11\,806C$

Finally, we need to convert the rating into the appropriate units.

$\frac{11\,806C}{3600\frac{C}{\text{A}\cdot\text{hr}}}=3.3\text{A}\cdot\text{hr}$

Our battery has a rating of $3.3\text{A}\cdot\text h\text r$.