Extra Credit 17

Q17.2.6

From the information provided, use cell notation to describe the following systems:

1. In one half-cell, a solution of Pt(NO₃)₂ forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO₃)₂ solution with all solute concentrations 1 M.
2. The cathode consists of a gold electrode in a 0.55 M Au(NO₃)₃ solution and the anode is a magnesium electrode in 0.75 M Mg(NO₃)₂ solution.
3. One half-cell consists of a silver electrode in a 1 M AgNO₃ solution, and in the other half-cell, a copper electrode in 1 M Cu(NO₃)₂ is oxidized.

S17.2.6

1. The first step is to determine the half reactions:

We know that \(\ce{Pt^2+}(aq)⟶\ce{Pt}(s)\) because since (NO₃) is a spectator ion and does not participate in the overall reaction, we can leave this out in the half reaction. We can tell that this reaction is a reduction because we have a cation going to a solid of neutral charge, ie the charge of the Pt is decreasing and the half reaction is gaining electrons. Therefore this half reaction goes on the right for cell notation.

Then we have \(\ce{Cu}(s)⟶\ce{Cu^2+}(aq)\) and we know that this is oxidation because the Cu(s) is losing electrons and increasing in charge. The oxidation reduction takes place in the anode and thus goes on the left side of the cell notation.

-Notice that the electrons cancel out in the two half reactions-

The next step is to know how to separate the half reactions. First, we must indicate the phase change that takes place as the half reactions move forward. We indicate the phase change between (aq) and (s) with a single line, |. Then, we must indicate the separation of the half reactions themselves. We do this by putting a || between the two reactions to denote the salt bridge. Remember to include the molarities in the final notation. When writing the half reactions always write anode on the left and cathode after the salt bridge.

\[\ce{Cu}(s)│\ce{Cu^2+}(aq, 1.0M)║\ce{Pt^2+}(aq, 1.0M)│\ce{Pt}(s)\]

2. In this problem, it is given that the cathode contains the gold electrode in a .55M solution, so we know that this reaction will be the reduction of \(\ce{Au^3+}(aq)⟶\ce{Au}(s)\). Again, we can ignore the NO₃ because it is a spectator.

Then they tell us that the anode is a magnesium electrode in a .75M solution, so we know that this reaction will be the oxidation of \(\ce{Mg}(s)⟶\ce{Mg^2+}(aq)\).

The overall reaction is:

\[\ce{Mg}(s)│\ce{Mg^2+}(aq, .75M)║\ce{Au^3+}(aq, .55M)│\ce{Au}(s)\]

3. Lastly, the problem tells us that a copper electrode in 1 M Cu(NO₃)₂ is oxidized, so we know that this reaction takes place at the anode and goes on the left of the cell notation. The half reaction is \(\ce{Cu}(s)⟶\ce{Cu^2+}(aq)\).

Because we already know the oxidation reaction, we know that the half cell that consists of a silver electrode in a 1 M AgNO₃ solution is must be undergoing the reduction half reaction and will be written on the right of the notation. The reduction half reaction is \(\ce{Ag^+}(aq)⟶\ce{Ag}(s)\) and the overall reaction is:

\[\ce{Cu}(s)│\ce{Cu^2+}(aq, 1M)║\ce{Ag^+}(aq, 1M)│\ce{Ag}(s)\]

Q19.1.15

The standard reduction potential for the reaction \(\ce{[Co(H_{2}O)_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(H_{2}O)_{6}]^{+2}}(aq)\) is about 1.8 V. The reduction potential for the reaction \(\ce{[Co(NH_{3})_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(NH_{3})_{6}]^{+2}}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H₂O)₆]²⁺ and/or [Co(NH₃)₆]²⁺, can be oxidized to the corresponding cobalt(III) complex by oxygen.

S19.1.15

The first step is to solve for the standard reduction potential of the oxidation of oxygen by setting up the equation (which we obtain from the Important Standard Electrode (Reduction) Potentials table found here):

\(O _{2}(g)+ 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2} (aq)\) \(E^o= 0.7V\)

However, we need this equation to be written for oxidation, not reduction. Therefore we can switch the two sides of the equation in order to see the standard oxidation potential. This allows us to flip the Eº value as well and it then becomes a negative value.

\(H_{2}O_{2} (aq)\rightarrow O _{2}(g)+ 2H^{+}(aq) + 2e^{-}\) \(E^o= -0.7V\)

To calculate the Eº_cell, we can now add the two values together because we are using the oxidation value that takes place at the anode instead of the reduction value. Reduction always occurs at the cathode and oxidation occurs at the anode.

A helpful way to remember this is the phrase OARC where oxidation is at the anode and reduction as at the cathode

\[E^{o}_{cell}= E^{o}_{cathode} + E^{o}_{anode}\]

1. First, we plug in \(E^o= -0.7V\) from the oxidation of oxygen into the first equation to see if the complex ion can be oxidized to the cobalt(III) complex: \[\ce{[Co(H_{2}O)_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(H_{2}O)_{6}]^{+2}}(aq)\]\[E^{o}=1.8 V + (- 0.7V) = 1.1V\]

This equation yields a positive Eº value and this means that the reaction is spontaneous because we get a negative ∆G° when we plug this value into the equation \[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \]

2. Next, we plug in \(E^o= -0.7V\) to the second equation to see if this complex ion can be oxidized to the cobalt(III) complex: \[\ce{[Co(NH_{3})_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(NH_{3})_{6}]^{+2}}(aq)\] \[E^{o}=0.1V + (-0.7V) = -0.6V\]

This equation yields a negative Eº value, meaning that the reaction is nonspontaneous because we get a positive ∆G° when we plug this value into the equation \[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \]

Q19.3.7

Explain how the diphosphate ion, [O₃P−O−PO₃]⁴⁻, can function as a water softener that prevents the precipitation of Fe²⁺ as an insoluble iron salt.

Q12.4.7

What is the half-life for the first-order decay of carbon-14? \[ {^{6}_{14}}C \rightarrow {^{7}_{14}}N + e^- \] The rate constant for the decay is 1.21 × 10⁻⁴ year⁻¹.

S12.4.7

1. To determine the half life of a compound t_1/2 we have to know 3 things:
   1. The order of the reaction
   2. The rate constant, k, or the equation used to solve for k
   3. Concentration
2. The problem tells us that the reaction is first order, and they give us k. If they had not had given us k, we would be able to determine the value based on the following rules:
   1. For a zero order reaction A → products, rate = k
   2. For a first order reaction A → products, rate = k[A]
   3. For a second order reaction A → products, rate = k[A]²
3. The next step in determining the half-life of a compound is to know which t_1/2 equation to use:
   1. For a zero order reaction A → products, \( \mathrm{t_{1/2}} = \mathrm{(\frac{[A]_o}{2k})}\)
   2. For a first order reaction A → products, \( \mathrm{t_{1/2}} = \mathrm{(\frac{ln2}{k})}\) = \( \mathrm{t_{1/2}} = \mathrm{(\frac{.693}{k})}\)
   3. For a second order reaction A → products, \( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{k[A]_o})}\)

4. This is a first order reaction, so we use the the respective equation for first order 1/2 life \( \mathrm{t_{1/2}} = \mathrm{(\frac{.693}{k})}\)

   1. \( \mathrm{t_{1/2}} = \mathrm{(\frac{.693}{1.21e^-4})}\)=5727 years

Q21.2.2

Write the following isotopes in nuclide notation. e.g., \[ ^{14}_{6}\text{C} \]

1. oxygen-14
2. copper-70
3. tantalum-175
4. francium-217

S21.2.2

A nuclide is a distinct kind of atom or nucleus characterized by a specific number of protons and neutrons.

The nuclide notation is: \(\ce{^{A}_{Z}X}\) , where A= the number of nucleons (protons + neutrons, is different in isotopes), Z= number of protons, and X is the elemental symbol for the corresponding number of protons (atomic number).

- Notice that you can subtract these two numbers from each other to calculate the number of neutrons indivdially, this can be helpful in calculations you may make later in this course.

1. "oxygen-14" means that this isotope of oxygen has a nucleon number of 14. The atomic number remains the same because the problem did not tell us otherwise, so the notation is:

\[ ^{14}_{8}\text{O} \]

2. "copper-70" means that this isotope of copper has a nucleon number of 70. The atomic number remains the same so the notation is:

\[ ^{70}_{29}\text{Cu} \]

3. "tantalum-175" means that this isotope of copper has a nucleon number of 175. The atomic number remains the same. Notation is:

\[ ^{175}_{73}\text{Ta} \]

4. "francium-217" means that this isotope of copper has a nucleon number of 217. The atomic number remains the same. Notation is:

\[ ^{217}_{87}\text{Fr} \]

Q21.5.5

Describe the components of a nuclear reactor.

S21.5.5

A nuclear reactor is a device in which nuclear reactions are generated from unstable isotope, and the chain reaction is controlled to release large amount of steady heat, thus producing energy. A chain reaction occurs when the neutrons released in fission (splitting of an atom's nucleus) collide with at least one other nuclei, causing the fission of another nuclei.

This link to a video will help to visualize how these mechanisms work:

https://www.nei.org/Knowledge-Center...-Reactors-Work

Nuclear reactors have 5 components:

1. Nuclear fuel: an isotope that is able to be fissioned must be supplied in copious amounts in order to sustain a controlled chain reaction. The fuel must have a higher concentration of U-235 than what is found in nature.

2. A moderator: This is an essential tool used to slow neutrons produced by nuclear reactions so that they aren't flying wildly and can be absorbed by the fuel and controlled to cause a chain reaction, increases efficiency of energy. Since neutrons are moving so fast, they cannot cause fission without being slowed and absorbed by the fuel. This is the job of the moderator, to slow the neutrons.

3. A coolant: The coolant is necessary to carry heat from the fission reaction to an external boiler and turbine, where it is transformed into electricity. Often times, two overlapping coolant loops are used in order to counteract the transfer of radioactivity from the reactor to the primary coolant loop. An example of a coolant is water used in US nuclear power plants.

4. A control system: This system is made up of control rods placed between fuel rods to absorb neutrons and is used to adjust the number of neutrons keeping the rate of the chain reaction at a safe and manageable level. Control systems adjust fission rates of the nuclear fuel.

5. A shield and containment system: This is perhaps the most important component as it protects workers from radiation produced by the nuclear fission reactions taking place, and to withstand the high pressures resulting from such high-temperature reactions. The shield and containment is so important because even when shut down, the decay products are very radioactive and the operating reactor is extremely hot. The containment system has 3 parts:

1. The reactor vessel = a thick, steel shell that helps the moderator to absorb as much of the radiation produced by the reactor.

2. A main shield with high-density concrete

3. A shield of lighter materials that protects operators from gamma and X-rays

Q20.4.3

What is the relationship between electron flow and the potential energy of valence electrons?

If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?

S20.4.3

In order to understand the relationship between electron flow and potential energy of valence electrons, we must first understand how Galvanic cells work:

1. Galvanic cells utilize the electric energy produced from electron transfer of a redox reaction in order to perform work. These cells are spontaneous, anode to cathode.

2. In order to capture this electron flow, the oxidation and reduction reactions must be separated by a salt bridge ( which maintains neutrality in cell so that it is continous over time) but also connected by a wire, which gives electrons something to flow through. This flow, called a current, is then sent through a circuit if one wishes to power an electrical device.

3. As electrons flow from the oxidation half-cell (beaker) to the reduction half-cell a negative charge builds up in the reduction half-cell while a positive charge builds up in the oxidation half-cell.

Understanding this, we can see that when there is a difference in potential energy of valence electrons, ie when substance A's valence electrons have a higher potential energy than those of substance B, this creates a desire for electrons to flow through the circuit from anode to cathode (from the direction of higher potential to lower potential, similar to osmotic movement), and the electrons will flow from A to B.

Q20.7.1

What advantage is there to using an alkaline battery rather than a Leclanché dry cell?

S20.7.1

Before analyzing the advantage of these two types of batteries, we must first understand the three different types of batteries:

1. Primary batteries. These are the common types of batteries that come to mind when people think of batteries. As a primary cell is used, chemical reactions in the battery use up the chemicals that generate the power, which is the reason that these batteries are not rechargeable. These batteries make up 90% of the market.

2. Secondary batteries. This is an electrical battery that can be charged, discharged into a load, and recharged many times. It is composed of one or more electrochemical cells and are typically more expensive than primary batteries. An example is a lead-acid storage battery in cars.

3. Fuel cells. These are not "true" batteries because they are not self-contained but they are nonetheless added as a third category.

Now that we know that both are primary batteries, we narrow our comparison to within this first category:

a. Leclanché cells are just another word for the cells used in the common dry-cell, zinc-carbon battery. This cell is made up of an outer zinc container which is the "anode," while the cathode is a central carbon rod.

• cathode (reduction):

\[2MnO_{2(s)} + 2NH^+_{4(aq)} + 2e^− \rightarrow Mn_2O_{3(s)} + 2NH_{3(aq)} + H_2O_{(l)} \label{Eq1}\]

• anode (oxidation):

\[Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{Eq2}\]

The Zn(s) oxidized to form Zn²⁺ ions at the anode and these ions react with NH₃ formed at the cathode and Cl^- ions present in solution.

• overall:

\[2MnO_{2(s)} + 2NH_4Cl_{(aq)} + Zn_{(s)} \rightarrow Mn_2O_{3(s)} + Zn(NH_3)_2Cl_{2(s)} + H_2O_{(l)} \label{Eq3}\]

The dry cell is inexpensive to manufacture but it is not efficient in producing electrical energy because only the relatively small fraction of the MnO₂ that is near the cathode is reduced and only a small fraction of the zinc cathode is consumed as the cell discharges. In addition, dry cells have a limited shelf life because the Zn(s) anode reacts spontaneously with NH₄Cl in the electrolyte, and dry cells cannot work properly in all conditions for example in cold temperatures the efficiency of the cell decereases immensly.

Diagram of a Leclanché cell:

b. Alkaline batteries

The alkaline battery is just a Leclanché cell that can operate under alkaline (basic) conditions. The half-reactions that occur in an alkaline battery are:

• cathode (reduction)

\[2MnO_{2(s)} + H_2O_{(l)} + 2e^− \rightarrow Mn_2O_{3(s)} + 2OH^−_{(aq)} \label{Eq4}\]

• anode (oxidation):

\[Zn_{(s)} + 2OH^−_{(aq)} \rightarrow ZnO_{(s)} + H_2O_{(l)} + 2e^− \label{Eq5}\]

• overall:

\[Zn_{(s)} + 2MnO_{2(s)} \rightarrow ZnO_{(s)} + Mn_2O_{3(s)} \label{Eq6}\]

This battery works better at low temperatures compared to the Leclanché cell which is why some prefer it even though it might be heavier than other batteries. The alkaline battery is also better for the environment because it doesn't cause as much destruction when thrown away. In addition, the alkaline cell has twice the amount of energy density as the Leclanché, allowing it to produce the same amount of voltage while having a longer shelf life.

Diagram of the alkaline cell: