Extra Credit 17

Q17.2.6

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

a. In one half-cell, a solution of \(Pt{(NO_3)}{_2}\) forms Pt metal, while in the other half-cell, Cu metal goes into a \(Cu{(NO_3)}{_2}\) solution with all solute concentrations 1 M.

b. The cathode consists of a gold electrode in a 0.55 M \(Au{(NO_3)}_{3}\) solution and the anode is a magnesium electrode in 0.75 M \(Mg{(NO_3)}{_2}\) solution.

c. One half-cell consists of a silver electrode in a 1 M \(Ag{(NO_3)}\) solution, and in the other half-cell, a copper electrode in 1 M \(Cu{(NO_3)}{_2}\) is oxidized.

S17.2.6

In order to do cell notation, you must identify the anode, cathode, and any phases changes that occur in the redox reaction.

An example of cell notation for the redox reaction \(Zn(s) +Cu^{2+}(aq)\Rightarrow Zn^{2+}(aq) + Cu(s)\) in standard conditions is \(\underset{anode}{Zn(s) | Zn^{2+}(aq,1M)} || \underset{cathode}{Cu^{2+}(aq,1M)| Cu(s)}\)

Where

• The anode and cathode are separated by a two bars which represent a salt bridge.
• The anode is placed on the left and the cathode is placed on the right.
• Phases changed within a half-cell are written as a single bar.
• Concentrations and phase of species can be written in the parentheses after the phase notation (s, l, g, or aq)

Let us start with problem (a):

In one half-cell, a solution of \(Pt{(NO_3)}{_2}\) forms Pt metal, while in the other half-cell, Cu metal goes into a \(Cu{(NO_3)}{_2}\) solution with all solute concentrations 1 M.

\(Pt{(NO_3)}{_2}\) and \(Cu{(NO_3)}{_2}\) are solutions, but every part of them do not necessarily take part in redox chemistry. We have to break them down to their oxidized and reduced counterparts. Consider these two reactions:

\[{Pt(NO_3)_2}\rightarrow {Pt^2}{^+}+2NO_3^-\]

\[{Cu(NO_3)_2}\rightarrow {Cu^2}{^+}+2NO_3^-\]

Since \(NO_3^{-}\) is a spectator ion (because the ion exists in the same form on both sides of the chemical reaction) it takes no part in the reaction and it can be ignored when writing the reaction in cell notation, however it does help us determine the charge on the platinum ion and the copper ion. So we use \({Pt^2}{^+}\) and \({Cu^2}{^+}\) to be a part of the cell notation. The \({Pt^2}{^+}\) ion forms the Pt metal and is reduced (going from oxidation state of +2 to 0 as it goes from a charged ion to a neutral) while the Cu metal goes to \({Cu^2}{^+}\) in the \(Cu{(NO_3)}{_2}\) and is thus oxidized (going from oxidation state of 0 to +2). Since Pt is the location of reduction and Cu is the location of oxidation, Cu is the anode and Pt is the cathode. Another way to know which is the anode and which is the cathode is from our knowledge of anodes and cathodes. We know that cathodes gain mass from ions in solution and anodes lose ions to solution.

We form cell notation starting with the anode: the Cu metal which will oxidize and form \({Cu^2}{^+}\) at 1M. The \(\parallel\) symbol represents the salt bridge between the anode and cathode and serves to maintain electric neutrality within the cell. The aqueous solutions are written directly before (the anode) and after (the cathode) the salt bridge, with their corresponding concentrations in Molarity (M). Following the salt bridge is the cathode, where \({Pt^2}{^+}\) will be reduced to and form the metal Pt. Taking this information into consideration, the final cell notation is written below:

\[Cu(s)|{Cu^2}{^+}(1M)\parallel{Pt^2}{^+}(1M)|Pt(s)\]

For problem (b):

The cathode consists of a gold electrode in a 0.55 M \(Au{(NO_3)_3}\) solution and the anode is a magnesium electrode in 0.75 M \(Mg{(NO_3)_2}\) solution.

Again \(NO_3^-\) is a spectator ion and is not written in the cell notation. The cathode and anode are already given to us. In the case of the anode, \(Mg(s)\) is oxidized to form \(Mg^{2+}\) since Mg has a +2 oxidation state in the \((NO_3^-){_2}\) solution. In the cathode, \(Au^{3+}\) that forms from the \({(NO_3)}{_2}\) solution are reduced to form \(Au(s)\). To get the cell notation as:

\[Mg(s)|Mg^{2+} (.55M)\parallel Au^{3+} (.75M)|Au(s)\]

Note the concentrations for cathode and anode solutions are not necessarily 1 M. These conditions are typical for that of a cell with "non-standard" conditions.

For problem (c):

One half-cell consists of a silver electrode in a 1 M \(AgNO_3\) solution, and in the other half-cell, a copper electrode in 1 M \(Cu(NO_3)_{2}\) is oxidized.

From the question we know that the copper solution is oxidized and is thus and anode and using deductive reasoning we can conclude the the silver electrode is the cathode. In the anode \(Cu(s)\) is oxidized to form \(Cu^{2+}\), and in the cathode \(Ag^+\) is reduced to form \(Au(s)\). So the cell notation is:

\[Cu(s)|{Cu^{2+}}(1 M)\parallel {Ag^+}(1 M)|Ag(s)\]

Q19.1.15

The standard reduction potential for the reaction \([Co{(H_2O)}_{6}]^{3+}(aq)+e^−\rightarrow[Co{(H_2O)}_{6}]^{2+}(aq)\) is about 1.8 V. The reduction potential for the reaction \([Co{(NH_{3})}_{6}]^{3+}(aq)+e^−\rightarrow[Co{(NH_{3})}_{6}]^{2+}(aq)\) 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

We have two reactions with coordination complexes given, each with their own \({E^{\circ}}\) which is the standard reduction potential:

\[[Co(H{_2}O)_6]^{3+} (aq)+ e^-\rightarrow[Co(H{_2}O)_6]^{2+} (aq)\] with a \({E^{\circ}}\) of 1.8 V

\[[Co({NH}_3)_6]^{3+} (aq)+ e^-\rightarrow[Co({NH}_3)_6]^{2+} (aq)\] with a \({E^{\circ}}\) of .1 V

In order to determine whether \([Co(H{_2}O)_6]^{2+} (aq)\) and \([Co({NH}_3)_6]^{2+} (aq)\) can be oxidized to their +3 counterparts in the presence of oxygen, we need to find the \(E_{cell}\) of the reaction between each ion and Oxygen.

First consider this equation needed to find the \(E_{cell}\): \[{E^{\circ}}_{cell} = {E^{\circ}}(cathode) - {E^{\circ}}(anode)\].

In this case, \([Co(H{_2}O)_6]^{2+} \) or \([Co({NH}_3)_6]^{2+} \) is functioning as the anode because they are oxidized and oxygen is functioning as the cathode.

The standard reduction potential for oxygen can be expressed using the following equation as found on this chart:

\[O_2 (g)+4H^+ (aq) + 4e^- \rightarrow 4H_2O (l)\] with a \({E^{\circ}}\) of 1.23V

In the case of \([Co(H{_2}O)_6]^{2+} \), the \({E^{\circ}}_{cell}\) would be \(1.23-1.8 = -.57 V\). = \(E^o_{cell}\)

A negative \({E^{\circ}}_{cell}\) means the reaction will not be spontaneous and thus will not occur without a current being added to it. We determine spontaneity by looking at the equation \({ΔG}\) = - \({nF}\text{E}^°_\text{cell}\) . \({ΔG}\) must be negative in order for the reaction to be spontaneous. If \(E_{cell}\) is negative, the overall \({ΔG}\) will result in a positive number, which means the reaction will not be spontaneous.

In the case of \([Co({NH}_3)_6]^{3+}\), the \({E^{\circ}}_{cell}\) would be \(1.23-.1 = 1.13 V\). = \(E^o_{cell}\)

A positive \({E^{\circ}}_{cell}\) means the reaction will be spontaneous and thus will occur with or without the presence of a current.

Therefore, because the the reaction of \([Co({NH}_3)_6]^{3+}\) and oxygen would be spontaneous, this complex can be oxidized to the corresponding cobalt(III) complex by oxygen (without the input of current).

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.

S19.3.7

Consider the scenario of a water ion complex \([Fe{(H_2O)}_{6}]^{2+}\) that precipitates \(Fe^{2+}\) as the insoluble salt. The presence of a bidentane ligand such as diphosphate would cause a reaction like below to occur: \[[Fe{(H_2O)}_{6}]^{2+}+3{P_2O_7}^{4-}\rightarrow[Fe{(P_2O_7)}_3]^{10-}+6H_2O\]

This reaction is entropically favored, as there are a higher number of species as products than as reactants. The resulting \([Fe{(P_2O_7)}_3]^{10-}\) complex is more stable as it involves a stronger bidentane ligand. This would reduce the amount of \(Fe^{2+}\) that precipitate since the ligands are bound to the metal stronger. This "phenomena" is explained by the chelating effect, where polydentate ligands have multiple areas to "grab" onto a metal, resulting in a tighter bond with the metal. Since chelation is entropically driven, the resulting bond with the bidentate ligand will be more stable and the iron will be less likely to precipitate out as a salt. This causes the water to be "softened".

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

We are given that the decay of carbon-14 is first order and that the rate constant (k) is \(1.24 \times 10^{-4}\) per year.

The half life of a first order reaction is given by: \[t_{1/2}=\frac{ln(2)}{k}\]

Where \(t_{1/2}\) is the half life.

By substituting k into our equation we get \[\frac{ln(2)}{1.24 \times 10^{-4}/year}=5.58 \times10^{3}\ years\].

Therefore, the half life of carbon-14 decay is around \(5.58x10^{3}\)years.

Q21.2.2

Write the following isotopes in nuclide notation (e.g., " \(_6^{14}C\) ")

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

S21.2.2

Nuclide notation is expressed in this general form: \[^N_ZA\]

Where A is the element, N is the element's mass number (# of neutrons + protons), and Z is the element's atomic number (# of protons that corresponds to the specific element). Use the periodic table to find an element's atomic number. The mass number is given with the name of the isotope. (example carbon-12 has a mass number of 12). Notice that among isotopes of an element, the N value may change by a few numbers but the Z value stays constant.

1. oxygen-14 has a mass number of 14 and an atomic number of 8, thus its nuclide notation is \(^{14}_{8}O\)
2. copper-70 has a mass number of 70 and an atomic number of 29. \(^{70}_{29}Cu\)
3. tantalum-175 has a mass number of 175 and an atomic number of 73. \(^{175}_{73}Ta\)
4. francium-217 has a mass number of 217 and an atomic number of 87. \(^{217}_{87}Fr\)

Q21.5.5

Describe the components of a nuclear reactor.

S21.5.5

You can read more about nuclear reactors here.

In order to provide nuclear energy, there must be a sufficient amount of a fissionable isotope available to provide a chain fission reaction. A chain fission reaction is process where neutrons released in fission produce an additional fission in at least one further nucleus. When this process is controlled, large amounts of energy can be produced. This is the fuel source. There must also be a moderator to make sure the neutrons produced from one fission reaction are slow enough to collide with another nucleus. The coolant allows the energy from the nuclear reaction to be used as electricity. The control rods consist of other materials, such as boron cadnium and hafnium, which are used to control the rate of fission of the fuel source by absorbing slow neutrons . The shield and containment system protects workers from radiation and moderates the high pressure caused the coolant or other water.

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

Electrons with high potential energy want to repel from each other and thus want to decrease potential energy (increasing bond distance will decrease the relative potential energy of the system). Thus, the electron transfer will occur from substance A with high potential energy to the substance B with lower potential energy, as the electrons in A repel and goes to substance B.

Q20.7.1

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

S20.7.1

Alkaline batteries work in a basic environment, but function like Lelanché cells. An alkaline battery will not spontaneously react (Lelanché cells will have zinc react with the amine chloride solution causing corrosion) and has a higher fraction of the zinc react, allowing a more consistent 1.5 V current.

Other advantages of using alkaline batteries as opposed to Leclanché dry cells are that alkaline batteries are less likely to leak and thus have a longer shelf life. They also function better than Lelanché cells in low temperatures. Alkaline batteries have a higher energy density than dry cells that allows the battery to last longer while providing the same amount of energy.