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.

A17.2.6

In general, cell notation is written with oxidation on the left, with a single line representing phase change. Reactants are written on the left side of the single line and products on the right. Reduction is written on the right side of the double line which denotes the salt bridge used to maintain neutrality. Reactants are written on the left side of the single line and products on the right.

1. For the first example, we can see that Pt(NO₃)₂ goes Pt, which means that Platinum goes from a charge of +2 to a charge of 0. This is reduction because Pt gained 2 electrons. Secondly, Cu metal which has a charge of 0 goes to Cu(NO₃) with a charge of +2. This represents oxidation because Copper lost an electron.

\(\ce{Cu^2+}(s) | \ce{Cu(NO3)2} (s, 1M) || \ce{Pt(NO3)2} (s, 1M) | \ce{Pt}(s)\)

2. The question automatically tells us that Au is reduced because it's on the cathode side. Au(NO₃)₂ is written before Au because it goes from having a charge of +2 to 0. Cu oxidizes at the anode, and goes from having a charge of 0 to +2.

\(\ce{Mg^2+}(aq) | \ce{Mg(NO3)2} (s, 0.75M) || \ce{Au(NO3)2} (s, 0.55M) | \ce{Au^3+}(aq)\)

3. The question tells us that Copper has been oxidized, meaning it will go from having a charge of 0 to a charge of +2. That means that Silver will go from having a charge of +1 to a charge of 0 because it has been reduced.

\(\ce{Cu}(s) | \ce{Cu(NO3)2} (s, 1M) || \ce{AgNO3} (s, 1M) | \ce{Ag^+}(aq)\)

Q19.1.15

The standard reduction potential for the reaction [Co(H₂O)₆]³⁺ (aq)+e^-⟶[Co(H₂O)₆]²⁺(aq) is about 1.8 V. The reduction potential for the reaction [Co(NH₃)₆]³⁺(aq)+e^-⟶[Co(NH₃)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

First of all, we know that the standard reduction potential for oxygen is 1.23V. In order to find which will be oxidized, we will need to reverse the redox reaction for oxygen because oxidation reactions only occur when coupled with reduction reactions. This in turn causes the E^º to also become negative. Therefore, it will be written as:

\(\ce{2H2O}(l)⟶\ce{O2}(g)+\ce{4H^+}(aq)+\ce{4 \ e^-}\) E^º = -1.23V

The given reactions will be written as is:

Equation 1: \(\ce{[Co(NH3)6]^3+}(aq)+\ce{e^-}⟶\ce{[Co(NH3)6]^2+}\) E^º= 0.1V

Equation 2: \(\ce{[Co(H2O)6]^3+}(aq)+\ce{e^-}⟶\ce{[Co(H2O)6]^2+}\) E°= 1.8V

Now, both equations must be multiplied by 4 in order to balance the number of electrons. However, this will not affect the cell potential because E^º is a intensive property. They come our to be written as:

Equation 1: \(\ce{4[Co(NH3)6]^3+}(aq)+\ce{4 \ e^-}⟶\ce{4[Co(NH3)6]^2+}\) E^º= 0.1V

Equation 2: \(\ce{4[Co(H2O)6]^3+}(aq)+\ce{4 \ e^-}⟶\ce{4[Co(H2O)6]^2+}\) E°= 1.8V

Then we are going to subtract to find the cell potential using:

E^º= cathode - anode

For equation 1, our cathode value is 0.1V while Oxygen is -1.23V. Therefore once you subtract the 2 values, you'll get 1.33V.

For equation 2, our cathode value is 1.8V and Oxygen's is -1.23V. Therefore, your cell potential value will turn out be -0.57V.

Now, we see that one value is positive while the other is negative. To decide which will be oxidized, we need to see which is spontaneous. We know that our cell potential value must be positive in order for deltaG to be negative. That means the first equation can be oxidized by oxygen.

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.

A19.3.7

The reason the diphosphate ion can act as a water softener is because of high electronegativity and it acting as a chelating ligand. When the Fe²⁺ comes in contact with the ion, it will try to bond with Oxygen because it has a negative charge and Fe²⁺ has a positive charge, which will cause ionic bonding to occur, therefore Fe(OH)₂ will be formed. Ionic bonding is very strong because of the transfer of electrons among the two ions. Since it's a very strong type of bond, when it comes into contact with water, the iron will precipitate in water, letting it act a good water softener.

Q12.4.7

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

A12.4.7

In order to calculate the half life for the first order decay of carbon-14 we use the half life equation pertaining to first order rate laws:

t_1/2= \(\frac{0.693}{k}\)

The given is the k constant and you are trying to find t_1/2. To solve for t_1/2, you will divide 0.693 by 1.21 × 10⁻⁴ year⁻¹, which will give you a half life of 5727.3 years.

t_1/2= \(\frac{0.693}{1.21\times 10^{-4}} = 5727.3 \ years\)

Q21.2.2

Write the following isotopes in nuclide notation.

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

A21.2.2

The formula for nuclide notation is written as \(\ce{^{A}_{Z}X}\), where A represents the mass number(sum of protons and neutrons) found on the bottom of the elements symbol on the periodic table. Z is the number of protons which can be found by looking at the atomic number of the element while X is simply the elements symbol.

1. We know that Oxygen has a mass number of 14 which is why its written in place of "A". From the periodic table, we can see that O has an atomic number of 8, which represents the number of protons, so it's written:

\(\ce{^{14}_{8}O}\)

2. Were given that Copper has a mass number of 70 and an atomic number of 8, therefore it's written as:

\(\ce{^{70}_{29}Cu}\)

3. 175 is written before 73 because it accounts for the mass of protons and neutrons while 73 just accounts for protons, which is defined as the atomic number.

\(\ce{^{175}_{73}Ta}\)

4. Francium has a mass number of 175 while there are 87 protons, so it's written as:

\(\ce{^{217}_{87}Fr}\)

Q21.5.5

Describe the components of a nuclear reactor.

A21.5.5

A nuclear reactor is what controls and initiates the chain reaction involved with electricity generation. They catalyze the nuclear fission process by introducing uranium atoms with neutrons, generating heat from the high kinetic energy.

The components required for a nuclear reactor are:

1. Nuclear Fuels: Consists of a fissionable isotope, usually Uranium-235 because of its high melting point, providing a sufficient chain reaction.
2. Control Rods: These are made with neutron absorbing material which is inserted or removed depending on the rate of the reaction, therefore controlling the rate of fission. The rods absorb neutrons in order to stop the reaction from occurring. The rods are usually composed of Steel and Boron or Cadmium because of the ability to absorb neutrons.
3. Moderator: This is the material in the core which slows down the neutrons without absorbing them to cause fission. The reason the neutrons may be fast moving is because of the fission process and we want slow neutrons for an ideal chain reaction. Modern reactors include heavy water, light water, carbon dioxide, beryllium, or graphite.
4. Coolant: The coolant absorbs and transmits heat from the reactor in order to create steam and turn the turbines and prevent the reactor from reaching high temperatures.
5. Containment System: The structure around the reactor made of concrete and steel, keeping the radioactive gases and liquids inside, absorbing the radiation produced by the reactor.

Image of a Nuclear Reactor:

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?

A20.4.3

The relationship between electron flow and the potential energy of valence electrons is electrons will flow from high energy to lower areas of energy, which is a similar concept to diffusion. Because A has a higher potential energy compared to B, electrons will flow from A to B in the galvanic cell. This means that substance A is located at the anode while Substance B is located at the cathode because at the cathode, electrons are gained.

Q20.7.1

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

A20.7.1

A Leclanche dry cell is actually a “wet cell” battery containing an electrolytic solution which allows electricity to be conducted. The solution is composed of manganese dioxide, ammonium chloride, and zinc. Zinc is oxidized into Zn²⁺ , while the manganese is reduced into manganese trioxide. An alkaline battery is a type of Leclanche dry cell which functions under alkaline, or basic conditions. It’s dependant upon the oxidation of zinc into zinc oxide and reduction of manganese dioxide into manganese trioxide . Similarly, both are primary batteries, meaning they can’t be recharged. However, the advantage of an alkaline battery is it has a better shelf life compared to the dry cell, even with the same voltage of 1.5V. A Leclanche dry cell has a poor shelf life because the zinc metal will react with the ammonium ions even without a electric current passing through. If the dry cell begins to react with air, then there won’t be enough zinc metal to react with, which will affect the amount of electric current that will pass through. This would make the Leclanche cell an inefficient cell.