Extra Credit 27

Writing in black was done by Aradhna Grover. Edits in red were done by Chloe Celniker, however changes to LATEX are still in black so that the code works.

Q17.3.6: Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br2(l) is the same as for Br2(aq).

\[\ce{Pt}(s)│\ce{H2}(g)│\ce{H+}(aq)║\ce{Br2}(aq)│\ce{Br-}(aq)│\ce{Pt}(s)\]

It is important to understand before solving this problem that a Galvanic cell has 2 aspects to it, the anode and the cathode. In the cell, electrons move from the anode to the cathode and, subsequently, the substance at the anode becomes oxidized and the substance at the cathode becomes reduced. Oxidized means that electrons are lost and reduced means that electrons are gained.

The first step to solving this problem is to figure out the overall reaction. The way that notation above is read is that everything on the right of the double line is pertains to the cathode and everything on the left pertains to the anode. A single line signifies a phase change and the \(\ce {Pt}(s)\)on either side signifies the metal that is used on the actual cathode and anode. This is an inert electrode that passes electrons but does not partake in the reaction. Another metal is used in this situation because the substances that are reacting are (g) and (aq) and the galvanic cell needs a solid metal to function.

Once again, you know that the hydrogen is oxidized from the gaseous form to \(\ce{H+}(aq)\) as it is at the anode and that Bromide in its liquid form is reduced to \(\ce{Br-}(aq)\)

Thus, equation of the overall reaction can be set up as follows.

\(\ce{H2}(g)+\ce{Br2}(aq)⟶\ce{2H^+}(aq)+\ce{2Br^-}(aq)\). Two moles of electrons are transferred.

To solve for the standard cell potential at 25 degrees Celcius you have to use the equation

\(E^\circ_{\ce{cell}} = E^\circ_{cathode}-E\circ_{anode}\)

To find the \(E^\circ\) at the anode and the cathode, you have to look at the standard reduction reactions taking place at the anode and the cathode.

The reduction potential at the anode is:

\(\ce{2H^+}(aq)+\ce{2e^-}⟶\ce{H2}(g)\) \(E^\circ_={0.00V}\)

The reduction potential equation at the cathode is

\(\ce{Br2}(aq)+\ce{2e^-}⟶\ce{2Br^-}(aq)\) \(E^\circ_={1.07V}\)

You then take the values of \(E^\circ\) at the anode and the cathode and you plug them into the standard equation given above:

\(E^\circ_{\ce{cell}} = {1.07V}-{0.00V}={1.07V}\)

Since the value of \(E^\circ\) is positive, you know that the value of ∆G^O will be negative. This indicates that the reaction will be spontaneous.

I agree with this solution.

Q19.1.25: Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M₃O₄ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M₂O₃, to permit estimation of the metal’s two oxidation states.)

The first step to solving this problem is looking at the rules of Oxidizing states for various elements:

https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State

The main rules that will be used in these problems will be the oxidation state rule 6 which states that oxidation state for Oxygen is (-2) and rule 2 which is that the total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. Solving these problems requires simple algebra. The oxidation states of both elements in the compound is equal to zero, so set the unknown oxidation of the element that is not oxygen to a variable \({x}\), and the oxidation state of Oxygen equal to \({-2}\). Then multiply both oxygen states by the number of atoms of the element present. Add the values together, set the equation equal to zero and solve for \({x}\).

1. \(\ce{Sc2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2{x}}={0}⟶{x}={Sc}={+3}\) \(Sc^{3+}\)
2. \(\ce{TiO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Ti}={+4}\) \(Ti^{4+}\)
3. \(\ce{V2O5}={5{(-2)}}+{2{x}}={0}⟶{-10}+{2{x}}={0}⟶{x}={V}={+5}\) \(V^{5+}\)
4. \(\ce{CrO3}={3{(-2)}}+{x}={0}⟶{-6}+{x}={0}⟶{x}={Cr}={+6}\) \(Cr^{6+}\)
5. \(\ce{MnO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Mn}={+4}\) \(Mn^{4+}\)
6. \(\ce{Fe3O4}=\ce{FeO}·\ce{Fe2O3}=\)

\(\ce{FeO}={-2}+{x}={0}⟶{x}={Fe}={+2}\) \(Fe^{2+}\)

\(\ce{Fe2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Fe}={+3}\) \(Fe^{3=}\)

(One Fe Atom has an oxidation state of +2 and the other 2 Fe atoms have an oxidation state of +3)

7. \(\ce{Co3O4}=\ce{CoO}·\ce{Co2O3}=\)

\(\ce{CoO}={-2}+{x}={0}⟶{x}={Co}={+2}\) \(Co^{2+}\)

\(\ce{Co2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Co}={+3}\) \(Co^{3+}\)

(One Co Atom has an oxidation state of +2 and the other 2 Co atoms have an oxidation state of +3)

8. \(\ce{NiO}={-2}+{x}={0}⟶{x}={Ni}={+2}\) \(Ni^{2+}\)

9. \(\ce{Cu2O}={-2}+{2{x}}={0}⟶{-2}+{2x}={0}⟶{x}={Cu}={+1}\) \(Cu^{1+}\)

I agree with this solution, with minor edits to LATEX and page formatting.

Q12.4.18: Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.

In order to solve this, the following equation is necessary. This equation details the half-life of a first order reaction:

\(\ln\dfrac{[\textrm A]_0}{[\textrm A]}=\lambda(t)\)

\({[\textrm A]_0}\) is the initial value of the amount of carbon present

\({[\textrm A]}\) is the final amount of the carbon present after a certain amount of time (t)

\({\lambda}\) is the rate constant and is found through the equation \(t_{1/2}=\dfrac{0.693}{\lambda}\)

You are given the half life or \(t_{1/2}\) of carbon which is 5730 years. (You don't have to convert it to some other unit of time because the final question also requests an answer in terms of years). You are also given \(\dfrac{[\textrm A]_0}{[\textrm A]}\) which is the percentge of carbon-14 expected, 93.79%.

1) Find the rate constant. You plug in the value for half-life that is given into the equation \(t_{1/2}=\dfrac{0.693}{\lambda}\) and then solve.

\({5730}=\dfrac{0.693}{\lambda}\)

\({\lambda}={0.000121}\)

2) Plug the value for \({\lambda}\) and the percentage of carbon-14 expected back into the equation \(\ln\dfrac{[\textrm A]_0}{[\textrm A]}=\lambda(t)\) to get:

\(\ln\dfrac{1}{0.9379}={0.000121t}\)

3) Solve for \({t}\)

\({0.064112}={0.000121t}\)

\({t}={530.103\,years}\)

This tells you that King Richard III died around 530 years ago. Since it is currently 2017, King Richard III died in the year 1487. If the question was written a few years ago, then the value of 1487 is consistent with the record that King Richard III died in 1485 (http://news.nationalgeographic.com/n...ensic-science/).

I agree with this solution, with minor edits to LATEX. I used the equation \(ln(\frac {N_t}{N_0})=-\lambda(t)\), but I got the same answer.

Q21.3.2: Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?

Alpha radiation/decay is a nucleus of the element Helium or He²⁺. The alpha particle itself is composed of 2 protons and 2 neutrons. The charge of 2+ is because there are no electrons to balance out the charge of the two protons. This would not be a nucleus if there were electrons to balance the charge. ß^- and ß⁺ are massless charged particles, not nuclei. Gamma radiation is electromagnetic energy, not even a particle, so it cannot be a nucleus either.

Q21.7.4: A scientist is studying a 2.234 g sample of thorium-229 (t_1/2 = 7340 y) in a laboratory.

1. What is its activity in Bq?
2. What is its activity in Ci?

The equation used to solve this problem is \(A =\lambda(N)\). This states that the activity of a sample is directly related to the number of atoms of the radioactive isotope or \({N}\)

1) Find the value of \({\lambda}\)

\({\lambda}\) is the rate constant and is found through the equation \(t_{1/2}=\dfrac{0.693}{\lambda}\)

The equation gives you the half life in years, and \({\lambda}\) is found in \(\ce{s^{-1}}\), meaning that years will have to be converted to seconds. There are 2.31474x10¹¹ seconds in 7340 years. \({\lambda}\) can be in any unit of time, raised to the -1. One should use the easiest measurement of time so that mistakes are minimal. However, the unit of Bq is decays per second, so converting to seconds in this step is logical. Plug this value into the equation and you get:

\(\lambda=\dfrac{0.693}{2.31474x10^{11} seconds}={2.99385x10^{-12} {sec^{-1}}}\)

2) To find N you take the mass in grams (it is given) and divide it by the molar mass of the isotope, to ultimately get a value in moles.

\(N=\dfrac{2.234g}{229 g/mol}={0.009755 mol}\)

3) Now that you have moles and you want an answer in atoms (that is the unit of N) you can multiply the value of moles by 6.02x10^{23} (Avogadro's number) to get the number of atoms or N, which is the value you are looking for:

\({(0.009755mol)}{(6.02x10^{23} atoms/mol)}={5.87279x10^{21} atoms}\)

4) Plug the value for N and \({\lambda}\) into \(A =\lambda(N)\) and solve for Bq or decays per second

\(A = {(2.99385x10^{-12} {sec^{-1}})}{5.87279x10^{21} atoms}={1.75823x10^{10} \,decays\,per\,second\,or\,Bq}\)

5) To find in Curies, just divide the value in Bq by 3.7x10¹⁰ as there are 3.7x10¹⁰Bq in 1 Curie

\(\dfrac{1.75823x10^{10} \,decays\,per\,second\,or\,Bq}{3.7x10^{10} Bq}={0.475196 Ci}\)

I agree with this solution, with minor edits to LATEX.

Q20.4.17 : The standard cell potential for the oxidation of Pb to Pb²⁺ with the concomitant reduction of Cu⁺ to Cu is 0.39 V. You know that E° for the Pb²⁺/Pb couple is −0.13 V. What is E° for the Cu⁺/Cu couple?

To solve for the standard cell potential at 25 degrees Celcius you have to use the equation:

\(E^\circ_{\ce{cell}} = E^\circ_{cathode}-E\circ_{anode}\)

To find the \(E^\circ\) at the anode and the cathode, you have to look at the standard reduction reactions taking place at the anode and the anode. Note: anode is where the oxidation occurs (Pb to Pb²⁺⁾ and cathode is where the reduction occurs (Cu⁺ to Cu)

You then take the values of \(E^\circ\) and \(E\circ_{anode}\), and you plug them into the standard equation given above to solve for \(E^\circ_{cathode}\):

\({0.39V}=E^\circ_{cathode}-{-0.13V}\)

\(E^\circ_{cathode}={0.26V}\)

Because the value for \(E^{o}_{cell}\) is positive, we know that the reduction potential for Cu⁺/Cu has to be greater than -0.13 V. The value we obtain, 0.26 V, fulfills the requirement.

I agree with this solution, however the experimentally determined E^o_cathode for Cu⁺/Cu is 0.52 V (https://chem.libretexts.org/Referenc...tials_by_Value).

Q20.9.2: How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?

Generally when determining the strength of various oxidants and reductants, an individual could simply look to the E° values of the substance. This could be done by looking at the chart that details the standard reduction potentials for various substances. It also details the oxidative and reductive strengths for various substances. When looking at standard reduction potentials, positive values mean that a compound is more spontaneous to reduce than H⁺/H₂ and have high reductive strength. These are good oxidizing agents. Compounds with negative standard reduction potentials are less spontaneous to reduce than H⁺/H₂ and have high oxidative strength. These are good reducing agents. Reduction potentials are relative values that can be compared between any two compounds.

However, the strength of various oxidants and reductants can also be compared by look at an electrolytic cell. Electrolytic cells work on the basis of being a chemical system through which an electrical current is driven. Just like a Galvanic Cell, electrolytic cells drive oxidation and reduction reactions at a cathode and anode. However, the reaction is not spontaneous. To determine the strengths of various oxidants and reductants, it is important to understand the stoichiometry behind the reaction that is taking place, and know the number of moles of electrons that are being moved through the cell system. One can then use this stoichiometric factor along with the corresponding time that has passed since the reaction began and the voltage of the current that was used by the reaction (as this reaction is, once again, not spontaneous) to determine the amount of the moles of electrons that is transferred between the electrodes. (Note: the electrons were passed from the anode to the cathode). We can multiply the time of the reaction by the current of electricity added (amp/sec) to find the total number of coulombs. We use the relationship between 1 mole of electrons and 96,495 coulombs to find the moles of electrons passed from the anode to the cathode. Using the stoichiometric factor between moles of electrons and moles of solid product, and then the molar mass of solid product, we can find the mass of product that reduced onto the cathode via the electrolytic process. To make quantitative comparisons between various oxidants and reductants, one can compare the amount of product formed on the cathode and compare the amounts of product transferred in different reactions. To be able to experimentally determine this, we keep one half of the cell the same in all trials as a control. We then change the concentration or identity of the of the compounds in the other half cell and compare the mass of product formed on the cathode. If we keep the oxidation half as a control, we can determine the strength of reductants, and if we keep the reduction half as a control we can determine the strength of oxidants.

Q14.1.2: If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?

Chemical kinetics is the area of chemistry that explores speeds, rates, and, subsequently, the rates of various reactions. The subject explores how rapidly chemical reactions occur, rather than the products they form. However, kinetics does focus on any intermediates that form during a reaction, and the disappearance and reappearance of catalysts. This is important because rates of various chemical reactions span a wide range: a couple seconds, such as explosions, to millions of years, such as the formation of diamonds from carbon or radioactive decay. It is necessary to study chemical reaction kinetics in an industrial setting, initially, in order to understand the danger that comes with chemical reactions. The study of chemical kinetics can inform individuals of the hazards that come with reactions, such as radioactivity and explosions. A reaction could also make dangerous or corrosive intermediates that should be avoided. This will inform individuals to follow necessary safety procedures. It is also important to study kinetics in order to determine if the reaction will happen on a time scale that is profitable for industry. A pencil company would go bankrupt if its source of graphite was from decomposing diamonds, which takes millions of years. The study of chemical kinetics can also provide industrial companies with information to innovate solutions for global issues. For example, chemical kinetics can be used to study catalysts that have the ability to speed up necessary reactions and synthesize new materials, or create products on a time scale that is reasonable for industrial production. In this situation, chemical kinetics would also help workers understand how long it takes reactions to occur and with what intensity. This will allow them to know things like when it is safe to mix various substances and how to tell if a reaction is proceeding at the correct rate. Synthetic catalysts could potentially help people who have cell disorders and create nonfunctional enzymes. If scientists study kinetics and how catalysts work, they may be able to create substitute enzymes and save lives. Some other questions to investigate are as follows: What determines the rate at which steel rusts? The activation energy determines the rate. How can we raise the activation energy so that the rate of corrosion is slower? What fuel helps an automobile and how does this process occur? What can we learn from biological enzymes and cell respiration combustion that we can apply to fuel efficiency and fossil fuel combustion? Could we someday use sugars as an everyday fuel source? What designs for products will allow for the most profitability while keeping safety in mind?