Extra Credit 11

17.1.10

Why must the charge balance in oxidation-reduction reactions?

There is no net change in the number of electrons in an oxidation-reduction reaction because electrons given off the oxidation half reaction are taken up by another species in the reduction half reaction. The charge must balance in oxidation-reduction reactions because they consist of half reactions with a reduced half and an oxidized half that occur simultaneously.

Oxidation state is the amount of electrons lost or gain during a chemical reaction. Oxidation reactions also generates electrons. In a redox reaction, the neutrality must be maintained to obstruct violation of the law of conservation. This is why a salt bridge is required to maintain the neutrality. So the charge must balance for an equal amount of electrons to be transferred between the cathode and anode to maintain the charge balance.

The Law of the Conservation of Mass states that matter is neither created nor destroyed in a chemical reaction. This is why when a chemical reaction takes place, its chemical equation is balanced. This takes into account the amount of electrons that are donated by the oxidized substance and the amount of electrons that are taken in by the reduced substance. The numbers of electrons cancel out as mass is conserved.

19.1.9

Why is the formation of slag useful during the smelting of iron?

Slag is a substance formed as a byproduct of iron ore or iron pellets melting together in a blast furnace. Slag is also the byproduct that is formed when a desired metal has been separated from its raw ore. It is important to note that slag from steel mills is created in a manner that reduces the loss of the desired iron ore. The \(\ce{CaSiO3}\)slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to \(\ce{O2}\), which would oxidize the \(\ce{Fe}\) back to \(\ce{Fe2O3}\). Since Fe has a low reduction potential of -0.440 this means it has a high oxidation potential so it would easily oxidize in the presence of O2. This would allow the maximum product of iron to be obtained in the end of smelting.

19.3.1

Determine the number of unpaired electrons expected for \(\ce{[Fe{(NO2)}6]}^{-3}\) and for \(\ce{[FeF6]}^{-3}\) in terms of crystal field theory.

\(\ce{[Fe{(NO2)}6]}^{-3}\)

Given: Chemical Formula of complex ions

Asked for: number of unpaired electrons

1. For the first species, Fe has a charge of 3+ because \(\ce{NO2}\) has a -1 charge and 6(-1)+3=-3 which is the overall charge of the complex ion so the electron configuration would be [Ar]3d5 after removing 3 electrons, the first from the 4s orbital and then from the d orbital.

2. It is octahedral because it has 6 \(\ce{NO2}\) ligands attached to the transition metal, Fe. Octahedral has splitting levels of eg and t2g levels

3. \(\ce{NO2}\) is a strong field ligand so it is low spin since P<deltaO. This means the pairing energy is lower than the splitting energy. Thus, the compound would rather pair at a lower level than jump to fill a higher level. Therefore it goes to low spin. When adding 5 electrons to the diagram in low spin formation you find there is 1 unpaired electron.

\(\ce{[FeF6]}^{-3}\)

1. F has a charge of -1 so Fe would have a charge of 3+ also since 6(-1)+3=-3 which is the overall charge of the complex ion.

2. The electron configuration is [Ar]3d5 after removing 3 electrons from the 4s orbital first then the 3d orbital

3. F is a weak field ligand so it would have a high spin formation since P>deltaO. The electrons would rather move up to next E level (eg) instead of pairing up together at a lower level (t2g). This is because less E output is favored. After adding 5 electrons in high spin formation, you find there are 5 unpaired electrons.

12.4.1

Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of A at varying times.

Graphs for the different orders of chemical reactions can allow an individual to accurately determine the order of a cell. The graphs of the various are only linear at the following specifications. In turn, these specifications, if met, indicate the order of a reaction. The linear graph of [A] vs t indicates the the 0 order. The linear graph of ln[A] vs. t indicates a first order reaction. The linear graph of 1/[A] vs. t indicates the reaction has an order of 2. Remember to keep in mind that these linear graphs are always straight lines when indicating the order. In other words, if you get a straight line for those graphs it indicates the reaction's order.

The k of a chemical reaction can be determined by m=-k for the 0 and 1st order and m=k for the 2nd order. If you were given two concentrations at t1 and t2 you would use the corresponding graphs that gave you straight lines and find the slope of those lines given the concentrations and times.

12.7.4

For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:

For the reaction diagrams, the lower the activation energy means the reaction is catalyzed therefore in the first set of diagrams, b is the catalyzed reaction and for the second set, b is also the catalyzed reaction.

A catalyst is a substance that lowers the activation energy required for a reaction to take place and, in turn, increases the rate of the reaction. Thus, to determine which of the pair has been catalyzed, you have to look at which reaction has the least amount or decreased amount of Ea or Activation Energy. Both b diagrams show a reaction with a lower activation energy than the one before.

21.4.27

Write a balanced equation for each of the following nuclear reactions:

1. bismuth-212 decays into polonium-212

Given: nuclear decay reactant and product

Asked for: balanced nuclear reactions

Both atomic and mass number have to be equal on both sides of the equation. Polonium has an atomic number of 84 therefore a negative beta particle which has a mass number of 0 and -1 charge would need to be emitted from bi-212 which has an atomic number of 83. 84+(-1)=83=83 on both sides and 212=212 so this is right.

₈₃²¹²Bi--> ₈₄²¹²Po + _-1⁰e

If you add the mass and atomic numbers they are equal on both sides.

2. beryllium-8 and a positron are produced by the decay of an unstable nucleus

Given: products of a positron emission reaction

Asked for: the reactant needed to produce Be-8 and a positron

A positron is a positively charged beta ray with mass number of 0 and charge of +1 and Be has an atomic number of 4 so 4+1=5 and Boron has this atomic number so it would be

⁸₅B --> ⁸₄Be + ⁰₁e

3. neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239

Given: reaction between Np-239 and U-238 with a neutron and chain reaction product Pt-239

Asked for: necessary particles needed for these reactions to occur.

They give you ₉₂²³⁸U + ₀¹n ---> ₉₃²³⁹Np. so you would have to add an electron or beta particle to the product side so the sum of the atomic numbers are equal it would be ₉₂²³⁸U + ₀¹n ---> ₉₃²³⁹Np + _-1⁰e, to convert this to Plutonium, Np emits a negative beta particle to make the sum equal for both atomic number and mass number so it would be ₉₃²³⁹Np--> ²³⁹₉₄Pu + ⁰_-1e

4. strontium-90 decays into yttrium-90

Given: Products and reactants of a nuclear reaction

Asked for: necessary particles for this to take place.

yttrium has an atomic number of 39 so Sr would emit a beta particle to make the sums equal on both sides since a beta particle has a mass number of 0 and a -1 charge, and it would be

⁹⁰₃₈Sr ---> ₃₉⁹⁰Y + ⁰_-1e

(Important notes to solve nuclear reaction questions:

• - to balance a nuclear reaction, the sum of all the mass numbers and atomic numbers, respectively and accordingly, on either side of the arrow must equal each other.
• - the mass number is the value on top
• - atomic numbers are the value on the bottom
• - the transformation of particles follows conservation laws such as the law of conservation of mass)

20.3.13

For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.

1. \[\ce{Zn}(s)│\ce{Zn^2+}(aq)║\ce{H+}(aq)│\ce{H2}(g)│\ce{Pt}(s)\]
2. \[\ce{Ag}(s)│\ce{AgCl}(s)│\ce{Cl-}(aq)║\ce{H+}(aq)│\ce{H2}(g)│\ce{Pt}(s)\]
3. \[\ce{Pt}(s)│\ce{H2}(g)│\ce{H+}(aq)║\ce{Fe2+}(aq)│\ce{Fe3+}(g)│\ce{Pt}(s)\]

Given: cell diagram for redox reactions

Asked for: half reactions and overall reaction

For number 1, the reduction potential of Zn is -0.76 from the standard reduction table and H has a potential of 0 so the higher the reduction value the better oxidizing agent so it will be reduced and be in the cathode. H has a higher potential so it is in the cathode and will be reduced. The lower one will be oxidized so it is in the anode.

1. reduction: \(\ce{2H+}(aq)+\ce{2e^-}⟶\ce{H2}(aq)\); cathode; the charges have to be balanced so 2 electrons are needed to make the left side 0 charge which equals the right side.

oxidation: \(\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2e^-}\); anode; 2 electrons needed to make the right side charge 0 to be equal to the left side.

overall: \(\ce{Zn}(s)+\ce{2H^+}(aq)⟶\ce{Zn^2+}(aq)+\ce{H2}(aq)\); cancel the electrons and rewrite the remaining products and reactants

Using the same logic as the first, AgCl has a reduction potential of 0.2223 and H has a value of 0 so the higher one, AgCl will be reduced and will be at the cathode and H at the anode.

2. reduction: \(\ce{AgCl}(s)+\ce{e^-}⟶\ce{Ag}(s)+\ce{Cl^-}(aq)\); cathode; one electron needed to make the left side a -1 charge to match the right side.

\(\ce{2AgCl}(s)+\ce{2e^-}⟶\ce{2Ag}(s)+\ce{2Cl^-}(aq)\) To cancel the electrons we multiple the reduction by 2

oxidation: \(\ce{H2}(g)⟶\ce{2H^+}(aq)+\ce{2e^-}\); anode; add 2 electrons to make the right side 0 to match the left side.

overall: \(\ce{2AgCl}(s)+\ce{H2}(g)⟶\ce{Zn^2+}(aq)+\ce{H2}(aq)\)

The reduction value for Fe3+ is 0.771 and for H its 0 so Fe3+ will be reduced and be at the cathode and H at the anode since a higher reduction potential equals better oxidizing agent.

(Note to Dr. Larsen: I think that in this cell diagram indicates that that the AgCl (s) half reaction is at the anode and the \(\ce{H2}(g)\) is at the cathode. As per the instructions provided by libretext in the lecture below, when writing a cell diagram for a galvanic (voltaic cell) the anode (what is oxidized) is always shown on the left and the cathode (what is reduced) is shown on the right. (There is a link for the libretext page given below). The question should be edited to fit this standard or the idea that the cell diagram is not always anode to cathode, when left to right respectively, in order for students to understand that this example breaks from commonly taught diagrams. This could be done with a hint. Either way, I think it would help students understand this better if this concept was clarified in the question.)

https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells

3. reduction: \(\ce{Fe^3+}(aq)+\ce{e^-}⟶\ce{Fe^2+}(aq)\) ; cathode; add one electron to make the charges equal on both sides

\(\ce{2Fe^3+}(aq)+\ce{2e^-}⟶\ce{2Fe^2+}(aq)\) multiply both sides by 2 so the electrons will cancel on both sides of the equation

oxidation: \(\ce{H2}(g)+\ce{e^-}⟶\ce{2H^+}(aq)+\ce{2e^-}\) ; anode;

overall: \(\ce{2Fe^3+}(aq)+\ce{H2}(g)⟶\ce{2H^+}(aq)+\ce{2Fe^2+}\)

20.5.26

Ideally, any half-reaction with E° > 1.23 V will oxidize water as a result of the half-reaction \[O_2(g) + 4H^+(aq) + 4e^− → 2H_2O(l)\]

1. Will FeO₄2⁻ oxidize water if the half-reaction for the reduction of Fe(VI) → Fe(III) is \(\ce{FeO4^2-}(aq)+\ce{8H^+-}+\ce{3e^-}⟶\ce{Fe^3+}(aq)+\ce{4H2O}(aq)\); E° = 1.9 V?

1.9 is greater than 1.23 therefore, according to the given information, it will oxidize water.

1. What is the highest pH at which this reaction will proceed spontaneously if [Fe³⁺] = [FeO₄²⁻] = 1.0 M and \(P_\mathrm{O_2}\)= 1.0 atm?

Couldn't get this....

When this reaction is balanced the equation is as follows. In order to get this equation you have to realize that water is oxidized at the anode and iron is reduced at the cathode. You have to multiply the water half reaction by 3 and the iron reaction by 4 so that the electrons balance out. After this is all done, the equation looks like the following.

\(\ce{4FeO4^2-}(aq)+\ce{20H^+-}+\ce{6H2O}(l)⟶\ce{4Fe^3+}(aq)+\ce{16H2O}(aq)+\ce{3O2}(g)\)

To solve this equation to determine the highest pH possible one has to use the Nernst Equations and use the concentration and the values for the pressure of gas given above.

\[E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0257 V}}{n}\right)\ln Q \label{Eq4}\]

The following equation determines the value of \(E^\circ_{\ce{cell}}\)

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

\(E^\circ_{\ce{cell}} = {1.9V}-{1.23V}={0.67V}\)

For the reaction to be spontaneous, the E cell has to be positive. This means that the value of everything after the (-) sign on the right side of the equation, has to be less than 0.67.

To find Q, you just have to set up the equilibrium constants with the products on top and reactants on the bottom. Important reminders for this are that only substances that are aqueous or gaseous are included in Q. Also remember to raise the concentration or the Pressure to the value of their coefficient. Since the values for concentrations of the various iron substances and the pressure for oxygen are given as 1, they all cancel out and the only substance that we are left with is H+ concentration. Since H+ is on the products side, Q is equal to 1 over \(\ce{H^+}\) raised to the twenty. Also note that n in this case is equal to 12. Plug the values for Q,n, and \(E^\circ_{\ce{cell}}\) into the Nernst equation given above.

\({1.67} > {(0.002142)}ln{\dfrac{1}{H+^{(20)}}}\)

Solve this equation to get the H+ concentration isolated and plug the value for H+ into the equation for pH which is \[pH = -\log[H^+]\]

The pH value you should receive is 6.93

Side note to Larsen: I just want a B in the class, I'm at a B- now (82.68%), hope this can boost me :)