Extra Credit 7

Question 17.1.6

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions.

1. H₂O₂+Sn²⁺+⟶H₂O+Sn⁴⁺
2. PbO₂+Hg⟶Hg₂²⁺+Pb²⁺
3. Al+Cr₂O₇^2-⟶Al³⁺+Cr³⁺

Answer

oxidized: (a) Sn²⁺; (b) Hg; (c) Al

reduced: (a) H₂O₂; (b) PbO₂; (c) Cr₂O₇^2-

oxidizing agent: (a) H₂O₂; (b) PbO₂; (c) Cr₂O₇^2-

reducing agent: (a)Sn²⁺ ; (b) Hg; (c) Al

Solutions

1. You can tell that Sn is oxidized since it goes from a 2+ charge to a +4 charge. This means that it lost electrons in order to become more positively charged and oxidation is losing electrons. This means that H₂O₂ is going to be reduced and is since the oxygen gains electrons. The oxidized species is also the reducing agent and vice versa since in order for something to gain electrons it must take them from something else and vice versa

2. Hg is oxidized since its oxidation number goes from 0 to 2 meaning it lost two electrons which are then given to Pb to go from an oxidation number of 4 to 2.

3. Al goes from an oxidation state of 0 to 3, meaning it lost 3 electrons which are then given to Cr₂O₇^2- moving Cr's oxidation number down from 6 to 3.

Question 19.1.5

Which of the following elements is most likely to be used to prepare La by the reduction of La₂O₃: Al, C, or Fe? Why?

Explanation

Al³⁺(aq)+3e^- ---> Al(s) E^º= -1.66V

While iron is only....

Fe³⁺(aq) + 3e^----> Fe(s) E^°= -0.036V

• Aluminum will be more spontaneous than Iron and will have a more negative ΔG. You can also tell this since the more negative the E^º the better the substance will act as a reducing agent

Answer

Al

Question 19.2.7

Name each of the compounds/ions

1. [Co(en)₂(NO₂)Cl]⁺
2. [Co(en)₂Cl₂]⁺
3. [Pt(NH₃)₂Cl₄]
4. [Cr(en)₃]³⁺
5. [Pt(NH₃)₂Cl₂]

Answers

1. Chlorobisethylenediaminenitrocobaltate(III) ion

2. Dichlorobisethylenediaminecobaltate(III) ion

3. diamminetetrachloroplatinate(IV)

4. Tris-ethylenediaminechromate(III)

5. diamminedichloroplatinate (II)

Solution

When naming the compund you put the ligands first in alphabetical order. You then add the prefixes that correlate to the corresponding number of ligands in the compound in front of them (di for two tri for three etc.). When you have a complex ligand which already has a prefix like this in it (like en) you will use bis or tris instead. The central atom comes last and you also must find the oxidation state of the atom. You do this by adding up the charges of the ligands and then deducing what the atoms charge must be to achieve the overall charge of the complex. For example number 1 has a +1 charge but there are 2 NO₂ which have a -1 charge (en is nuetral) so what charge must we add to -2 to make the overall charge +1 → +3

Question 12.3.20

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N₂O₅, dissolved in chloroform, CHCl₃, is 6.2 × 10⁻⁴ min⁻¹.

2N₂O₅⟶ 4NO₂+O₂

What is the rate of the reaction when [N₂O₅] = 0.40 M?

Solution

1.First step is to create a rate law for the reaction. Since it is stated that is is first order you know that the rate is equal to the concentration of the reactant times the rate constant. (you can also deduce the overal order using the units of the constant and vice versa)

rate=k[N₂O₅]

2. From here you can just plug in the given numbers and solve for the rate

rate=(6.2x10^-4min^-1)(0.4 M)

rate= 2.5x10^-4M * min^-1

Question 12.6.11

The reaction of CO with Cl₂ gives phosgene (COCl₂), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:

• Cl₂(g)⇌2Cl(g) (fast, k₁ represents the forward rate constant, k₋₁ the reverse rate constant)
• CO(g)+Cl(g)⟶ COCl(g) (slow, k₂ the rate constant)
• COCl(g)+Cl(g)⟶COCl₂(g) (fast, k₃ the rate constant)

1. Write the overall reaction.
2. Identify all intermediates.
3. Write the rate law for each elementary reaction.
4. Write the overall rate law expression.

Solution

1. To write the overall reaction you have to identify the intermediates and leave them out. The easiest way to do this is to write out all the products and reactants and cross out anything that is on both sides.

Cl₂(g) + CO(g) + 2Cl(g) +COCl(g) ⇒ 2Cl(g) + COCl(g) + COCl₂(g)

In this you will cross out the 2Cl(g) molecules and the COCl(g). What is left after that is the overall reaction.

Cl₂(g) + CO(g) ⇒ + COCl₂(g)

2. For part two you will just list the intermediates that you crossed out.

Cl and COCl are intermediates

3. Each rate law will be the rate equal to the rate constant times the concentrations of the reactants

reaction 1

(forward) rate=k₁[Cl₂] ( reverse) rate=k_-1[Cl]

reaction 2

rate=k₂[CO][Cl]

Reaction 3

rate=k₃[COCl][Cl]

4. The overall rate law is based off the slowest step (step #2), since it is the rate determining step, but Cl is present in that rate law so we have to replace it with an equivalent that does not contain an intermediate. To do this you use the equilibrium since the rates are the same you can set up the rate laws of the forward and reverse equal to each other.

k₁[Cl₂] = k_-1[Cl]

[Cl]= k₁[Cl₂]/k_-1

rate=k₂[CO]k₁[Cl₂]/k_-1

rate=k^°[CO][Cl₂]

Steps to replacing and intermediate

1. Set the forward and reverse reaction equal to each other using separate constants
2. Solve for the intermediate using algebra
3. Plug into the rate determining formula
4. All the k's will be condensed into a K prime constant

Question 21.4.23

Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this ²³⁹Pu was the capture of neutrons by ²³⁸U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 10⁹ years ago?

Solution

The half life of plutonium is 24,100 years and a concentration will be effectively zero after ten half lives. Since 241,000 years is much smaller than 4.7x10⁹ it could not have been trapped in the solar system all this time

Question 20.3.11

Sulfate is reduced to HS⁻ in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?

Answer

Reduction: 3x (SO₄^2- + 9H⁺ + 8e^- → HS^- + 4H₂O)

Oxidation: C₆H₁₂O₆ + 12H₂O → 6HCO₃^- + 30H⁺ + 24e^-

Overall: C₆H₁₂O_{6 +} 3SO₄^2-→ 6HCO₃^- +3H⁺ + 3HS^-

Solution

1. Reduction is gaining election so sulfate gains electrons and hydrogen ions to create HS^- and water. You have to balance this equation though by adding water to one side to balance the number of oxygen atoms and hydrogen ions to balance out the hydrogen from the water on the other side. After this there should be an equal amount of each element on every side and then you add electrons (in this case on the reactants side since its reduction) in order to balance out the charges on both sides.

2. You then will do the same thing with the oxidation reaction except the electrons will be on the products side since the glucose is loosing electrons.

3. You will then want to multiply the equations so that when you add them together the electrons will cancel out on both sides. In this problem you only need to multiply the reduction by three to make the electrons on the products 24 like in the reactants to be able to add them together.

4. Combine both reactions and cancel out anything that is on both sides. In this reaction there was 27 hydrogen ions on the reactants side and 30 on the products, so you are left with 3 on the products side in the final reaction.

Question 20.5.22

Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:

Mn³⁺(aq) + e⁻ → Mn²⁺(aq) E°=1.51 V

Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺(aq) + e⁻ E°=0.95 V

1. What is E° for the disproportionation reaction?
2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
3. How could you prevent the disproportionation reaction from occurring?

Solutions

1. To find the E° you have to add the two individual reactions together. This is because the disproportionation reaction is the combination of both reduction and oxidation (which is happening in these two reactions). So since the amount of electrons transferred in both of them is equal you can just add them together and get the correct Energy

E°= 1.51 V + 0.95 V = 2.46 V

2. It is more favored at a high PH since this means the concentration of hydrogen ions will be lower than at a low PH, which sounds counter intuitive, but the more acidic it will be the higher the amount of hydronium ions there will be. Due to Le Chateliers principle since hydrogen ions are a product of the reaction we want there to be a small concentration of them in order to drive the reaction forward .

3. You could prevent the reaction by having it be the opposite of the previous. If you put it in acidic conditions there will already be a high concentration of hydrogen ions and the forward reaction will go slowly. You could also get rid of reactants to push the reaction even more in the reverse direction and keep the dispoportionation from occurring.