Extra Credit 7

Q17.1.6

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem. Equations, taken from the previous problem, posted here:

Equation 1: H₂O₂ + Sn²⁺⟶ H₂O + Sn⁴⁺

Equation 2: PbO₂+ Hg ⟶ Hg₂²⁺ + Pb²⁺

Equation 3: Al + Cr₂O₇^2- ⟶ Al³⁺ + Cr³⁺

Solution 17.1.6

Step 1: The first step in determining which species undergoes oxidation and reduction, and which species acts as the reducing agent and the oxidizing agent, is to understand what these terms mean. We know that oxidation refers to the loss of electrons (which will correspond to an increase in the Oxidation Number), and reduction refers to the gain of electrons (which will correspond to an decrease in the Oxidation Number). We also know that the oxidizing agent is the species that causes the oxidation, with itself being reduced. We also know that the the reducing agent is the species that causes the reduction, with itself being oxidized.

Step 2: The next step is to assign oxidation numbers to each element in the equations.

Equation 1: H₂O₂ + Sn²⁺⟶ H₂O + Sn⁴⁺

• H₂O₂: The molecule is neutral so the sum total of the Oxidation Numbers are 0. The Oxidation Number of Hydrogen is +1, and there are 2 Hydrogen atoms, for a sum total of a +2 charge. In order to balance this +2 charge, the Oxidation Number of each Oxygen molecule is -1, and with 2 Oxygen atoms, the sum total is a charge of -2. Therefore the overall charge of the neutral molecule is 0.
• Sn²⁺: The Oxidation Number of an ion matches the charge of the ion. Therefore the Oxidation Number of Tin is +2 because the charge of the ion is +2.
• H₂O: The molecule is neutral so the sum total of the Oxidation Numbers are 0. The Oxidation Number of Hydrogen is +1. There are 2 Hydrogen atoms, so the overall charge from the Hydrogen atoms is +2. Therefore, in order for the sum total to be zero, the Oxidation Number of the Oxygen atom must be -2.
• Sn⁴⁺: The Oxidation Number of an ion matches the charge of the ion. Therefore the Oxidation Number of Tin is +4 because the charge of the ion is +4.

Equation 2: PbO₂+ Hg ⟶ Hg₂²⁺ + Pb²⁺

• PbO₂: The molecule is neutral so the sum total of the Oxidation Numbers are 0. We typically assign an Oxidation Number of +2 to Oxygen, and there are 2 Oxygen atoms, so the charge of Lead must be +4 in order for the molecule to be neutral.
• Hg: The Oxidation Number of a Neutral atom is 0, therefore the Oxidation Number of the Mercury atom is 0.
• Hg₂²⁺:The Oxidation Number of an ion matches the charge of the ion. The charge of this ion is +2, and there are two Mercury atoms making up this ion, therefore the Oxidation Number of each individual Mercury atom is +1.
• Pb²⁺: The Oxidation Number of an ion matches the charge of the ion. The charge of this ion is +2, and there is one Lead atom making up this ion, therefore the charge of the Lead ion is +2.

Equation 3: Al + Cr₂O₇^2- ⟶ Al³⁺ + Cr³⁺

• Al: The Oxidation Number of a Neutral atom is 0, therefore the Oxidation Number of the Aluminum atom is 0.
• Cr₂O₇^2-: The charge of the overall ion is -2. The Oxidation Number of each Oxygen is -2, with 7 Oxygen atoms for a total of -14. Therefore, the Oxidation Number of each of the two Chromium atoms must be +6, which together will be +12, with the -14 from the Oxygen, leaving -2, which is the overall charge of the ion.
• Al³⁺: The charge of the ion is +3, therefore the Oxidation Number is +3.
• Cr³⁺: The charge of the ion is +3, therefore the Oxidation Number is +3.

Step 3: Now, we look at the change in Oxidation Numbers of the elements to see which elements gained or lost electrons.

Equation 1: H₂O₂ + Sn²⁺⟶ H₂O + Sn⁴⁺

• H: No change in Oxidation Numbers.
• O: The Oxidation Number goes from -1 to -2 therefore it is gains electrons. Therefore H₂O₂ is reduced and acting as an oxidizing agent.
• Sn: The Oxidation Number goes from +2 to +4 therefore it loses electrons. Therefore, Sn²⁺ is oxidized and acting as a reducing agent.

Equation 2: PbO₂+ Hg ⟶ Hg₂²⁺ + Pb²⁺

• Pb: The Oxidation Number of the Lead goes from +4 to +2 therefore it gains electrons. Therefore, PbO₂ is reduced and acting as an oxidizing agent.
• Hg: The Oxidation Number of the Mercury goes from 0 to +1 therefore it loses electrons. Therefore, Hg is oxidized and acting as a reducing agent.

Equation 3: Al + Cr₂O₇^2- ⟶ Al³⁺ + Cr³⁺

• Al: The Oxidation Number of Aluminum goes from 0 to +3 therefore it loses electrons. Therefore, Al is oxidized and acting as a reducing agent.
• Cr: The Oxidation Number of Chromium goes from +6 to +3 therefore it gains electrons. Therefore, Cr₂O₇^2- is reduced and acting as an oxidizing agent.

Q19.1.5

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

Solution 19.1.5

In order to prepare La by the reduction of La₂O₃, we must pick an element, between Al, C and Fe, with the strongest reduction potential. To determine which element has the strongest reduction potential, we use compare the Standard Reduction Potentials of the elements in question. Comparing the Standard Reduction Potentials, we see that Aluminum has the most negative Standard Reuction Potential and therefore is the strongest reducing agent. We want to use the strongest reducing agent because we want to induce a reduction of La₂O₃, therefore Aluminum is most likely to be used to prepare La by the reduction of La₂O₃.

Q19.2.7

Name each of the compounds or ions given in Exercise Q19.2.5.

I have reprinted the compounds or ions here:

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

Solutions 19.2.7

When naming complex ions, be sure to follow IUPAC Rules. These include the following:

• Name the cation before anion.
• Name ligands before the metal ion.
• Be sure to use the modified ligand name.
• In the case of multiple ligands, name them in alphabetical order.
• Designate the number of ligands with proper prefixes.
• Designate the Oxidation State of the metal ion with roman numerals.
• If the complex ion has a negative charge, end it with -ate.
• Designate isomerism with the proper prefix.

1. [Co(en)₂(NO₂)Cl]⁺

• The metal ion is cobalt. It has an oxidation state of +3.
• There is a chloro, a nitro, and two ethylenediamine ligands.
• Therefore: chlorodiethylenediaminenitrocobalt (III) ion.

2. [Co(en)₂Cl₂]⁺

• The metal ion is cobalt. It has an oxidation state of +3.
• There are two chloro and two ethylenediamine ligands.
• Therefore: dichlordiethylenediaminecobalt (III) ion. Correction: Dichlorobis(ethylenediamine)cobalt (III) ion

3. [Pt(NH₃)₂Cl₄]

• The metal ion is platinum. The oxidation state is +4.
• There are four chloro ligands and two ammine ligands.
• Therefore: Diamminetetrachloroplatinum (IV).

4. [Cr(en)₃]³⁺

• The metal ion is chromium. The oxidation state is +3.
• There are three ethylenediamine ligands.
• Therefore: Triethylenediaminechromium (III). Correction: Tris(ethylenediamine)chromium (III)

5. [Pt(NH₃)₂Cl₂]

• The metal ion is platinum. The oxidation state is +2.
• There are two ammine ligands and two chloro ligands.
• Therefore: Diamminedichloroplatinum (II).

Q12.3.20

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 12.3.20

Step 1: The first step is to write the rate law. We know the general formula for for a first-order rate law. It is as follows: Rate=k[A]

Step 2: We now plug in [N₂O₅] in for [A] in our general rate law. We also plug in our rate constant (k), which was given to us. Now our equation looks as follows:

Rate=(6.2x10^-4 min^-1)[N₂O₅]

Step 3: We now plug in our given molarity. [N₂O₅]=0.4 M. Now our equation looks as follows:

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

Step 4: We now solve our equation. Rate=(6.2x10^-4 min^-1)(0.4 M)= 2.48x10^-4 M/min.

Q12.6.11

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 12.6.11

1. Write the Overall Reaction. We know the overall reaction involves, reactants and products with correct stoichiometry and phases. We also know that the overall reaction does not include intermediates. First we identify our reactants. These are CO gas and Cl₂ gas. Then we identify our products. This is just COCl₂ gas. Therefore, we can now build our overall reaction, which is as follows:

CO(g) + Cl₂(g) ⟶ COCl₂(g)

2. Identify all intermediates. First, lets define intermediate. An intermediate is any species involved in the mechanism but that is not in the overall reaction. Intermediates are not reactants or products. Intermediates can now be easily identified. They are as follows: Cl and COCl.

3. Write the rate law for each elementary expression. We define the rate law only in terms of reactants only and use the stoichometry of the elementary steps to determine which power to raise each concentration to.

• Cl₂(g)⇌2Cl(g) (fast, k₁ represents the forward rate constant, k₋₁ the reverse rate constant). This reaction is reversible, therefore we can define it in terms of both Cl₂ and Cl. Rate=-k₁[Cl₂] + k_-1[Cl]². At equilibrium, there is no change in [Cl₂], therefore we can say -k₁[Cl₂] + k_-1[Cl]²=0. This we can simplify to k_-1[Cl]²=k₁[Cl₂]. This we can simplify to [Cl]=√((k₁/k_-1)[Cl₂])
• CO(g)+Cl(g)⟶COCl(g) (slow, k₂ the rate constant). This is the slow step, and therefore is the rate determining step. We can write the rate law as follows: Rate=k₂[CO][Cl].
• COCl(g)+Cl(g)⟶COCl₂(g) (fast, k₃ the rate constant). We can write the rate law for this step as follows: Rate=k₃[COCl][Cl].

4. Write the overall rate law expression. First we know that the rate law for the overall reaction is the rate law for the slow step, which is step 2. Rate=k₂[CO][Cl]. However, we cannot leave intermediates in our overall rate law, thus we must use the simplified rate law from step 1. Pluging this in for [Cl], we get Rate=k₂[CO](√((k₁/k_-1)[Cl₂])). This is our overall rate law, because there are no intermediates.

Q21.4.23

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 239-Pu was the capture of neutrons by 238-U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 10⁹ years ago?

Solution 21.4.23

In order to explain why Plutonium-239 would not have been likely to have been trapped at the formation of the solar system 4.7x10⁹ years ago, we must show what percentage of Plutonium-239 would still exist if it had been trapped 4.7x10⁹ years ago. Nuclear decay is a first-order process, so we can use the equation ln(N/N₀)=-kt. We also make the following relation; fN=N₀, with f being the decimal that represents the amount of Plutonium-239 still remaining compared to the original amount.

First we will find k. To do this, we have the following equation: t_(t/2)=ln(2)/k. We can solve this for k=ln(2)/t_(t/2). We then plug in the half life of 239-Pu (which we found of Wikipedia), and solve for k: k=ln(2)/24110 years. k=.0000287493646 years^-1.

We now set uo our first order decay equation, plugging in k, t, and substituting in f for the fraction of Plutonium-239 still remaining. This is as follows:

ln(fN/N)=-(.0000287493646 years^-1)(4.7x10⁹ years). Solving for f, we get;

ln(f)=-135122.0136

f=e^{-135122.0136​}. Solving for f, we find f=0. Therefore, there is no Plutonium-239 left from the start of the solar system 4.7x10⁹ years ago.

Q20.3.11

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?

Solution 20.3.11

First, we must identify that this question is asking us to write two half reactions which we then will balance into an overall redox reaction.

Step 1: Write half- reactions.

We have SO₄^2- (sulfate) being reduced to HS^-.

Reduction: SO₄^2- ⟶ HS^-

We have C₆H₁₂O₆ (glucose) being oxidized to HCO^3-. Correction: Bicarbonate is actually HCO₃^-

Oxidation: C₆H₁₂O₆ ⟶ HCO^3- Correction: Oxidation: C₆H₁₂O₆ ⟶ HCO₃^-

Step 2: Balance all elements except H or O.

Reduction: SO₄^2- ⟶ HS^-

Oxidation: C₆H₁₂O₆ ⟶ 6HCO^3- Correction: Oxidation: C₆H₁₂O₆ ⟶ 6HCO₃^-

Step 3: Balance O by adding H₂O.

Reduction: SO₄^2- ⟶ HS^- + 4H₂O

Oxidation: C₆H₁₂O₆ ⟶ 6HCO^3- Correction: Oxidation: C₆H₁₂O₆+12H₂O ⟶ 6HCO₃^-

Step 4: Balance H by adding H⁺

Reduction: 9H⁺ + SO₄^2- ⟶ HS^- + 4H₂O

Oxidation: C₆H₁₂O₆ ⟶ 6HCO^3- + 6H⁺ Correction: Oxidation: C₆H₁₂O₆+12H₂O ⟶ 6HCO₃^-+30H⁺

Step 5: Balance the charge by adding electrons to the more positive side.

Reduction: 8e^- + 9H⁺ + SO₄^2- ⟶ HS^- + 4H₂O

Oxidation: C₆H₁₂O₆ ⟶ 6HCO^3- + 6H⁺ + 3e^- Correction: Oxidation: C₆H₁₂O₆+12H₂O ⟶ 6HCO₃^-+30H⁺​​​​​​​+24e^-

Step 6: Multiply the reactions by a whole number so that the number of electrons transferred match.

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

Reduction: 24e^- + 27H⁺ + 3SO₄^2- ⟶ 3HS^- + 12H₂O

Oxidation: 8(C₆H₁₂O₆ ⟶ 6HCO^3- + 6H⁺ + 3e^-)

Oxidation: 8C₆H₁₂O₆ ⟶ 48HCO^3- + 48H⁺ + 24e^- Correction: Oxidation: C₆H₁₂O₆+12H₂O ⟶ 6HCO₃^-+30H⁺+24e^-​​​​​​​

Step 7: Add the two half reactions and cancel the common terms.

24e^- + 27H⁺ + 3SO₄^2- + 8C₆H₁₂O₆ ⟶ 3HS^- + 12H₂O + 48HCO^3- + 48H⁺ + 24e^-

24e^- + 27H⁺ + 3SO₄^2- + 8C₆H₁₂O₆ ⟶ 3HS^- + 12H₂O + 48HCO^3- + 27H⁺ + 21H⁺ + 24e^-

Correction: 24e^- + 27H⁺ + 3SO₄^2-+ C₆H₁₂O6+12H₂O ⟶ 6HCO₃^-+30H⁺+24e^-​​​​​​​+ 3HS^- + 12H₂O

Here is the final balanced redox equation:

3SO₄^2- + 8C₆H₁₂O₆ ⟶ 3HS^- + 12H₂O + 48HCO^3-+ 21H⁺

Correction: Final balanced redox equation should be: 3SO₄^2-+ C₆H₁₂O₆ ⟶ 6HCO₃^-+3H⁺+ 3HS^-

Q20.5.22

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?

Solution 20.5.22

1. First, we must know our equation for the E°. It is as follows: E°_Cell = E°_Cathode - E°_Anode. We know that reduction occurs at the cathode and oxidation occurs at the anode. In the following equations, we must decide which one is the reduction half-reaction and which one is the oxidation half-reaction.

Mn³⁺(aq) + e⁻ → Mn²⁺(aq) In this equation Mn³⁺ is gaining an electron therefore it is being reduced therefore this is the reduction half-reaction which occurs at the cathode.

Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺(aq) + e⁻ The Oxidation Number of Mn in the MnO₂ molecule is +4. The Oxidation Number of Mn went from +3 to +4 therefore it is being oxidized. Therefore this is the oxidation half-reaction that occurs at the anode.

Now, we take the corresponding E° values and plug them into E°_Cell = E°_Cathode - E°_Anode in their corresponding spots. Therefore, our equation is as follows:

E°_Cell = 1.51V-0.95V=0.56V. This is the E° value for disproportionation reaction.

2. A low pH will have more H⁺ ions than a pH of 7 becuase pH is a negative logarithm. This will increase Q in the Nernst equation (E=E^o−(0.0592V/n)logQ) thus the second term will increase. This second term is being subtracted from the E° term, therefore the E term will decrease. Becuase this is happening in the oxidation half-reaction, E_Anode will decrease, therefore E_Cell will increase. A more positive E_Cell corresponds to a more negative ΔG. A more negative ΔG corresponds to a more thermodynamically favored process. Therefore, at a lower pH, the disproportionation will be more thermodynamically favored.

3. To prevent the disproportionation of Mn(III), another element, such as Cerium, with a more negative E°, can be added to the solution to act as a sacrificial anode, because it is more readily oxidized than Mn(III).