Extra Credit 6

Q19.1.4

Why are the lanthanoid elements not found in nature in their elemental forms?

Solution: Q19.1.4

Lanthanides are the chemical elements found in Row 6 of the periodic table between Groups 3 and 4. They follow lanthanum (La), element #57, which accounts for their family name. The lanthanides include the metals cerium (Ce), praseodymium (Pr), neodymium (Nd), promethium(Pm), samarium (Sm), europium (Eu), gadolinium (Gd), terbium (Tb), dysprosium (Dy), holmium (Ho), erbium (Er), thulium (Tm), ytterbium (Yb), and lutetium (Lu). One thing about lanthanides is that they are not so rare, but they are very alike. Most of the lanthanides occur together in nature, and they are very difficult to separate from each other. They are not found in their elemental form in nature because of where they are found and their characterisitics. They are found in the Earth's crust which makes it incredibly difficult to extract.

Q19.2.6

Name each of the compounds or ions given in Exercise Q19.2.3, including the oxidation state of the metal.

1. [Co(CO₃)₃]^3-
2. [Cu(NH₃)₄]²⁺
3. [Co(NH₃)₄Br₂]₂(SO₄)
4. [Pt(NH₃)₄][PtCl₄]
5. [Cr(en)₃](NO₃)₃
6. [Pd(NH₃)₂Br₂]
7. K₃[Cu(Cl)₅]
8. [Zn(NH₃)₂Cl₂]

Solution: 19.2.6

All compound names must be written in alphabetical order!

1. Given: [Co(CO₃)₃]^3-
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Co(CO₃)₃]^3- → (CO₃)₃ = 3- carbonato

Step 2: Write out prefixes and name of metal; add -ate at the end of the transition metal because the complex is an anion.

Tricarbonatocobaltate

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

[Co(CO₃)₃]^3- Co=x 3CO^-₃=3(−2)

3(−2)+x=−3 x=+3 So the oxidation number of Co is +3

Step 4: Combine the name and oxidation number into one name

Tricarbonatocobaltate (III) ion

2. Given: [Cu(NH₃)₄]²⁺
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Cu(NH₃)₄]²⁺ → (NH₃)₄ = 4-ammine

Step 2: Write out prefixes and name of metal

tetraamminecopper

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

[Cu(NH₃)₄]²⁺ Cu=x 4NH₃=4(0)

4(0)+x=2 x=+2 So the oxidation number of Cu is +2

Step 4: Combine the name and oxidation number into one name

tetraamminecopper(II) ion

3. Given: [Co(NH₃)₄Br₂]₂(SO₄)
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Co(NH₃)₄Br₂]₂(SO₄) → (NH₃)₄ = 4 - ammine , Br₂ = 2-bromo

Step 2: Write out prefixes and name of metal

Tetraamminedibromocobalt

Step 3: Determine the oxidation number of the ligand and anion

[Co(NH₃)₄Br₂]₂(SO₄) Co=x 8NH₃=4(0) Br₂=2(−1) SO₄=−2

2[4(0)+2(−1)+x]+−2=0 −6+2x=0 x=+3 So the oxidation number of Co is +3

Step 4: Combine the name and oxidation number into one name

tetraamminedibromocobalt(III) sulfate

4. Given: [Pt(NH₃)₄][PtCl₄]
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands in the cation and anion

[Pt(NH₃)₄][PtCl₄] → Cation: (NH₃)₄ = 4-ammine

Anion: Cl₄ = 4-chloro

Step 2: Write out prefixes and name of metal; add -ate at the end of the transition metal because the complex is an anion.

Cation: tetraamineplatinum

Anion: tetrachloroplatinate

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

Cation: [Pt(NH3)4] Pt=x 4NH₃=4(0)

4(0)+x=

Anion: [PtCl4] Pt=y 4Cl=4(−1)

4(−1)+y=

4(0)+ x + 4(-1)+ y= 0 x+y+−4=0 x+y=4

Pt has common oxidation states of +2 and +4. Using this, we see that Pt is equal in both the anode and cathode which must be equal to +2.

Step 4: Combine the name and oxidation number into one name

tetraammineplatinum(II) tetrachloroplatinate(II)

5. Given: [Cr(en)₃](NO₃)₃
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Cr(en)₃](NO₃)₃ → (en)₃ = 3-en

Step 2: Write out prefixes and name of metal

tris-(en)chromium

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

[Cr(en)₃](NO₃)₃ Cr=x 3(en)=3(0) 3NO₃=3(0)

3(0)+x=−3 x=+3 So the oxidation number of Cr is +3

Step 4: Combine the name and oxidation number into one name

tris-(ethylenediamine)chromium(III) nitrate

6. Given: [Pd(NH₃)₂Br₂]
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Pd(NH₃)₂Br₂] → (NH₃)₂ = 2-ammine , Br₂= 2-bromo

Step 2: Write out prefixes and name of metal

Diamminedibromopalladium

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

[Pd(NH₃)₂Br₂] Pd=x 2NH₃=2(0) 2Br=2(−1)

2(0)+2(−1)+x=0 x=+2 So the oxidation number of Pd is +2

Step 4: Combine the name and oxidation number into one name

diaminedibromopalladium(II)

7. Given: K₃[Cu(Cl)₅]
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

K₃[Cu(Cl)₅] → (Cl)₅ = 5-chloro

Step 2: Write out prefixes and name of metal with the suffix; add -ate at the end of the transition metal because the complex is an anion.

Pentachlorocuprate

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

K₃[Cu(Cl)₅] 3K=+3 Cu=x 5Cl=5(−1)

5(−1)+x+3(1)=0 x=+2 So the oxidation number of Cu is +2

Step 4: Combine the name and oxidation number into one name

potassium pentachlorocuprate(II)

8. Given: [Zn(NH₃)₂Cl₂]
Asked for: Name compound with oxidation state of metals

Strategy:

Step 1: Identify the ligands

[Zn(NH₃)₂Cl₂] → (NH₃)₂ = 2-ammine , Cl₂ = 2-chloro

Step 2: Write out prefixes and name of metal

Diamminedichlorozinc

Step 3: Determine the oxidation number of the ligand by counting charge and ion charge

[Zn(NH3)2Cl2] Zn=x 3NH=3(0) 2Cl=2(−1)

2(0)+2(−1)+x=0 x=+2 So the oxidation number is +2

Step 4: Combine the name and oxidation number into one name

Diamminedichlorozinc (II)

Q12.3.19

For the reaction Q⟶W+XQ⟶W+X, the following data were obtained at 30 °C:

1. What is the order of the reaction with respect to [Q], and what is the rate equation?
2. What is the rate constant?

Solution: Q12.3.19

1. What is the order of the reaction with respect to [Q], and what is the rate equation?

Step 1: Take ratio of the rate divided by concentration using this equation → rate/[Q]=k

6.68x10^-3/.17=.0393 1.04×10^-2/.212=.0491 2.94×10^-2/.357=.0824

None of the ratios above are the same. This means that this is NOT a first order reaction.

Step 2: Use 2nd order equation → rate/[Q]²=k

6.68x10⁻³/.172=.231 1.04×10⁻²/.2122=.231 2.94×10⁻²/.3572=.231

These ratio values are all the same, so this is a second order reaction.

The rate equation is: rate=k[Q]²

For a second order reaction, the units for the rate constant (k) is 1/Ms

2. The ratio value that we got is the rate constant: .231 1/Ms

Q12.6.10

Experiments were conducted to study the rate of the reaction represented by this equation.

2NO(g) + 2H₂(g) ⟶ N₂(g) + 2H₂O(g)

Initial concentrations and rates of reaction are given here.

Consider the following questions:

1. Determine the order for each of the reactants, NO and H₂, from the data given and show your reasoning.
2. Write the overall rate law for the reaction.
3. Calculate the value of the rate constant, k, for the reaction. Include units.
4. For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H₂ had been consumed.
5. The following sequence of elementary steps is a proposed mechanism for the reaction.

Step 1: NO + NO ⇌ N₂O₂

Step 2: N₂O₂ + H₂ ⇌ H₂O + N₂O

Step 3: N₂O + H₂ ⇌ N₂ + H₂O

Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

Solution: Q12.6.10

Step 1: Determine the order of NO by taking the change in concentration and comparing it to the change in rate.

0.002/.001 = 2 0.3x10^-4/1.2x10^-4 = 4.

The rate of the reaction quadruples and the concentration doubles. This tells us that this is a second order reaction.

Step 2: Determine the order of H2 by taking the change in concentration and comparing it to the change in rate.

0.002/0.001 = 2 3.6x10^-4/1.8x10^-4 = 2.

The concentration of hydrogen doubles and so does the rate. This tells us this is a first order reaction.

Step 3: Write the rate law for the entire equation.

Rate=K[H₂][NO]²

Step 4: Calculate the value of the rate constant, k, for the reaction. Include units.

In order to determine the rate constant, you must choose values from the chart given and plug them into the rate law equation.

1.8x10^-4 mol/L·min = k(.001 mol/L)(.006mol/L)²

1.8x10^-4 mol/L·min= k(3.6x10^-8 mol³/L³)

(1.8x10^-4 mol/L·min)/ (3.6x10^-8 mol³ L³)= 5000 mol^-2 L^-2 min^-1

k = 5.0 × 10³ mol⁻² L⁻² min⁻¹

Step 5: For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H₂ had been consumed.

[NO] and [H₂] have a 1:1 molar ratio.

If 0.001 moles of hydrogen decay after one half-life, then 0.001 moles of NO decay after one hydrogen half-life.

If the sample of NO was 0.006, then the remaining amount would be 0.006-0.001=0.005 M NO.

Step 6:

Step 1: NO + NO ⇌ N₂O₂

Step 2: N₂O₂ + H₂ ⇌ H₂O + N₂O

Step 3: N₂O + H₂ ⇌ N₂ + H₂O

Step II is the rate-determining step.

Step II: rate= k₂[N₂O₂][H₂]

N₂O₂ is an intermediate and does not appear in the overall reaction or overall rate law. Therefore, we must substitute N₂O₂ with the proper concentrations.

rate= k₁[NO]²/ k_-1[N₂O₂]

Rearrange the equation so it equal to [N₂O₂].

[N₂O₂]= k₁[NO]²/k_-1

Substitute [N₂O₂] for the new equation.

rate= k₂(k₁[NO]²/k_-1)[H₂]

The rate constants will combine to form regular k.

Overall rate law:

rate= k[NO]²[H₂]

This reaction corresponds to the observed rate law.

Combine steps 1 and 2 with step 3 (in Step 6), which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.

Q21.4.22

A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of ²³⁸U and 2.52 mg of ²⁰⁶Pb. Calculate the age of the ore. The half-life of ²³⁸U is 4.5 × 10⁹ yr.

Solution: Q21.4.22

Step 1: Determine the percentage of U in the given sample.

→ 5.37/(5.37+2.52)= 0.6806

→ 0.6806 x 100% = 68.06% (This is the percentage of uranium in the sample)

Step 2: Use the percentage found in the half life formula.

-Use 1 as the initial (A_o) mass (100% U)

-Use 0.6806 as the current (A) mass (68.06% left)

-Use 4.5x10⁹ years as the given half life (h)

-Solve for t

Use half life formula: A= A₀(.5 ^{t/(h^2)})

.6808 =.5^{t/((4.5x10^9)^2)}

Step 3: Solve for t

→ ln(.6808)=ln(.5^{t/((4.5x10^9)^2)})

→ −.38449=(t/((4.5x10^9)^2))*(ln(.5))

→ −.38449=(t/((4.5x10^9)^2)*(−.69315)

→ .554701=t/((4.5x10^9)^2)

→ t = 2,498,044,246 years

Q20.3.10

Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)₆]³⁻, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.

Corrosion of a iron nail with a coiled copper wire, in agar-agar medium with ferroxyl indicator solution (potassium hexacyanoferrate(III), indicator of iron ions, and phenolphthalein, indicator of hydroxide ions). Image used with permission (CC BY-SA 3.0; Ricardo Maçãs).

Solution: Q20.3.10

When an iron nail is placed in a gel that contains [Fe(CN)₆]³⁻, the gel around the nail begins to turn pink because the oxygen around the nail is being reduced when the nail is placed in the aqueous solution. The oxygen becomes OH^- (hydroxide). Hydroxide is a very strong base. Since phenolphthalein is a basic indicator and hydroxide is a strong base, the phenolphthalein turns pink.

Iron in this reaction is oxidized (loses electrons): 3Fe²⁺(aq) + 2Fe(CN)₆^3- (aq) → Fe₃[Fe(CN)₆]₂ (s)

The reduction of oxygen is what turns the indicator pink: O₂ (g) + 2H₂O (l) + 4e^- → 4 OH^- (aq)

Q20.5.21

The reduction of Mn(VII) to Mn(s) by H₂(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:

1. MnO₄−(aq) + e⁻ → MnO₄²⁻(aq); E° = +0.56 V (purple → dark green)
2. MnO₄²⁻(aq) + 2e⁻ + 4H⁺(aq) → MnO₂(s); E° = +2.26 V (dark green → dark brown solid)
3. MnO₂(s) + e⁻ + 4H⁺(aq) → Mn³⁺(aq); E° = +0.95 V (dark brown solid → red-violet)
4. Mn³⁺(aq) + e⁻ → Mn²⁺(aq); E° = +1.51 V (red-violet → pale pink)
5. Mn²⁺(aq) + 2e⁻ → Mn(s); E° = −1.18 V (pale pink → colorless)

1. Is the reduction of MnO₄⁻ to Mn³⁺(aq) by H₂(g) spontaneous under standard conditions? What is E°cell?
2. Is the reduction of Mn³⁺(aq) to Mn(s) by H₂(g) spontaneous under standard conditions? What is E°cell?

Solution: Q20.5.21

In order to solve this equation we must understand the Latimer diagram. The Latimer diagram shows the cell potential values for successive redox reactions shown in the diagram below. (The diagram is just an example and should not be followed for the questions below as the standard cell potentials are different).