Extra Credit 6

17.1.5B

Balance the following in acidic solution:

a) \(H_2O_2 + Sn^{+2} \rightarrow H_2O + Sn^{+4}\)

b) \(PbO_2 + Hg \rightarrow Hg_2^{+2} + Pb^{+2}\)

c) \(Al + Cr_2O_7^{-2} \rightarrow Al^{+3} + Cr^{+3}\)

a) 1) Separate the two half-reactions

\(Sn^{+2} \rightarrow Sn^{+4}\) \(H_2O_2 \rightarrow H_2O\)

2) Balance oxygen molecules with water molecules (H₂O)

\(Sn^{+2} \rightarrow Sn^{+4}\) \(H_2O_2 \rightarrow H_2O + H_2O\)

3) Balance hydrogen molecules with H⁺ ions

\(Sn^{+2} \rightarrow Sn^{+4}\) \(H_2O_2 + {2H^+} \rightarrow H_2O + H_2O\)

4) Balance any unequal charge with electrons (e^-)

\(Sn^{+2} \rightarrow Sn^{+4} + {2e^-}\) \(H_2O_2 + 2H^+ +{2e^-} \rightarrow H_2O + H_2O\)

5) Combine half-reactions

\(Sn^{+2} + H_2O_2 + 2H^+ +2e^- \rightarrow Sn^{+4} + 2e^- + H_2O + H_2O\)

6) Reduce and cancel out like terms

\(Sn^{+2} + H_2O_2 + 2H^+ \rightarrow Sn^{+4} + 2H_2O\)

b) 1) Separate the two half-reactions

\(Hg \rightarrow Hg_2^{+2}\) \(PbO_2 \rightarrow Pb^{+2}\)

2) Balance any elements other than oxygen or hydrogen

\(2Hg \rightarrow Hg_2^{+2}\) \(PbO_2 \rightarrow Pb^{+2}\)

3) Balance oxygen molecules with with water molecules (H₂O)

\(2Hg \rightarrow Hg_2^{+2}\) \(PbO_2 \rightarrow Pb^{+2}+2H_2O\)

4) Balance hydrogen molecules with H⁺ ions

\(2Hg \rightarrow Hg_2^{+2}\) \(PbO_2+4H^+ \rightarrow Pb^{+2}\)

5) Balance any unequal charge with electrons (e^-)

\(2Hg \rightarrow Hg_2^{+2}+2e^-\) \(2e^-+PbO_2+4H^+ \rightarrow Pb^{+2}\)

6) Combine half-reactions

\(2Hg + 2e^- + PbO_2 +4H^+\rightarrow Hg_2^{+2} +2e^- + Pb^{+2}\)

7) Reduce and cancel out like terms

\(2Hg + PbO_2 +4H^+\rightarrow Hg_2^{+2} + Pb^{+2}\)

c) 1) Write the half-reactions

\(Al \rightarrow Al^{+3}\) \(Cr_2O_7^{-2} \rightarrow Cr^{+3}\)

2) Balance any elements other than oxygen or hydrogen

\(Al \rightarrow Al^{+3}\) \(Cr_2O_7^{-2} \rightarrow 2Cr^{+3}\)

3) Balance oxygen molecules with with water molecules (H₂O)

\(Al \rightarrow Al^{+3}\) \(Cr_2O_7^{-2} \rightarrow 2Cr^{+3}+ 7H_2O\)

4) Balance hydrogen molecules with H⁺ ions

\(Al \rightarrow Al^{+3}\) \(14H^++Cr_2O_7^{-2} \rightarrow 2Cr^{+3}+ 7H_2O\)

5) Balance any unequal charge with electrons (e^-)

\(Al \rightarrow Al^{+3}+3e^-\) \(6e^-+14H^++Cr_2O_7^{-2} \rightarrow 2Cr^{+3}+ 7H_2O\)

6) Balance any unequal electrons (e^-) by multiplying the reactions to result in a common factor

\(2Al \rightarrow 2Al^{+3}+6e^-\) \(6e^-+14H^++Cr_2O_7^{-2} \rightarrow 2Cr^{+3}+ 7H_2O\)

7) Combine half-reactions

\(6e^-+14H^++Cr_2O_7^{-2}+2Al \rightarrow 2Cr^{+3}+ 7H_2O+2Al^{+3}+ 6e^-\)

8) Reduce and cancel out like terms

\(14H^++Cr_2O_7^{-2}+2Al \rightarrow 2Cr^{+3}+ 7H_2O+2Al^{+3}\)

19.1.4

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

Lanthanides are rarely found in their elemental forms because they readily give their electrons to other more electronegative elements, forming compounds that exist in a combined form instead of a pure elemental form. They have very similar chemical properties with one another. They are found deep within the earth and difficult to extract. They are the inner transition elements and have partially filled d orbitals. Because of this, they are very reactive and electropositive.

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⁻ (note that CO₃²⁻ is bidentate in this complex)
2. [Cu(NH₃)₄]²⁺
3. [Co(NH₃)₄Br₂]₂(SO₄)₃
4. [Pt(NH₃)₄][PtCl₄]
5. [Cr(en)₃](NO₃)₃
6. [Pd(NH₃)₂Br₂] (square planar)
7. K₃[Cu(Cl)₅]
8. [Zn(NH₃)₂Cl₂]

Simplified naming rules:

Cations are always named before anions.

Ligands are named before the metal atom or ion.

Ligand names are modified with an -o added to the root name of an anion. For neutral ligands, the name of the molecule is used (OH₂, NH₃, CO, and NO are exceptions).

The prefixes mono-, di-, tri-, tetra-, penta-, and hexa- are used to denote the number of simple ligands.

The prefixes bis-, tris-_, tetrakis-, etc, are used for more complicated ligands or ones that already contain di-, tri-, etc.

The oxidation state of the central metal ioin is designated by a Roman numeral in parenthesis.

When more than one type of ligand is present, they are named alphabetically. Prefixes do not affect the order.

If the complex ion has a negative charge, the suffix -ate is added to the name of the metal.

For complex ions, cis-, trans-, fac-, or mer- may be used to indicate spatial arrangement.

To find the oxidation number of the metal, count the charge of the ions and ligands and use the oxidation number to assign the correct value that leads to the overall charge.

1. There are three carbonate and one cobalt molecules in this compound.

2. There are three amine and one copper molecules in this compound.

3. The cation contains three amine, one cobalt, and two bromine molecules. The anion is sulfate.

4. The cation contains four amine and one platinum molecule, and the anion contains four chlorine and one platinum molecule.

5. There are three ethylenediamine and one chromium molecule in the cation and three nitrates make up the anion.

6. The cation is made of a palladium molecule with two amine and two bromine molecules.

7. The cation is solely potassium and the anion is made of five chlorine and one copper molecule.

8. The molecule is made of a zinc, two amine, and two chlorine molecules.

1. tricarbonatocobaltate(III) ion; +3 oxidation state
2. tetraaminecopper(II) ion; +2 oxidation state
3. tetraaminedibromocobalt(III) sulfate; +3 oxidation state
4. tetraamineplatinum(II) tetrachloroplatinate(II); +2 oxidation state
5. tris-(ethylenediamine)chromium(III) nitrate; +3 oxidation state
6. diaminedibromopalladium(II); +2 oxidation state
7. potassium pentachlorocuprate(II); +2 oxidation state
8. diaminedichlorozinc(II); +2 oxidation state

Q12.3.19

For the reaction Q⟶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?

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

1. The order of the reaction with respect to [Q] with 2. The ratio of the quotient of two initial concentrations was approximately 2. Dividing the rates of the corresponding concentrations also produces a quantity of 2. Therefore, the reaction is second order based on the corresponding data. corresponding data. When the concentration increases by 1.7, the rate changes by a factor of 2.8.

Second order: \(\frac{rate}{[Q]^2}=k\)

\(\frac{6.68x10^{-3}}{.17^2} = .231\) \(\frac{1.04 × 10^{-2}}{.212^2} = .231\) \(\frac{2.94 × 10^{-2}}{.357^2} = .231\)

2. The rate constant is about .23 according to the second order rate law:

\(\frac{rate}{[Q]^2}=k\)

\(\frac{6.68 x 10^{-2}}{[.170]^2}=.231\)

Q12.6.10

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.

1) To determine the order of H₂, take the change in concentration and compare it to the change in rate.

• As the concentration of hydrogen doubles, the rate doubles. This means it is first order.

To determine the order of NO, take the change in concentration and compare it to the change in rate.

• Because the rate quadruples has the concentration doubles, it is second order.

2) Raise the corresponding concentration to the power of its order.

Rate=K[H₂][NO]²

3) To determine the rate constant, choose values on the chart and plug them into the equation. As long as the values for the same experiment are used in the calculation, it will be the same for all the experiments' rate laws as it is a constant.

\(rate = k [NO]^2 [H_2]\)

\(.00018 = k [.006]^2 [.001]\)

\(k = 5000 M^{-2}s^{-1}\)

d.) [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 moles/L.

e.) Step 1: rate=k[NO]²

Step 2: rate= k[N₂O₂][H₂]. Because [NO]² is proportional to [N₂O₂], the rate law can be re-written as: rate=k[NO]²[H₂]

Step 3: rate = k[N₂O][H₂]. Because [N₂O₂] is proportional to [N₂O], and [NO]² is proportional to [N₂O₂], the rate law can be re-written as: rate=k [NO]²[H₂].

Step 2 is thus the rate-determining step. The stoichiometry of each step had a 1:1 ratio.

Solution: (a) Doubling [H₂] doubles the rate. [H₂] must enter the rate equation to the first power. Doubling [NO] increases the rate by a factor of 4. [NO] must enter the rate law to the second power. (b) Rate = k [NO]²[H₂]; (c) k = 5.0 × 10³ mol⁻² L⁻² min⁻¹; (d) 0.0050 mol/L; (e) Step II is the rate-determining step. If step I gives N₂O₂ in adequate amount, steps 1 and 2 combine to give 2NO+H₂⟶H₂O+N₂O₂NO+H₂⟶H₂O+N₂O . This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, 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\({^{238}_{92}U}\) and 2.52 mg of \({^{206}_{82}Pb}\). Calculate the age of the ore. The half-life of \({^{238}_{92}U}\) is 4.5 × 10⁹ yr.

First, find the mass percentage. Then use the half-life equation and solve for time, the variable that describes how old the ore will be.

\(\frac{5.37}{5.37+2.52} = 68.06% U\)

\(A = A_0 (.5^{\frac{t}{t_{1/2}^2}})\)

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

\(ln(.6808) = ln(.5^{\frac{t}{(4.5x10^9)^2}})\)

\(-.38449 = \frac{t}{(4.5x10^9)^2}ln(.5)\)

\(.554701 = \frac{t}{(4.5x10^9)^2}\)

t= 2.5 x 10⁹

The ore is 2.5 x 10⁹ years old.

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 and transferred from Extra Credit Questions (CC BY-SA 3.0; Ricardo Maçãs).

The indicator turns pink because the oxygen is reduced in the aqueous solution when the iron nail is placed in. The oxygen becomes OH^-, hydroxide, which is a very strong base. Because phenolphthalein is a basic indicator, and hydroxide is a strong base, the phenolphthalein turns pink.

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

The half reaction that turns the indicator pink though is the reduction of oxygen: 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?

E_cell= E_cathode-E_anode

If E°cell is positive, the reaction will be spontaneous because it implies that there is a release of Gibb's free energy, or a -∆G.

For the reduction of MnO₄^-

E°_cell=0.56V+2.26V+0.95V=+3.77V

Because E°cell is positive, the reduction is spontaneous.

For the reduction of Mn³⁺

E°_cell=1.51V-1.18V=+.33V

Both reactions are spontaneous.