Extra Credit 6

Q17.1.5B

Balance the following in acidic solution:

1. H₂O₂+Sn²⁺⟶H₂O+Sn⁴⁺
2. PbO₂+Hg ⟶ Hg₂²⁺+Pb^₂+
3. Al+Cr₂O₇²⁻⟶ Al³⁺+Cr³⁺

S17.1.5B

1. H₂O₂+Sn²⁺⟶H₂O+Sn⁴⁺

^{step 1: Split reaction into half-reactions}

^Reduction:H₂O₂⟶H₂O

^Oxidation:Sn²⁺⟶Sn⁴⁺

step 2: Balance elements (except Hydrogen and Oxygen)

^Reduction:H₂O₂⟶H₂O

^Oxidation:Sn²⁺⟶Sn⁴⁺

step 3:Balance O with H₂O

^Reduction:H₂O₂ ⟶ 2H₂O

^Oxidation:Sn²⁺⟶ Sn⁴⁺

step 4: Balance H with H⁺

^Reduction: 2H⁺+H₂O₂ ⟶ 2H₂O

^Oxidation:Sn²⁺⟶ Sn⁴⁺

step 5: Balance the charges

^Reduction: 2e^-+2H⁺+H₂O₂ ⟶ 2H₂O

^Oxidation:Sn²⁺⟶ Sn⁴⁺+2e^-

step 6: Multiply by some integer to make electrons (lost) = electrons (gained)

^Reduction: 2e^-+2H⁺+H₂O₂ ⟶ 2H₂O

^Oxidation:Sn²⁺⟶ Sn⁴⁺+2e^-

step 7: Add half equations

2H⁺+H₂O₂+Sn²⁺⟶ Sn⁴⁺+2H₂O

step 8: Check atom count and charge are balanced on both side of the equation.

2H⁺+H₂O₂+Sn²⁺⟶ Sn⁴⁺+2H₂O

2. PbO₂+Hg ⟶ Hg₂²⁺+Pb^₂+

^{step 1: Split reaction into half-reactions}

^Reduction:PbO₂⟶ Pb^₂+

^Oxidation:Hg⟶ Hg₂²⁺

step 2: Balance elements (except Hydrogen and Oxygen)

^Reduction:PbO₂⟶ Pb^₂+

^Oxidation:2Hg⟶ Hg₂²⁺

step 3:Balance O with H₂O

^Reduction:PbO₂⟶ Pb^₂++2H₂O

^Oxidation:2Hg⟶ Hg₂²⁺

step 4: Balance H with H⁺

^Reduction:4H⁺+PbO₂⟶ Pb^₂++2H₂O

^Oxidation:2Hg⟶ Hg₂²⁺

step 5: Balance the charges

^Reduction:4H⁺+2e^-+PbO₂⟶ Pb^₂++2H₂O

^Oxidation:2Hg⟶ Hg₂²⁺+2e^-

step 6: Multiply by some integer to make electrons (lost) = electrons (gained)

^Reduction:4H⁺+2e^-+PbO₂⟶ Pb^₂++2H₂O

^Oxidation:2Hg⟶ Hg₂²⁺+2e^-

step 7: Add half equations

2Hg+4H⁺+PbO₂⟶ Hg₂²⁺+Pb^₂++2H₂O

step 8: Check atom count and charge are balanced on both side of the equation.

2Hg+4H⁺+PbO₂⟶ Hg₂²⁺+Pb²⁺+2H₂O

3. Al+Cr₂O₇²⁻⟶ Al³⁺+Cr³⁺

^{step 1: Split reaction into half-reactions}

^Reduction:Cr₂O₇²⁻⟶ Cr³⁺

^Oxidation:Al⟶ Al³⁺

step 2: Balance elements (except Hydrogen and Oxygen)

^Reduction:Cr₂O₇²⁻⟶ 2Cr³⁺

^Oxidation:Al⟶ Al³⁺

step 3:Balance O with H₂O

^Reduction:Cr₂O₇²⁻⟶ 2Cr³⁺+7H₂O

^Oxidation:Al⟶ Al³⁺

step 4: Balance H with H⁺

^Reduction:Cr₂O₇²⁻+14H⁺⟶ 2Cr³⁺+7H₂O

^Oxidation:Al⟶ Al³⁺

step 5: Balance the charges

^Reduction:Cr₂O₇²⁻+14H⁺+6e^-⟶ 2Cr³⁺+7H₂O

^Oxidation:Al⟶ Al³⁺+3e^-

step 6: Multiply by some integer to make electrons (lost) = electrons (gained)

^Reduction:Cr₂O₇²⁻+14H⁺+6e^-⟶ 2Cr³⁺+7H₂O

^Oxidation:2(Al⟶ Al³⁺+3e^-)

2Al⟶ 2Al³⁺+6e^-

step 7: Add half equations

Cr₂O₇²⁻+14H⁺+2Al⟶ 2Cr³⁺+7H₂O+2Al³⁺

step 8: Check atom count and charge are balanced on both side of the equation.

Cr₂O₇²⁻+14H⁺+2Al⟶ 2Cr³⁺+7H₂O+2Al³⁺

Q19.1.4

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

S19.1.4

Lanthanoid elements are rare earth elements. They are also scarcely available because their nucleus decays into lighter elements since its size makes it unstable.

Q19.2.6

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

S19.2.6

1. tricarbonatocobaltate(III) ion;
2. tetraaminecopper(II) ion;
3. tetraaminedibromocobalt(V) sulfate;
4. tetraamineplatinum(II) tetrachloroplatinate(II);
5. tris-(ethylenediamine)chromium(III) nitrate;
6. trans- or cis- diaminedibromolead(II);
7. potassium pentachlorocuprate(II);
8. diaminedichlorozinc(II)

Q12.3.19

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

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

b.What is the rate constant?

S12.3.19

a. Q⟶W+X

rate=k[Q]^m

Substitute given values:

6.68 × 10⁻³M/s=k(0.170M)^m

1.04 × 10⁻²M/s=k(0.212M)^m

Combine equations and solve for order of [Q]

0.642 =(0.801)^m

^{m=2, the rate order of the reaction with respect to [Q] is 2}

rate=k[Q]²

b. substitute given values into rate=k[Q]² to find rate constant (k)

6.68 × 10⁻³M/s=k(0.170M)²

k=.231 M^-1s^-1

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:

a. Determine the order for each of the reactants, NO and H₂, from the data given and show your reasoning.

b. Write the overall rate law for the reaction.

c. Calculate the value of the rate constant, k, for the reaction. Include units.

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

e. 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.

S12.6.10

a. Determine the order for each of the reactants, NO and H₂, from the data given and show your reasoning.

rate=k[NO]^m[H₂]ⁿ

Substitute given values:

0.30 × 10⁻⁴M/min=k[0.0010M]^m[0.0060M]ⁿ

1.2 × 10⁻⁴M/min=k[0.0020M]^m[0.0060M]ⁿ

Combine equations and solve for m to find the order of NO

0.25=[0.5]^m; m=2; the order of NO is 2

Substitute given values:

1.8 × 10⁻⁴M/min=k[0.0060M]^m[0.0010M]ⁿ

3.6 × 10⁻⁴M/min=k[0.0060M]^m[0.0020M]ⁿ

Combine equations and solve for n to find the order of H₂

0.5=[0.5]ⁿ; n=2; the order of H₂ is 1

b. Rate = k [NO]²[H₂]

c. k = 5.0 × 10³ mol⁻² L² 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+N₂O₂. 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. The second step eliminates the intermediates so it is the rate determining step.

Q21.4.22

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

S21.4.22 (this question was left blank).

First step is to find the original amount of Uranium-238 in the sample by dividing the milligrams of lead with its molar mass to get the moles and then multiplying by uranium's molar mass:

Second step is to add the uranium masses to find the initial concentration, afterward use a first order equation to solve for time.

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).

S20.3.10 (This question was left blank)

Reduction half-reaction:

Oxidation half-reaction:

Overall reaction:

The oxygen is dissolved in the solution. The complex binds with the iron ion which gives the blue color:

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?

S20.5.21 (This question was left blank)

1. E°=0.56V+2.26V+0.95V=+3.77V The reaction is spontaneous with a standard condition of E°=+3.77V which makes G° negative; it is favorable.
2. E°=1.51V+(-1.18V)=+0.33V The reaction is spontaneous with a standard condition of E°=+0.33V. It is not as spontaneous as the previous because manganese has low reduction potential.