Extra Credit 6

Q17.1.5B

Balance the following in acidic solution:

a. \(H_{2}O_{2}+Sn^{2+}⟶H_{2}O+Sn^{4+}\)

b.\( PbO_{2}+Hg⟶Hg_{2}^{2+}+Pb^{2+}\)

c. \(Al+Cr_{2}O_{7}^{2-}⟶Al^{3+}+Cr^{3+}\)

Solution:

a.

step 1. Separate the half reactions to the oxidized(loss of electrons) half and the reduced(gain of electrons) half. Think of the acronym OILRIG.

Red: \(H_{2}O_{2}⟶H_{2}O\)

Ox: \(Sn^{2+}⟶Sn^{4+}\)

step 2: Balance other elements other than O and H

no changes

step 3: Add H₂O to balance the O

\(H_{2}O_{2}\)⟶\(H_{2}O+H_{2}O\)

\(Sn^{2+}⟶Sn^{4+}\)

step 4: Balance hydrogen by adding H⁺

/(H_{2}O_{2}+2H^{+}⟶H_{2}O +H_{2}O\)

\(Sn^{2+}⟶Sn^{4+}\)

step 5: Balance the charge of each equation with electrons (e^-). The charge must be equal on both sides of the reaction. Electrons add a negative charge.

\(H_{2}O_{2}+2H^{+}+2e^{-}⟶H_{2}O +H_{2}O\)

\(Sn^{2+}⟶Sn^{4+}+2e^{-}\)

step 6: Scale the equations so that they are equal. No changes made because both equations have two electrons.

\(H_{2}O_{2}+2H^{+}+2e^{-}⟶H_{2}O +H_{2}O\)

\(Sn^{2+}⟶Sn^{4+}+2e^{-}\)

step 7: Combine the two equations to form the overall balanced equation. Cancel out like terms and electrons.

\(H_{2}O_{2}+2H^{+}+Sn^{2+}⟶H_{2}O +H_{2}O +Sn^{4+}\)

b.

step 1. Separate the half reactions to the oxidized(loss of electrons) half and the reduced(gain of electrons) half.

Red: \(PbO_{2}⟶Pb^{2+}\)

Ox: \(Hg⟶Hg_{2}^{2+}\)

step 2: Balance other elements other than O and H

\(PbO_{2}⟶Pb^{2+}\)

\(2Hg⟶Hg_{2}^{2+}\)

step 3: Add H₂O to balance the O

\(PbO_{2}⟶Pb^{2+} +2H_{2}O\)

\(2Hg⟶Hg_{2}^{2+}\)

step 4: Balance hydrogen by adding H⁺

\(PbO_{2}+4H^{+}⟶Pb^{2+} +2H_{2}O\)

\(2Hg⟶Hg_{2}^{2+}\)

step 5: Balance the charge of each equation with electrons (e^-). The charge must be equal on both sides of the reaction. Electrons add a negative charge.

\(PbO_{2}+4H^{+}+2e^{-}⟶Pb^{2+} +2H_{2}O\)

\(2Hg⟶Hg_{2}^{2+}+2e^{-}\)

step 6: Scale the equations so that the equal. No changed made because both equations have two electrons.

\(PbO_{2}+4H^{+}+2e^{-}⟶Pb^{2+} +2H_{2}O\)

\(2Hg⟶Hg_{2}^{2+}+2e^{-}\)

step 7: Combine the two equations to form the overall balanced equation. Cancel out like terms and electrons

\(PbO_{2}+4H^{+} +2Hg⟶Pb^{2+} +2H_{2}O+Hg_{2}^{2+}\)

c.

step 1. Separate the half reactions to the oxidized(loss of electrons) half and the reduced(gain of electrons) half

Red: \(Cr_{2}O_{7}^{2-}⟶Cr^{3+}\)

Ox: \(Al⟶Al^{3+}\)

step 2: Balance other elements other than O and H

\(Cr_{2}O_{7}^{2-}⟶2Cr^{3+}\)

\(Al⟶Al^{3+}\)

step 3: Add H₂O to balance the O

\(Cr_{2}O_{7}^{2-}⟶2Cr^{3+}+7H_{2}O\)

\(Al⟶Al^{3+}\)

step 4: Balance hydrogen by adding H⁺

\(Cr_{2}O_{7}^{2-}+14H^{+}⟶2Cr^{3+}+7H_{2}O\)

\(Al⟶Al^{3+}\)

step 5: Balance the charge of each equation with electrons (e^-). The charge must be equal on both sides of the reaction. Electrons add a negative charge.

\(Cr_{2}O_{7}^{2-}+14H^{+} +2e^{-}⟶2Cr^{3+}+7H_{2}O\)

\(Al⟶Al^{3+} +3e^{-}\)

step 6: Scale the equations so that they are equal.

\(3Cr_{2}O_{7}^{2-}+42H^{+} +18e^{-}⟶6Cr^{3+} +21H_{2}O\)

\(6Al⟶6Al^{3+} +18e^{-}\)

step 7: Combine the two equations to form the overall balanced equation. Cancel out like terms and electrons

\(3Cr_{2}O_{7}^{2-}+42H^{+} +6Al⟶6Cr^{3+}+21H_{2}O+6Al^{3+}\)

Original solution wrong for c. Corrected above.

Q19.1.4

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

The lanthanoid elements are along the bottom of the periodic table as indicated:

They are not found in nature in their elemental forms because they have high levels of reactivity and sensitivity to contamination. Typically lanthanum, cerium, praseodymium, neodymium, and europium are highly reactive. When they are exposed to oxygen, they form as oxide coating. Lanthanoids also experience combustion readily in air which is why they are usually stored under oil (both for preventing oxidation and combustion). Therefore lanthanoids are usually in their combined form rather then their elemental form in nature.

Q19.2.6

Rules to follow for coordination complexes

1. Cations are always named before the anions.

2. Ligands are named before the metal atom or ion.

3. 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, with the exception of OH2, NH3, CO and NO.

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

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

6. The oxidation state of the central metal ion is designated by a Roman numeral in parentheses.

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

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

9. In the case of complex‐ion isomerism the names cis, trans, fac, or mer may precede the formula of the complex‐ion name to indicate the spatial arrangement of the ligands. Cis means the ligands occupy adjacent coordination positions, and trans means opposite positions just as they do for organic compounds. The complexity of octahedral complexes allows for two additional geometric isomers that are peculiar to coordination complexes. Fac means facial, or that the three like ligands occupy the vertices of one face of the octahedron. Mer means meridional, or that the three like ligands occupy the vertices of a triangle one side of which includes the central metal atom or ion.

1. tricarbonatocobaltate(III) ion; \([(CO_{3})Co]^{3-}\)
2. tetraaminecopper(II) ion; \([Cu(NH_{3})_{4}]^{2+}\)
3. tetraaminedibromocobalt(III) sulfate; \([(NH_{3})_{4}Br_{2}Co]_{2}SO_{4}\)
4. tetraamineplatinum(II) tetrachloroplatinate(II); \([(NH_{3})_{4}Pt][(Cl)_{4}Pt]\)
5. tris-(ethylenediamine)chromium(III) nitrate; \([Cr(en)_{3}](NO_{3})_{3}\)
6. diaminedibromopalladium(II); \([Pd(NH_{3})_{2}Br_{2}]\)
7. potassium pentachlorocuprate(II); \(K_{3}[(Cl)_{5}Cu]\)
8. diaminedichlorozinc(II); \([Zn(NH_{3})_{2}Cl_{2}]\)

Q12.3.19

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

[Q]initial (M)	0.170	0.212	0.357
Rate (mol/L/s)	6.68 × 10−3	1.04 × 10−2	2.94 × 10−2

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

The order of the reaction is 2 because both W and X have an order of one. In order to find the overall order of the reaction you must add 1+1=2.

2. What is the rate constant?

To solve for the rate constant you must use the equation:

rate=k[Q]^x

6.68x10^-3=k[.17]¹

k=.0393 Mol^-1L S^-1

*Original solution wrong (#2)

rate=k[Q]^x

6.68x10^-3=k[.17]²

k=0.2311 Mol^-1L S^-1

Q12.6.10

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

\[2NO_{(g)}+2H_{2(g)}⟶N_{2(g)}+2H_2O_{(g)}\]

Initial concentrations and rates of reaction are given here.

Experiment	Initial Concentration [NO] (mol/L)	Initial Concentration, [H2] (mol/L)	Initial Rate of Formation of N2 (mol/L min)
1	0.0060	0.0010	1.8 × 10−4
2	0.0060	0.0020	3.6 × 10−4
3	0.0010	0.0060	0.30 × 10−4
4	0.0020	0.0060	1.2 × 10−4

Consider the following questions:

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

To find the order for each reactant it is a multiple step process.

1. Look at the [NO] column and and find where the concentration changes.
2. Where it changes for [NO] look to the [H₂] column to and see where the concentration is constant.
3. Use those two values on the [NO] column to find the order. In this problem I used 0.001 and 0.002.
4. The order equation also needs a rate so use the correlating rate for the rows.
5. Now you can use the equation \([\frac{concentration 2}{concentration 1}]^{x}=\frac{rate 2}{rate 1}\). X is the order of the reactant we are solving for.
6. Plug in all values. \([\frac{0.002}{0.001}]^{a}=\frac{1.2x10^{-4}}{0.30x10^{-4}}\) Solve for a. a=2
7. [NO] has an order of 2.
8. Repeat the same process for H₂. The equation should be \([\frac{0.002}{0.001}]^{b}=\frac{3.6x10^{-4}}{1.8x10^{-4}}\). Solve for b. b=1
9. [H₂] has an order of 1.
10. Write the overall rate law for the reaction.

To find the overall rate law you need both the values of a and b we found previously. The rate law equation is rate=k[NO]^a[H₂]^b. The rate law for this reaction is rate=k[NO]²[H₂]¹. According to the rate law, this reaction is 3rd order. (We find this by summing up the exponents of each, 2+1 = 3)

1. Calculate the value of the rate constant, k, for the reaction. Include units. To find k we have to choose one set of values on the chart and put them into the equation. 1.8x10^-4=k[0.006]²[0.001]¹ k=5000 Mol^-2L S^-1
2. For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H₂ had been consumed. To solve for the concentration of NO write the rate equation rate=k[NO]²[H₂]¹ but half the concentration of H_2. 3.6x10^-4=5000[X]²[.001]¹ x=0.0085mol/L
3. The following sequence of elementary steps is a proposed mechanism for the reaction.

\[Step 1: NO+NO⇌N_2O_2\]

\[Step 2: N_2O_2+H_2⇌H_2O+N_2O\]

\[Step 3: N_2O+H_2⇌N_2+H_2O\]

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.

The rate determining step is the slowest step in the chemical reaction. The slowest step is step 3. Step 3 has the smallest initial rate of formation of 0.30x10^-4.The rate law for the reaction is the same as the observed rate law for the reaction. rate=k[NO]²[H₂]¹

Original solution wrong, last part of question.

Step 2 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 equals 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, which shows that the mechanism is consistent with the rate law.

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 × 10⁹yr.

1. Use the half life equation. \(t_{1/2}=\frac{ln(2)}{k}\). You are given the half life 4.5 × 10⁹yr and ln(2) can be calculated. Solve for k. 4.5x10^{9}=\frac{ln(2)}{k}

k=1.54x10^-10

2. Now that you have the k value you can use the integrated rate law \(ln\frac{A}{A_{0}}=-kt\). We have k and are looking for t. A and A₀ can be found by turning the mg of the elements into moles.

5.37mg \(^{238}_{92}U\)	1g	1 mol	2.256x10-5 moles \(^{238}_{92}U\)
	1000mg	238.03g \(^{238}_{92}U\)

2.52 mg of \(^{206}_{82}Pb\)	1g	1 mol	1.216x10-5 moles \(^{206}_{82}Pb\)
	1000mg	207.2g \(^{206}_{82}Pb\)

3. Use the moles of 2.256x10^-5 moles \(^{238}_{92}U\)as the A value and 2.256x10^-5 moles \(^{238}_{92}U\) as the A₀ value.

4. Plug all values into the \(ln\frac{A}{A_{0}}=-kt\) and solve for t. \(ln\frac{2.256x10^{-5} }{1.216x10^{-5}}=-1.54x10(t)\). t=1.455x10¹¹ years old

*Original solution wrong.

(moles of Pb and U correct, k value correct.)

• Plug and chug these numbers into the equation: N_Now=N_Orige^−λt and solve for t
• ln(2.26 x 10^-5 mol/3.48 x 10^-5 mol) = -1.54 x 10^-10 year^-1 (t)
• t = 2.8 x 10⁹ 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.

Phenolphthalein is used in many acid-base titrations because it is pH sensitive. The gel around the iron nail turns pink indicating that there is a basic reaction occurring in the area surrounding the nail.

Fe²⁺+2OH^-→Fe(OH)₂

4Fe(OH)₂+O₂+H_2O→4Fe(OH)₃

2Fe(OH)₃→Fe₂O₃H₂O+H₂O

*Original solution wrong.

• Reduction:
  • [Fe(CN)₆]³⁻+e⁻⟶[Fe(CN)₆]⁴⁻
• Oxidation:
  • Fe(s)+2OH−⟶Fe(OH)₂+2e⁻
• Overall reaction:
  • 2[Fe(CN)₆]³⁻+Fe(s)+2OH⁻⟶Fe(OH)₂+2[Fe(CN)₆]⁴⁻

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. MnO^₄-(aq) + e− → MnO₄^2-(aq); E° = +0.56 V (purple → dark green)
2. MnO^₄2-(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? The spontaneity of a reaction can be determined by looking at the E°cell value. The larger(more positive) the E°cell is the greater driving force it has causing it to be more spontaneous. The E°cell for this reaction can be determined by E°cell=E_cathode-E_anode. E°cell=0.56V-0.95V=-0.395V
2. Is the reduction of Mn3⁺(aq) to Mn(s) by H₂(g) spontaneous under standard conditions? What is E°cell? The reaction is spontaneous because spontaneous reactions have positive E° values. This is because the greater the value the larger the driving force is. E°cell for this reaction can be determined by E°cell=E_cathode-E_anode. E°cell=0.95V-−1.18−1.18 V=2.13V

Original solution wrong.

The E cell under standard for a series of reactions is determined by taking the sum until you reach the desired ion/element.

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 spontaneous.
2. E°=1.51V+(-1.18V)=+0.33V. The reaction is spontaneous with a standard condition of E°=+0.33V which makes G° negative; it is spontaneous.