Extra Credit 4

Q17.1.4

For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.

1. \[ \ce{Cl^-} \rightarrow \ce{Cl_2}\]
2. \[ \ce{Mn^2+} \rightarrow \ce{MnO_2} \]
3. \[ \ce{H_2} \rightarrow \ce{H^+} \]
4. \[ \ce{NO_3^-} \rightarrow \ce{NO}\]

Answer 17.1.4

1. First, determine the oxidation number of chlorine on both sides of the equation. The chloride ion has a -1 charge and chlorine gas has an oxidation number of 0. The oxidation number is increasing, which implies the loss of electrons, therefore oxidation is occurring.

2. Manganese (II) has a oxidation number of +2. Manganese in MnO₂ has an oxidation number of +4. The oxidation number of manganese is increasing, which is the loss of electrons. Therefore, oxidation is occurring.

3. First, determine the oxidation number of hydrogen on both sides of the equation. Hydrogen is in its elemental state as a gas, so its oxidation number is 0. The hydrogen ion has a charge of +1, so therefore its oxidation number is +1. Since the loss of electrons is occurring, oxidation is taking place.

4. Nitrogen has a +5 oxidation number in the polyatomic ion NO₃^-. Nitrogen has an oxidation number of +2 in NO. Since the oxidation number is decreasing, reduction is occurring because nitrogen is gaining electrons.

Q19.1.2

Write the electron configurations for each of the following elements and its ions:

1. \[ \ce{Ti} \]
2. \[ \ce{Ti^2+} \]
3. \[ \ce{Ti^3+} \]
4. \[ \ce{Ti^4+} \]

Answer 19.1.2

1. First, locate Ti on the periodic table. Its atomic number is 22, which suggests Ti has 22 electrons in its unionized state. Then, go to the nearest noble gas, which is Argon (with atomic number 18). To write the electron configuration, start with the nearest noble gas (Argon in this case) and then add the outer shell electrons. In this case Ti has 4 outer shell electrons since 22-18 is 4. Two of these occupy the 4s orbital and the other two occupy the 3d orbital, so therefore the configuration is as follows: \[ \ce{[Ar] 4s^2 3d^2} \]
2. First, locate Ti on the periodic table. Its atomic number is 22, which suggests Ti has 22 electrons in its unionized state. Then, go to the nearest noble gas, which is Argon (with atomic number 18). To write the electron configuration, start with the nearest noble gas (Argon in this case) and then add the outer shell electrons. In this case there are 2 outer shell electrons since Ti has 4 outer shell electrons, and when Ti is ionized to Ti²⁺, it loses two electrons. So, in order to write the electron configuration for Ti²⁺, it is best to write the electron configuration for Ti: \[ \ce{[Ar] 4s^2 3d^2} \] and then subtract 2 electrons. Electrons are taken from the s orbitals before the d orbitals (because they are higher in energy in comparison to the d) leaving the configuration for Ti²⁺ as: \[ \ce{[Ar] 3d^2} \]
3. First, locate Ti on the periodic table. Its atomic number is 22, which suggests Ti has 22 electrons in its unionized state. Then, go to the nearest noble gas, which is Argon (with atomic number 18). To write the electron configuration, start with the nearest noble gas (Argon in this case) and then add the outer shell electrons. In this case there is 1 outer shell electron since Ti has 4 outer shell electrons, and when Ti is ionized to Ti³⁺, it loses three electrons. So, in order to write the electron configuration for Ti³⁺, it is best to write the electron configuration for Ti: \[ \ce{[Ar] 4s^2 3d^2} \] and then subtract 3 electrons. Electrons are taken from the s orbitals first (because they are higher in energy in comparison to the d) and then the d orbitals before the d orbitals which gives the configuration: \[ \ce{[Ar] 3d^1} \]
4. First, locate Ti on the periodic table. Its atomic number is 22, which suggests Ti has 22 electrons in its unionized state. Then, go to the nearest noble gas, which is Argon (with atomic number 18). To write the electron configuration, start with the nearest noble gas (Argon in this case) and then add the outer shell electrons. In this case there are 0 outer shell electrons since Ti has 4 outer shell electrons, and when Ti is ionized to Ti⁴⁺, it loses all 4 electrons. So, in order to write the electron configuration for Ti⁴⁺, it is best to write the electron configuration for Ti: \[ \ce{[Ar] 4s^2 3d^2} \] and then subtract 4 electrons. Since all the valence electrons are ionized to get Ti⁴⁺, the electron configuration for the Ti⁴⁺ ion is as follows: \[ \ce{[Ar] } \]

Q19.2.4

Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.

1. [Pt(H₂O)₂Br₂] (square planar)
2. [Pt(NH₃)(py)(Cl)(Br)] (square planar, py = pyridine, C₅H₅N)
3. [Zn(NH₃)₃Cl]⁺ (tetrahedral)
4. [Pt(NH₃)₃Cl]⁺ (square planar)
5. [Ni(H₂O)₄Cl₂]
6. [Co(C₂O₄)₂Cl₂]³⁻ (note that C2O2−4C2O42− is the bidentate oxalate ion, −O2CCO−2−O2CCO2−)

Answer 19.2.4

a. [Pt(H₂O)₂Br₂]: First, identify the ligands. In this case there are 2 aqua ligands and 2 bromo ligands. So, this suggests that there are both cis and trans isomers. And, since this complex is square planar, no optical isomer exists because a plane of symmetry exists.

b. [Pt(NH₃)(py)(Cl)(Br)]: First identify the ligands. This complex has an ammine, pyridine, chloro, and bromo ligands. Since there 4 different ligands, there are no cis or trans isomers. There is also no optical isomer for this complex since it is square planar and has a plane of symmetry.

c. [Zn(NH₃)₃Cl]⁺: First, identify the ligands present. In this complex, there are 3 ammine ligands and 1 chloro ligand. Since this is a tetrahedral complex, no cis or trans isomers exist since each ligand is 109.5 degrees from the central metal atom. This complex cannot have an optical isomer because it is a tetrahedral complex with 3 ligands that are alike.

d. [Pt(NH₃)₃Cl]⁺: First, identify the ligands. Here, there are three ammine ligands and one cloro ligand, so no cis or trans because there are 3 identical ligands and it is a square planar complex. No optical isomers possible because it is a square planar complex, so therefore a plane of symmetry exists.

e. [Ni(H₂O)₄Cl₂]: First identify the ligands. This complex has four aqua ligands and two chloro ligands, so therefore it is octahedral in geometry. Cis and trans isomers exist since there are only two chloro ligands present. No optical isomers exist since there are planes of symmetry.

f. [Co(C₂O₄)₂Cl₂]³⁻: First, identify the ligands present. This complex has two oxalate ligands and two chloro ligands. Oxalate is a bidentate ligand, so this complex is octahedral in geometry. Both cis and trans isomers exist since the oxalate ligands can attach across or next to each other. The cis complex is an optical isomer since its mirror image is non-superimposable.

Q12.3.17

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:

\[ \ce{H_2}(\mathit{ g}) + \ce{2NO}(\mathit{ g}) \rightarrow \ce{N_2O}(\mathit{ g}) + \ce{H_2O}(\mathit{ g})\]

Determine the rate equation, the rate constant, and the orders with respect to each reactant from the following data:

Answer 12.3.17

When the concentration of H₂ doubles and the concentration of NO is held constant, the rate of reaction increases by a factor of two, which suggests that the order with respect to [H₂] is 1. When the concentration of NO doubles and the concentration of H₂ is held constant, the rate of reaction increases by a factor of 4, suggesting that the order with respect to [NO] is 2. So, the overall rate law for the reaction is: rate=k[H₂][NO]². The overall order of the reaction is third order (first order in [H₂] and second order in [NO] which adds to 3). By plugging in the data from the table above into the rate equation and isolating the k value, the rate constant can be determined. The rate constant k is 0.09 L²mol^-2s^-1. The units for k are L²mol^-2s^-1 because the reaction is third order overall.

Q12.6.8

Nitrogen(II) oxide, NO, reacts with hydrogen, H₂, according to the following equation:

\[ \ce{2NO} + \ce{2H_2} \rightarrow \ce{N_2} + \ce{2H_2O}\]

What would the rate law be if the mechanism for this reaction were:

\[ \ce{2NO} + \ce{H_2} \rightarrow \ce{N_2} + \ce{H_2O_2}(\mathit{ slow})\]

\[ \ce{H_2O_2} + \ce{H_2} \rightarrow \ce{2H_2O}(\mathit{ fast})\]

Answer 12.6.8

The rate law of the mechanism is determined by the slow step of the reaction. Since the slow step is an elementary step, the rate law can be drawn from the coefficients of the chemical equation. So therefore, the rate law is as follows: rate=k[NO]²[H₂]. Since both NO and H₂ are reactants in the overall reaction (therefore are not intermediates in the reaction), no further steps have to be done to determine the rate law.

Q21.4.20

What is the age of mummified primate skin that contains 8.25% of the original quantity of ¹⁴C?

Answer 21.4.20

The integrated rate law for this process is as follows: \[ \text{ Fraction remaining}= 0.5^{t\over t_{1/2}}\] where t is the time elapsed and t_1/2 is the half life. The experimentally determined half life of Carbon-14 is 5730 years. The percentage of carbon-14 remaining can be rewritten as the fraction remaining: 8.25%=(8.25/100)=0.0825. By entering the known quantities into the equation above and solving for the time elapsed, it is determined that the age of the mummified primate skin is 20,625 years old.

Q20.3.8

Consider the following spontaneous redox reaction: \[ \ce{NO_3^-}(\mathit{ aq}) + \ce{H^+}(\mathit{ aq}) + \ce{SO_3^2-}(\mathit{ aq})\rightarrow \ce{SO_4^2-}(\mathit{ aq}) + \ce{HNO_2}(\mathit{ aq})\]

1. Write the two half-reactions for this overall reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?
3. Which electrode is negatively charged, and which is positively charged?

Answer 20.3.8

1. First, split the overall reaction into two half reactions. Next, balance O by adding H₂O, and then balance H by adding H⁺. Then, balance the charge/oxidation numbers with electrons. To check your work, make sure both half reactions simplify to the given reaction.

Reduction (equation 1): \[ \ce{2e^-}+ \ce{NO_3^-}(\mathit{ aq}) + \text{ 3H}^+(\mathit{ aq}) \rightarrow \ce{HNO_2}(\mathit{ aq})+ \ce{H_2O}(\mathit{ l}) \]

Oxidation (equation 2): \[ \ce{H_2O}(\mathit{ l})+ \ce{SO_3^2-}(\mathit{ aq}) \rightarrow \ce{SO_4^2-}(\mathit{ aq})+ \ce{2H^+}(\mathit{ aq})+\ce{2e-}\]

2. Equation 1 (reduction) would occur in the cathode. Equation 2 (oxidation) would occur in the anode.

3. Since this reaction is carried out in a galvanic cell (as referenced in part 2 of the question), the anode is negatively charged, and the cathode is positively charged.

Q20.5.19

Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.

1. Write the balanced chemical equation for the oxidation of Fe²⁺ by Ce⁴⁺.
2. Plot the data and then locate the endpoint.
3. How many millimoles of Fe²⁺ did the solution being titrated originally contain?

Answer 20.5.19

1. \[ \ce{Fe^2+}+ \ce{Ce^4+} \rightarrow \ce{Fe^3+} + \ce{Ce^3+} \]

2. From the graph below, the endpoint can be determined. In potentiometric titrations, the endpoint is characterized by a steep increase in voltage. In this case, the greatest increase is seen between 9 and 10 mL, so the endpoint of this titration is approximately 9.5 mL.

3. This reaction took approximately 9.5 mL to reach the endpoint. Since the solution was 0.1 M in Ce⁴⁺, approximately 0.95 millimoles of Ce⁴⁺ were consumed during this reaction. From the balanced equation in part 1, it is known that the reaction of Fe²⁺ and Ce⁴⁺ occurs in a one-to-one ratio. So, therefore there were approximately 0.95 millimoles of Fe²⁺ in the original solution.