Extra Credit 19

Q 19.6A

Given the following voltaic cell diagram and using the standard gold electrode as the reduction half reaction, and metal \(\ce{M}\) as an oxidizing half reaction, find the \(\mathrm{E^\circ}\) reduction half reaction:

\(\begin{align}
& \mathrm{Au^{3+}(1\,M) +3e^- \rightarrow Au(s)} \quad && \mathrm E^\circ= +1.52 \\
& \mathrm{M^{3+}(1\,M) + 3e^-} \quad && \mathrm E^\circ= ?\end{align}\)

a. \(\mathrm{In \quad \hspace{5 pt} E^\circ = 1.858\: V}\)

b. \(\mathrm{Al \quad \hspace{3.5 pt}E^\circ = 3.20\: V}\)

c. \(\mathrm{La \quad E^\circ = 3.90\: V }\)

d. \(\mathrm{U \quad \hspace{5.5 pt}E^\circ = 3.18\: V}\)

Ans 19.6A

The question is already given \(\mathrm{E^\circ}\) reduction half reaction. The \(\mathrm{E^\circ}\) reduction half reaction is \(\mathrm{E^\circ_{cathode}}\) which is \(\ce{Au}\).

\(\mathrm{E^\circ_{cell}= E^\circ_{cathode} - E^\circ_{anode}}\)

\(\mathrm{E^\circ_{cell}= +1.52 - E^\circ_{anode}}\)

\(\mathrm{E^\circ_{anode}= E^\circ_{cathode} - E^\circ_{cell}}\)

a. \(\ce{In}\)

\(\mathrm{1.858 = +1.52- E^\circ_{anode}}\)

\(\mathrm{1.858 + E^\circ_{anode} = 1.52}\)

\(\mathrm{E^\circ_{anode}= -0.338\,V}\)

b. \(\ce{Al}\)

\(\mathrm{3.20=1.52 - E^\circ_{anode}}\)

\(\mathrm{3.20 + E^\circ_{anode} = 1.52}\)

\(\mathrm{E^\circ_{anode}= -1.68\,V}\)

c. \(\ce{La}\)

\(\mathrm{3.90 = 1.52 - E^\circ_{anode}}\)

\(\mathrm{3.90 + E^\circ_{anode} = 1.52}\)

\(\mathrm{E^\circ_{anode}= -2.38\,V}\)

d. \(\ce{U}\)

\(\mathrm{3.18 = 1.52 - E^\circ_{anode}}\)

\(\mathrm{3.18 + E^\circ_{anode}=1.52}\)

\(\mathrm{E^\circ_{anode}= -1.66\,V}\)

Q 19.59A

By referring back to figure 19-20, explain what would happen at each individual circumstance

1. zinc is wrapped around the head and tip of the iron nail

2. a hole is poked at the center of an iron nail

3. the nail is completely covered with copper

Ans 19.59A

1. If zinc is wrapped around the head and tip of the iron nail, through cathodic protection, zinc would get oxidized first, since it is more active. The zinc would then protect the nail from oxidation.

2. If a hole is poked at the center of an iron nail, oxidation would occur even more because there would now be another head and tip. With more strained regions, the air would have more access to the nail and be able to oxidize the nail more easily.

3. If the nail is completely covered with copper, the entire nail will be oxidized since copper isn't a sacrificial anode. Copper is less reactive than iron, so it won't protect it from corrosion.

Q21.3D

Write the correct names for the following:

1. \(\ce{[Fe(OH)(H2O)3(NH3)2]^2+}\)

2. \(\ce{[Cu(ONO)2(NH3)4]}\)

3. \(\ce{[Pt(H2O)2(NH3)2][PtCl6]}\)

4. \(\ce{[Fe(ox)2(H2O)(NH3)]-}\)

5. \(\ce{Ag2[HgCl4]}\)

Ans 21.3D

Some rules of naming coordination compound are

1. Always name the cation before the anion.

2. Name the ligands first, in alphabetical order, then the transition metal atom or ion.

3. Greek prefixes are used to designate the number of each type of ligand in the complex ion, such as bi- for two and tri- for three.

4. The oxidation state of the metal in the complex is given as a Roman numeral in parentheses.

Thus, the answers to this question are...

1. diamminetriaquahydroxoiron(III) ion

2. tetraamminedinitritocopper(II)

3. diamminediaquaplatinum(II) hexachloroplatinate(IV)

4. ammineaquadioxalatoferrate(III) ion

5. silver tetrachloromercurate(II)

Q24.11A

The initial rate of the reaction \(\ce{A + B \rightarrow C + D}\) is determined for different initial conditions, with the results listed in the table.

1. What is the order of reaction with respect to \(\ce{A}\) and to \(\ce{B}\)?

2. What is the overall reaction order?

3. What is the value of the rate constant, \(\ce{k}\)?

Ans 24.11A

1. \(\mathrm{Reaction\, 1 = 3.35 \times 10^{-4} = k \times [0.185]^m [0.144]^n}\)

\(\mathrm{Reaction\, 2 = 1.35 \times 10^{-3} = k \times [0.185]^m [0.288]^n}\)

\(\mathrm{Reaction\, 3 = 6.75 \times 10^{-4}= k \times [0.370]^m [0.144]^n}\)

\(\mathrm{Reaction\, 4 = 2.70 \times 10^{-3}= k \times [0.370]^m [0.288]^n}\)

Order of Reaction with respect to B:

Find where \(\ce{A}\) is constant and \(\ce{B}\) changes (Reaction 1 and Reaction 2)

Divide: \(\mathrm{\dfrac{Reaction\, 2}{Reaction\, 1} = \dfrac{1.35 \times 10^{-3}}{ 3.35 \times 10^{-4}} = \dfrac{[0.185]^m [0.288]^n}{[0.185]^m [0.144]^n}}\)

\(\mathrm{\dfrac{Reaction\, 2}{Reaction\, 1}= 4 = 2^n \rightarrow n=2}\)

The reaction is second order with respect to \(\ce{B}\).

Order of Reaction with respect to A:

Find where \(\ce{B}\) is constant and \(\ce{A}\) changes (Reaction 3 and Reaction 1)

Divide: \(\mathrm{\dfrac{Reaction\, 3}{Reaction\, 1} = \dfrac{6.75 \times 10^{-4}}{3.35 \times 10^{-4}} = \dfrac{[0.370]^m [0.144]^n}{[0.185]^m [0.144]^n}}\)

\(\mathrm{\dfrac{R3}{R1} = 2 = 2^m}\)

\(\mathrm{m = 1}\)

The reaction is first order with respect to \(\ce{A}\)

2. Overall reaction order = reaction order of A + reaction order of B

1^st + 2^nd = 3^rd order reaction

The overall reaction order is a 3^rd order reaction

3. To find the rate constant, k, simply plug in the reaction orders into the equation for reaction 1 and solve for k.

\(\mathrm{3.35 \times 10^{-4} = k \times [0.185][0.144]^2}\)

\(\mathrm{k=8.7 \times 10^{-2}}\)

Q 24.51A

The following observations of a reaction’s rate constant have been made: at \(\mathrm{T=325\,K}\), \(\mathrm{k=3.2E\,\textrm{-6}M^{-1}s^{-1}}\); at \(\mathrm{T=456\,K}\), \(\mathrm{k=2.8E\,\textrm{-5}M^{-1}s^{-1}}\). What is the activation energy of this reaction?

Ans 24.51A

\(\mathrm{\ln\dfrac{k_1}{k_2}=\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)=\ln\dfrac{3.2E-6M^{-1}s^{-1}}{2.8E-5M^{-1}s^{-1}}=\dfrac{E_a}{8.3145\,J\,mol^{-1}K^{-1}}\left(\dfrac{1}{456\,K}-\dfrac{1}{325\,K} \right)}\)

\(\mathrm{-2.2=-1.06\,mol\,J^{-1}(E_a)}\)

\(\mathrm{E_a=2.1E4\,J\,mol^{-1}}\)

Q 25.27A

As the museum authenticator, you have been given the task of determining which era a recently donated piece of pottery is from. If, through radiocarbon dating, you find that the disintegration rate is 13 dis min-1 g-1, what is the age of the pottery piece?

Ans 25.27A

Original rate of disintegration: 15 dis min-1 g-1 and \(\mathrm{\textrm{half-life} = 5730\, y}\)

\(\mathrm{\lambda = \dfrac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}}\)

\(\mathrm{\ln\dfrac{13\, dis/min}{15\, dis/min}=-\lambda t = -1.21\times10^{-4}\, y^{-1} (t)}\)

\(\mathrm{t=1182\, years}\)

Q 18.6

Which of the following ions will oxidize \(\ce{Br-}\) ion to \(\ce{Br2}\)?

1. \(\ce{Pb^2+}\)

2. \(\ce{H+}\)

3. \(\ce{Au^3+}\)

4. \(\ce{MnO4-}\)

Ans 18.6

\(\ce{MnO4-}\) will oxidize \(\ce{Br-}\) ion to \(\ce{Br2}\).

As you go up in the standard reduction potential table, it indicates of cathode (reduction); stronger oxidizing agent and as you go down, it indicates of anode (oxidation); stronger reducing agent. Thus the only one that place higher than \(\ce{Br2}\) is \(\ce{MnO4-}\). This means \(\ce{MnO4-}\) (aq) will oxidize \(\ce{Br-}\) ion to \(\ce{Br2}\) in aqueous solution, for example.

\(\mathrm{E^\circ_{cell}}\) = \(\mathrm{E^\circ_{cathode} - E^\circ_{anode}}\)

\(\mathrm{E^\circ_{cell}}\) = 1.5V - 1.09V

\(\mathrm{E^\circ_{cell}}\) = + 0.41V

Q 21.4.3

Half-lives for the reaction A + B → C were calculated at three values of [A]₀, and [B] was the same in all cases. The data are listed in the following table:

Does this reaction follow first-order kinetics? On what do you base your answer?

Ans 21.4.3

No, the reaction does not follow first-order kinetics. The reaction is second order for A because the half-life decreases as the reactant concentration increases which follows the second order half-life equation, t_1/2 = 1/k[A₀].