Extra Credit 27

Q17.3.6

Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br₂(l) is the same as for Br₂(aq).

\[Pt(s) | H_2(g)| H^+ (aq) || Br_2 (aq) | Br^-(aq) | Pt(s)\]

Given: cell diagram; under standard condition

Asked for: overall reaction, standard cell potential and spontaneity

Strategy: Identify the cathode and the anode of the galvanic cell using the cell diagram give. To identify the cathode, determine the reaction that gains electrons. To identify the anode, determine the reaction that loses electrons. Write the half-reactions of oxidation and reduction, then refer to the chart to find the standard reduction potential of two half reactions. Calculate the standard cell potential using the formula E^°_cell = E^°_cathode - E^°_anode. After the standard cell potential is calculated, compared the value to 0: if the value is greater than 0, this reaction is spontaneous because it indicates that the Gibbs free energy of the reaction is less than 0.

Solution: Since Pt(s) are written at the both side of the cell diagram, it is the inert electrode of the galvanic cell. H₂(g) and H⁺(aq) are at the anode because they were written on the left side of the salt bridge (and because oxidation occurred), and Br₂(aq) and Br^-(aq) are in the cathode (because reduction occurred). For emphasis, oxidation takes place at the anode and reduction takes place at the cathode.

Anode:

\[H_2(g) → 2H^+(aq) + 2e^-\]

\[E^°_{anode} = 0V\]

Cathode:

\[Br_2(aq) + 2e^-→ 2Br^-(aq)\]

\[E^°_{cathode}=1.087V\]

Overall:

\[H_2(g) + Br_2(aq) → 2H^+(aq) + 2Br^-(aq)\]

\[E^°_{cell}=E^°_{cathode}-E^°_{anode}=1.087V-0V=1.087V\]

Since

\[E^°_{cell}=1.087V>0\]

the overall reaction is spontaneous at standard conditions.

Q19.1.25

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M₃O₄ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M₂O₃, to permit estimation of the metal’s two oxidation states.)

1. Sc₂O₃
2. TiO₂
3. V₂O₅
4. CrO₃
5. MnO₂
6. Fe₃O₄
7. Co₃O₄
8. NiO
9. Cu₂O

Given: chemical formulas of oxides of different transition metals

Asked for: oxidation states of the transition metals

Strategy: Using the fact the sum of all oxidation states of all elements in a compound is 0, and the oxidation states of Oxygen (-2), calculate the oxidation states of the transition metals.

Solution:

Set the unknown oxidation states of transition metals in each oxides to be "x" in all cases. Write an algebraic expression and solve to determine what x is.

1. \[2x+3*(-2)=0; x = 3\]

Ans: Sc³⁺

2. \[x+2*(-2)=0; x = 4\]

Ans: Ti⁴⁺

3. \[2x+5*(-2)=0; x = 5\]

Ans: V⁵⁺

4. \[x+2*(-3)=0; x = 6\]

Ans: Cr⁶⁺

5. \[x+2*(-2)=0; x = 4\]

Ans: Mn⁴⁺

6. \[Fe_3O_4=FeO+Fe_2O_3\]

\[FeO: x+1*(-2)=0; x = 2\] Ans: Fe²⁺

\[Fe_2O_3: 2x+3*(-2)=0; x = 3\] Ans: Fe³⁺

7. \[Co_3O_4 = CoO + Co_2O_3\]

\[CoO: x+1*(-2)=0; x = 2\] Ans: Co²⁺

\[Co_2O_3: 2x+3*(-2)=0; x = 3\]Ans: Co³⁺

8. \[x+1*(-2)=0; x = 2\]Ans: Ni²⁺

9. \[2x+1*(-2)=0; x=1\] Ans: Cu⁺

Q12.4.18

Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.

Given: amount of Carbon-14 left; half life of carbon

Asked for: what year did King Richard III die, in other words, how old is King Richard III skeleton (asking for time)

Strategy: Using the half life and amount of carbon-14, find how old the skeleton is today, then trace back to what year he died.

Solution: Since the decay of carbon-14 is a first order reaction, we can use the integrated form of reaction law to derive an equation that relates the time it takes for carbon-14 to decay from 100% to 93.79% and its half time.

Equation 1: integrated form of 1st order rate law, where [C] is the amount of carbon after a period of time and [C_o] is the initial carbon amount, and k is the reaction rate constant

\[ln\frac{[C]}{[C_o]}=-kt\]

Plug this quantity in Equation 1 \[k=\frac{ln2}{t_{1/2}}\]

Equation 2:

\[ln\frac{[C]}{[C_o]}=-ln2 * \frac{t}{t_{1/2}}\]

Raised both side to "e" power to cancel the logarithm function to get Equation 3

\[\frac{[C]}{[C_o]} = 0.5^{\frac{t}{t_{1/2}}}\]

Since about 93.79% of the carbon-14 expected in living tissue, the fraction between amount of carbon-14 to the initial amount of carbon-14 is 0.9379.

Plug in

\[\frac{C}{C_o}=0.9379 ; t_{1/2} = 5730 yrs\]

into Equation 3

\[0.9379 = 0.5 ^{\frac{t}{5370}}\]

\[t=496.7 yrs ≈ 497 yrs\]

\[2017-497=1520\]

Therefore, King Richard III die in year 1520.

Q21.3.2

Which of the various particles that may be produced in a nuclear reaction(α particles, β particles, and so on) are actually nuclei?

Given: various particles

Asked for: which particles that are produced are actually nuclei

Strategy: Using the nature of each particle, i.e., the mass number and proton number, analyze which particles are nuclei.

Solution:

α particle is a Helium atom with +2 charge:

\[\alpha : ^4_2He\]

There are two types of β particles: β^- and β⁺.

\[\beta^- : ^0_{-1}e\]

\[\beta^+: ^0_1e\]

Gamma:

\[\gamma: ^0_0\gamma\]

Neutron:

\[^1_0 n\]

Since only α particle has both protons and neutrons (2 protons, 2 neutrons), α particle is the only actual nuclei.

Q21.7.4

A scientist is studying a 2.234 g sample of thorium-229 (t_1/2 = 7340 y) in a laboratory.

1. (a) What is its activity in Bq?
2. (b) What is its activity in Ci?

Given: mass of the sample thorium-229 and its half-life

Asked for: The activity in Bq and activity in Ci.

Strategy: The activity of a radioactive element describes it rate of radioactive decay and can be calculated by the formula A=\lambda * N, where \lamda is the decay constant and N stands for number of atoms. The decay constant is not given in the question, but we can calculate it using the half-life of the element, since t_1/2 = ln2/(\lamda). The result we get using the combination of these formulas should have the units of Bq, which is 1 dis/sec. We can further convert Bq into Ci, since we know that 1 Ci = 3.7e10 dis/sec.

Solution:

Half-life of a radioactive element can be calculated by:

\[t_{1/2}= \frac{ln2}{\lambda}\]

\[\lambda = \frac{ln2}{t_{1/2}} \Rightarrow\ Equation 1\]

Plug this quantity into Equation 1

\[t_{1/2}=7340 yrs\]

We get:

\[\lambda = \frac{ln2}{7340} = 9.44*10^{-5} yrs^{-1}\]

The activity of a radioactive element can be calculated by:

\[A= \lambda * N\]

\[N = \frac{mass}{molar mass} * Avogadro Number = \frac{2.234g}{229\frac{g}{mol}} * 6.02*10^{23} \frac{atoms}{mol} = 5.87*10^{21} atoms\]

Plug in the decay constant lambda and N:

\[A= 9.44*10^{-5} yrs^{-1} * 5.87*10^{21} atoms = 5.546*10^{17} \frac {atoms}{year}\]

Calculate the relationship between years and seconds:

\[\frac{365 days}{1 year} * \frac {24 hrs}{1 day} * \frac {60mins}{1hr} * \frac{60sec}{1hr} = 3.15*10^7 \frac{sec}{year}\]

To get activity in Bq, we need to take A [=]atoms/year divided by how many seconds in a year:

\[A=\frac {5.546*10^{17} \frac{atoms}{year}} {3.15*10^7 \frac{sec}{year}}=1.76*10^{10} \frac{atoms}{sec} = 1.76*10^{10} Bq\]

Since 1Ci = 3.7e10 Bq,

\[A=\frac{1.76*10^{10} Bq}{3.7*10^{10} \frac{Bq}{Ci}} = 0.475 Ci\]

Q20.4.17

The standard cell potential for the oxidation of Pb to Pb²⁺ with the concomitant reduction of Cu⁺ to Cu is 0.39 V. You know that E° for the Pb²⁺/Pb couple is −0.13 V. What is E° for the Cu⁺/Cu couple?

Given: The redox reaction of Pb/Pb²⁺ and Cu⁺/Cu, cell potential of Pb/Pb²⁺ and overall cell potential under standard condition

Asked for: The cell potential of reduction of Cu⁺/Cu

Strategy: Distinguish between the cathode and anode in this question. Since oxidation (loss of electrons) takes place at the anode, and reduction (gain of electrons) takes place at the cathode, the half reaction of Pb/Pb²⁺ takes place in anode. Use the relationship of the overall cell potential and the cell potential of the cathode/anode (E^°_cell = E^°_cathode - E^°_anode) to calculate the cell potential at the cathode (Cu⁺/Cu).

Solution:

\[Anode: Pb\to\ Pb^{2+} + 2e^-\ ; E^°_{anode} = -0.13V\]

\[Cathode: Cu^{+} + e^-\to\ Cu ; E^°_{cathode} = xV\]

\[Overall: Pb+2Cu^+ \to\ Pb^{2+} +2Cu\]

\[E^°_{overall} =E^°_{cathode}-E^°_{anode}=0.39 V = x V - (-0.13)V\]

\[x+0.13 =0.39\]

\[x = 0.26 V\]

Therefore, E^°_cathode = 0.26 V

Q20.9.2

How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?

Given: an electrolytic cell

Asked for: how to compare the strength of different oxidants and reductants given an electrolytic cell

Strategy: Refer to the standard reduction potential chart, find the standard reduction potentials for each of the half-reactions.

Solution:

When a certain compound is considered as a "reductant", it means this compound will lose electrons during a redox reaction and will be oxidized. Therefore, the half reaction of the reductant being oxidized should have the form of :

\[X \to\ X^+ + e^-\]

When certain oxidation half reaction has a large absolute value of standard reduction potential, for example, 3.00 V, the compound being oxidized can be considered a strong reductant. However, since the chart is organized as reduction potential, there's should be a negative sign in front of every oxidation reaction. Therefore, when we are consulting the standard reduction potential chart, an oxidation with a more negative value is considered to be a strong reductant.

Similarly, a half reaction with big value of standard reduction potential and also a positive sign in front of the value is considered a good oxidants.

In all: a large positive number value correlates with strong oxidants and a large negative number correlates with strong reductants.

Q14.1.2

If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?

Given: A particular reaction, the chemical kinetics of the reaction

Asked for: Why the chemical kinetics of the reaction is important to an industrial facility

Strategy: Consider the relationship between the rate of reaction and chemical kinetics and how important reaction rate is to industrial production

Solution:

Not all chemical reactions happens within one elementary step. In fact, most chemical reactions compose of several elementary steps, and some of the elementary steps are fast, while some aren't. "Fast" means certain elementary steps have large reaction rates, while slow steps have small reaction rates. The reaction rate of the overall reaction depends on the slowest rate among all the elementary steps. One can understand it as: when we're driving a car, heading to place B from place A, the time it takes to travel will increase if there's some traffic between A and B. What's more, there can be more than one proposed kinetics of a certain steps, with completely different elementary steps between the reactants and the products. Therefore, it is important to study the chemical kinetics of that particular reaction so that the intermediates, reaction rates of intermediates and the overall rate of reaction are well known. This way, the facility can arrange time and money accordingly to each step and maximize profits.