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).

\[\ce{Pt}(s)│\ce{H2}(g)│\ce{H+}(aq)║\ce{Br2}(aq)│\ce{Br-}(aq)│\ce{Pt}(s)\]

S17.3.6

The given voltaic cell components, Hydrogen and Bromine, must first be located in the list of Standard reduction potentials. From this information, the element which is being oxidized and reduced can be determined. The higher half reaction on the standard reduction potential chart is the oxidation reaction. This is because it is a better reducing agent. In this case, Hydrogen is being oxidized and Bromine is being reduced. Oxidation occurs at the anode, therefore the reduction potential for Hydrogen is E°_anode while the reduction potential of Bromine is equal to E°_cathode. From this information, the cell potential, E°_cell, and the overall reaction can be determined.

Using the half reactions:

anode half reaction, E°=0: \[2H^{+}_{(aq)} + 2e^{-} \rightleftharpoons H_{2(g)}\]

cathode half reaction, E°=1.087: \[Br_{2(aq)} + 2e^{-} \rightleftharpoons 2Br^{-}_{(aq)}\]

The values for E° can be substituted into the following equation and solved:

\[E°_{cell} = E°_{cathode} − E°_{anode}\]

\[E°_{cell}=(1.087 V)-(0 V)\]

\[E°_{cell}=1.087 V\]

To find the overall reaction, the reaction which takes place at the anode must be reversed. It can be rewritten as:

\[H_{2(aq)} \rightleftharpoons2H^{+}_{(aq)} + 2e^{-}\]

Because the two reactions have the same number of electrons, they can be added. Combining the two reaction gives the overall reaction:

\[Br_{2(aq)} + 2e^{-} \rightleftharpoons 2Br^{-}_{(aq)}\]

\[+H_{2(aq)} \rightleftharpoons2H^{+}_{(aq)} + 2e^{-}\]

Overall: \[Br_{2(aq)}+H_{2(g)} \rightleftharpoons2H^{+}+2Br^{-}\]

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

S19.1.25

When determining oxidation states, you should begin by writing the oxidation state for a known element from the oxidation state rules. In the following examples, this element is Oxygen. It may be helpful to review the oxidation state rules. From there, set up an algebraic equation using the overall charge of the compound and the number of atoms involved and solve for the unknown oxidation state. In the following examples, the overall charge is zero so the equation is set equal to zero.

\[Overall charge=(Number of atoms)(oxidation state of atom)+(number of atoms)(oxidation state of atom)\]

\(a. Sc_{2}O_{3}\)

Because oxygen has a -2 charge when bound and the overall charge of the compound is zero, the oxidation state of the metal can be calculated.

\[Overall charge=(number of Sc atoms)(oxidation state)+(number of Oxygen atoms)(oxidation state)\]

\(0=2x+3(-2)\)

\(x=+3\)

Therefore, the oxidation state is Sc⁺³

This same method can be followed for each compound given.

\(b. TiO_{2}\)

\(0=2(-2)+(x)\)

\(x=+4\)

\(Ti^{+4}\)

\(c. V_{2}O_{5}\)

\(0=5(-2)+2(x)\)

\(x=+5\)

\(V^{+5}\)

\(d. CrO_{3}\)

\(0=3(-2)+x\)

\(x=+6\)

\(Cr^{+6}\)

\(e. MnO_{2}\)

\(0=2(-2)+x\)

\(x=+4\)

\(Mn^{+4}\)

\(f. Fe_{3}O_{4}\)

This compound can be rewritten to determine the two oxidation states present. The resulting compounds are:

\(FeO\) and \(Fe_{2}O_{3}\)

For FeO:

\(0=-2+x\)

\(x=+2\)

\(Fe^{+2}\)

For Fe₂O₃:

\(0=3(-2)+2(x)\)

\(x=+3\)

\(Fe^{+3}\)

\(g. Co_{3}O_{4}\)

This compound can be rewritten as:

\(CoO\) and \(Co_{2}O_{3}\)

For CoO:

\(0=-2+x\)

\(x=+2\)

\(Co^{+2}\)

For Co₂O₃:

\(0=3(-2)+2(x)\)

\(x=+3\)

\(Co^{+3}\)

\(h. NiO\)

\(0=-2+x\)

\(x=+2\)

\(Ni^{+2}\)

\(i. Cu_{2}O\)

\(0=-2+2(x)\)

\(x=+1\)

\(Cu^{+1}\)

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.

S12.4.18

For this question, the time for the sample to decay to 93.79% of it's original carbon composition must be determined and then subtracted from the current year to find the year in which the sample is from. Using the half life of Carbon, solve for the decay constant in the decay equation:

\[N(t)=N_{o}e^{-kt}\]

\[1/2=e^{-5730k}\]

\[k=0.000121\]

Using this value of k and the given information for the amount of remaining carbon in the sample, solve for the time elapsed.

\[93.78=100e^{-0.000121t}\]

\[t= 530.732 years\]

Subtracting this time period from the current year results in the year of death of the king:

\[2017-530.7= year 1486\]

Q21.3.2

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

S21.3.2

An alpha particle is a nuclei. This is because an alpha particle is actually a helium nucleus. It may be helpful to review the section on nuclear reactions to familiarize yourself with the types of decay.

Q21.7.4

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

1. What is its activity in Bq?
2. What is its activity in Ci?

S21.7.4

1. To determine activity, use the formula:

\[activity=\dfrac{ln2}{t_{½}} m\]

Where t_1/2 is the half life and m is the mass of the sample. This will result in the unit of grams per year.

\[activity=\dfrac{ln2}{7340} 2.234g\]

\[activity=0.000094\]

This value is in grams per year, however, and the question asks for the answer to be in Bq. Becquerel are defined as the amount of sample which undergoes deacy per second. To get to Bq, perform the following stoicheometric conversions:

\[\dfrac{grams}{year} \dfrac{year}{365 days} \dfrac{1 day}{24 hours} \dfrac{1 hour}{60 min} \dfrac{1 min}{60 sec} \dfrac{1 mol}{229 grams} \dfrac{6.022x10^{23} molecules}{mole}=7.8745x10^{9}\]

\[7.8745x10^{9}Bq\]

2. For part 2, the answer from part one must be converted to Curie. One Ci is equal to \(3.7x10^{10}\) Bq.

\[7.8745x10^{9} Bq x \dfrac{1 Ci}{3.7x10^{10}}\]

\[=0.2128 Ci\]

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?

S20.4.17

From the question, it is known that copper is being reduced, therefore lead is being oxidized. Oxidation occurs at the anode so the given value of -0.13 V can be plugged in for E°_anode in the equation for cell potential. Using the given overall cell potential, solve for E°_cathode.

\[E°_{cell} = E°_{cathode} − E°_{anode}\]

\[E°_{anode}=-0.13 V\]

\[E°_{cathode}=X\]

\[E°_{cell}=0.39 V\]

\[0.39 V=X-(-0.13 V)\]

\[X=0.26 V\]

Q20.9.2

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

S20.9.2

An electrolytic cell could be used to determine the relative strengths of oxidants and reluctant by measuring the cell potential of voltaic cells. By taking measurements of the voltage of constructed voltaic cells, conclusions can be drawn about the strength of the oxidants using the equation \(E°_{cell} = E°_{cathode} − E°_{anode}\). It may be helpful to review the section on comparing strength of oxidants and 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?

S14.1.2

Through a study of kinetics, the time required for a reaction to proceed can be determined. In an industrial facility, a reaction must occur within a relatively short amount of time. If a reaction were to take a year to proceed, it would not be practical to use in a facility and an alternate, faster reaction would be preferential.