Extra Credit 24

Q17.3.3

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

Cu(s)│Cu²⁺(aq)║Au³⁺(aq)│Au(s)

SOLUTION

1) Determining the overall reaction:

a) identify half reactions

In order to determine the overall reaction, we can separate the reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. From the cell diagram, the left hand side represents the oxidation reaction, which occurs at the anode, and the right hand side represents the reduction reaction, which occurs at the anode. ( I think we can leave this part out, but i guess it's important for calculating delta E )

oxidation: Cu(s) ⟶ Cu²⁺(aq) + 2e^- (anode)

reduction: Au³⁺(aq) + 3e^- ⟶ Au(s) (cathode)

b) multiply the oxidation and reduction equations by appropriate coefficients so both contain the same number of electrons. This is important because a balance of electron is needed for redox reactions to proceed.

oxidation (x3): 3Cu(s) ⟶ 3Cu²⁺(aq) + 6e^-

reduction (x2): 2Au³⁺(aq) + 6e^- ⟶ 2Au(s)

c) add the two equations and cancel the electrons

3Cu(s) + 2Au³⁺(aq) + 6e^-⟶ 3Cu²⁺(aq) + 2Au(s) + 6e^-

which then yields

3Cu(s) + 2Au³⁺(aq) ⟶ 3Cu²⁺(aq) + 2Au(s)

2) Determining the standard cell potential

E°_cell = E°_cathode − E°_anode

a) half-reactions that actually occur in the cell and their corresponding electrode potentials (found at Table P1) are as follows:

anode: Cu(s) ⟶ Cu²⁺(aq) + 2e^- E°_anode = 0.3419 V

cathode: Au³⁺(aq) + 3e^- ⟶ Au(s) E°_cathode = 1.52 V

b) plugging into the standard cell potential equation

E°_cell = E°_cathode − E°_anode

= 1.52 V − 0.3419 V

= 1.178 V; this means that the overall reaction will be spontaneous *because delta G will be negative due to delta E being positive by observing the eqation

ΔG° = -nFE°_cell

This equation indicates that the more positive E°_cell is, the more negative ΔG° is and therefore, more spontaneous. In this case, again, because E°_cell is positive, the reaction is spontaneous.

Q19.1.22

Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.

Co(NO₃)₂(s) ⟶ Co₂O₃(s) + NO₂(g) + O₂(g)

SOLUTION

a) Write the unbalanced chemical equation for the reaction, showing the reactants and the products

Co(NO₃)₂(s) ⟶ Co₂O₃(s) + NO₂(g) + O₂(g)

b) Assign oxidation states to all atoms in the reactants and the products and determine which atoms change oxidation state

*To do this, we must follow the basic principles of oxidation state. Elements by itself does not have a charge, oxygen within a compound has -2 charge. We can then use this property of oxygen gas to determine the oxidation states of the elements in the compounds.

+2 -2 +3 +4 0

Co(NO₃)₂(s) ⟶ Co₂O₃(s) + NO₂(g) + O₂(g)

+5 -2 -2

c) Write separate equations for oxidation and reduction, showing

(i) the atom(s) that is (are) oxidized and reduced

oxidation: Co(NO₃)₂(s) ⟶ Co₂O₃(s)

H₂O(l) ⟶ O₂(g)

reduction: Co(NO₃)₂(s) ⟶ NO₂(g)

(ii) the number of electrons accepted or donated by each and showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms

oxidation: 2Co(NO₃)₂(s) ⟶ Co₂O₃(s) + 2e^-

2H₂O(l) ⟶ O₂(g) + 4e^-

reduction: Co(NO₃)₂(s) + 2e^- ⟶ 2NO₂(g)

and thus combining the reaction with the same reactant

2Co(NO₃)₂(s) +2e^- ⟶ Co₂O₃(s) + 4NO₂(g)

d) Balance the charge by adding protons (H⁺(aq)) and balance the oxygen atoms by adding H₂O(l) molecules to one side of the equation

oxidation: 2H₂O(l) ⟶ O₂(g) + 4e^- + 4H⁺

reduction: 2Co(NO₃)₂(s) + 2e^- + 2H⁺(aq) ⟶ Co₂O₃(s) + 4NO₂(g) + H₂O(l)

e) Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons (in order to satisfy electrons gained = electrons lost). In this case, a total of 4 electrons will be transferred

reduction(x2): 4Co(NO₃)₂(s) + 4e^- + 4H⁺(aq) ⟶ 2Co₂O₃(s) + 8NO₂(g) + 2H₂O(l)

f) Add the two equations and cancel the electrons and any like terms

oxidation: 2H₂O(l) ⟶ O₂(g) + 4e^- + 4H⁺

reduction: 4Co(NO₃)₂(s) + 4e^- + 4H⁺(aq) ⟶ 2Co₂O₃(s) + 8NO₂(g) + 2H₂O(l)

overall: 4Co(NO₃)₂(s) ⟶ 2Co₂O₃(s) + 8NO₂(g) + O₂(g)

Q12.4.14

There are two molecules with the formula C₃H₆. Propene, CH₃CH=CH₂CH₃CH=CH₂, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:

When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of 5.95 × 10⁻⁴ s⁻¹. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?

SOLUTION

1) Determining half-life of this reaction

a) Identify the order of the reaction

Given that the rate constant,k of the reaction is 5.95×10⁻⁴ s⁻¹ with the units being s⁻¹, it can be deduced that the reaction is first order. *Furthermore, because this reaction is a first order reaction, we can ignore factors such as concentrations because a first order reaction rate does not depend on the concentration of the reactants.

b) Half-life of a first order reaction is

\({t_{1/2}}=\frac{ln2}{k}\)

\({t_{1/2}}=\frac{ln2}{(5.95×10^{-4} s^{-1})}\)

\({t_{1/2}}= 1160 s\)

2) Determining the fraction of cyclopropane remaining after 0.75h

\(ln\frac{A_t}{A_0}=-kt\)

The fraction of cyclopropane remaining is equivalent to the ratio \(\frac{A_t}{A_0}\) therefore

\(lnx = -kt\)

where x denotes the fraction of cyclopropane

a) Since k, the rate constant is in units of seconds, we must convert the time given to seconds

\((0.75h)(\frac{60min}{1h})(\frac{60s}{1min})\)

2700s

b) Using the information given

\(lnx = -kt\)

lnx = -(5.95×10^-4 s^-1)(2700s)

lnx = -1.6065

e ^{ln x} = e^-1.6065

x = 0.20059

= 20%

Q21.2.9

Which of the following nuclei lie within the band of stability?

1. chlorine-37
2. calcium-40
3. ²⁰⁴Bi
4. ⁵⁶Fe
5. ²⁰⁶Pb
6. ²¹¹Pb
7. ²²²Rn
8. carbon-14

SOLUTION

Figure 24.1: The Relationship between Nuclear Stability and the Neutron-to-Proton Ratio.

Most elements have isotopes. For stable isotopes, *a plot referred to as the Nuclear Belt of Stability shows a ratio of neutrons to protons that will deem the isotope stable.

a) Isotopes with atomic number (Z) ≤ 20 and with a neutron (n) to proton (p) ratio of about 1 are more likely to be stable

b) Isotopes with atomic number (Z) < 82, have one or more stable isotopes with exceptions being technetium (Z = 43) and promethium (Z = 61) which do not have any stable isotopes

i) even number of protons and even numbers of neutrons is most likely to be stable

ii) odd numbers of protons and odd numbers of neutrons is most likely to be unstable

iii) limit of 209 nucleons in stable nucleus

c) Isotopes with atomic number (Z) > 83 are unstable

Looking at the graph:

1. chlorine-37; p = 17, n = 20, stable (within band of stability)
2. calcium-40; p = 20, n = 20, stable (within band of stability)
3. ²⁰⁴Bi; p = 83, n = 121, stable (within band of stability and under the limit of max nucleons)
4. ⁵⁶Fe; p = 26, n = 30, stable (within band of stability)
5. ²⁰⁶Pb; p = 82, n = 124, stable (within band of stability)
6. ²¹¹Pb; p = 82, n = 129, unstable, radioactive (above band of stability)
7. ²²²Rn; p = 86, n = 136, unstable, radioactive (above band of stability and atomic number > 82)
8. carbon-14; p = 12, n = 2, unstable, radioactive (below band of stability and ratio way below 1)

Q21.7.1

If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:

1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)

SOLUTION

An unstable nucleus may emit particles such as alpha particles, beta particles, and gamma particles to attain a stable configuration.

An alpha particle (α) is a helium atom that relatively easy to stop and is the least penetrating radioactive decay.

\(_2^4He\)

A beta particle (β) is an electron that has high energy that is released during the radioactive decay of the unstable nucleus.

\(_{-1}^0e\)

A gamma ray (γ) is part of the electromagnetic particles that have high energy and are emitted by nuclei in an excited state attempting to return to their ground state.

\(_0^0γ\)

Figure 24.2: The Ability of Different Types of Radiation to Pass Through Material, from Least to Most Penetrating.

1. cobalt-60 (a strong γ emitter used for irradiation)

Cobalt-60 is a source of gamma transmitters. Gamma rays are more penetrating than either alpha and beta radiation, it can *penetrate the skin and damage cells, go through water and concrete and can only be stopped by a thick layer of lead.

2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)

Molybdenum-99 is a source of beta emitters. Beta particles have moderate ionizing radiation but are however far more powerful than an alpha emission. Although it's more powerful than alpha emission, *it's still far less energized in comparison to a gamma emission and will be stopped by a thin sheet of metal.

Q20.4.10

For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?

1. measuring the potential of a Cl⁻/Cl₂ couple
2. measuring the pH of a solution
3. measuring the potential of a MnO₄⁻/Mn²⁺ couple

SOLUTION

1. Measuring the potential of a Cl⁻/Cl₂ couple

In a reaction typically evolving Cl₂ gas, one would use an inert electrode such as a graphite electrode to act as an electrode for the oxidation or reduction site. Such electrodes only serve as a source or sink for electrons without playing a chemical role in the reaction itself. An inert electrode has a standard electrode potential of zero, and thus is needed to calculate cell potentials using different electrodes or *different concentrations.

2. Measuring the pH of a solution

Measuring the pH requires a glass electrode (ion-selective electrode made of a doped glass membrane) and a reference electrode (stable electrode with a maintained potential typically a precious metal such as platinum with a standard electrode potential of zero). The value of the E°_cell can be calculated by measuring the voltage generated between the two electrodes by obtaining the electromotive force caused by the difference in pH between the reactions.

3. Measuring the potential of a MnO₄⁻/Mn²⁺ couple

In this reaction, both MnO₄⁻ and Mn²⁺ are aqueous in nature and therefore will include an inert electrode such as a graphite electrode to serve as the site for where reduction happens, as Mn is reduced from an oxidation number of +7 to +2. Again, an inert electrode transfers electrons rather than exchanging ions with the solution and does not participate in the reaction itself. More so, an inert electrode has a standard electrode potential of zero and is needed to calculate cell potentials using *different values of the half-reactions at each electrode.

Q20.4.11

Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

SOLUTION

a) Determining the half reactions

The reaction can be separated into two parts: oxidation—in which the atoms of one element lose electrons—and reduction—in which the atoms of one element gain electrons. *We can see that copper goes from an oxidation state of 2+ to an oxidation state of 0, which illustrates that the oxidation number went down, indicating that it got reduced (reduction reaction). On the other hand, hydrogen goes from an oxidation state of 0 to an oxidation state of 1+, which illustrates that the oxidation number went up, indicating that it got oxidized (oxidation reaction). The balanced half reactions are as they follow,

oxidation: H₂(g) ⟶ 2H⁺(aq) + 2e^-

reduction: Cu²⁺(aq) + e^- ⟶ Cu(s)

b) Constructing the cell diagram

In a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. However in this reaction, because hydrogen gas is involved, a standard hydrogen electrode must be used, with a platinum electrode acting as the anode for this reaction. Thus,

Pt(s) | H⁺(aq) || Cu²⁺(aq) | Cu(s)

Q20.8.2

What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?

SOLUTION

A sacrificial layer is a layer of metal which undergoes oxidation more than the metal that it protects. It is a type of corrosion control done through the application of thin metal layers that have lower standard electrode potential values (higher levels in the electrochemical series). An example of this is galvanized steel (zinc coated). Depending on the anode reaction, the more active sacrificial layer dissolves and as the electrons released by that layer (typically of zinc) flow to that part of metal that is protected, turning it into a cathode, and thus preventing corrosion of the metal. *In other words, a piece of metal with that is more reductive will be oxidized first.