Extra Credit 24

Q21.2.9

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

S21.2.9

In order to determine the stability for each of the following element, their nuclei must have a optimal neutron-to-proton ratios. For instance ¹H and ³He have neutron-to-proton ratios that are less than 1, ⁴He, ¹⁰B, and ⁴⁰Ca have neutron-to-proton ratios that are equal to 1, and all other stable nuclides have higher neutron-to-proton ratios that increase to 1.5. We must consider that all other stable nuclei do not have atomic numbers that are greater than 83 and stable nuclei do not have odd numbers of both p⁺ and n⁰. Nuclei with magic numbers (2, 8, 20, 28, 50, 82, or 126) of p⁺, n⁰, or both are stable.

Step 1: Calculate the \[\frac{n^{0}}{p^{+}}\] ratio and determine whether it's in the optimal range of 1 to 1.5

a. chlorine-37 (atomic #= 17; 20 neutrons)

\[\frac{20}{17}\] = 1.17 (stable)

b. calcium-40 (atomic #= 20; 20 neutrons)

\[\frac{20}{20}\] = 1 (stable)

Calcium-40 contains the magic number (20) of the neutron and proton, which makes it even more stable.

c. ²⁰⁴Bi (atomic #= 83; 121 neutrons)

\[\frac{121}{83}\] = 1.46 (stable)

Heavier elements like ²⁰⁴Bi should have a higher neutron-to-proton ratio that approaches 1.5.

d. ⁵⁶Fe (atomic #= 26; 30 neutrons)

\[\frac{30}{26}\] = 1.15 (stable)

e. ²⁰⁶Pb (atomic #= 82; 124 neutrons)

\[\frac{124}{82}\] = 1.48 (stable)

f. ²¹¹Pb (atomic #= 82; 129 neutrons)

\[\frac{129}{82}\] = 1.57 (unstable; 1.57 > 1.5)

g. ²²²Rn (atomic #= 86; 136 neutrons)

\[\frac{136}{86}\] = 1.58 (unstable; 1.58 > 1.5)

h. carbon-14 (atomic #= 6; 8 neutrons)

\[\frac{8}{6}\] = 1.33 (unstable)

This is because Carbon-14 undergoes beta decay and its neutron-to-proton ratio is higher than that expected for the stability for an element with this mass.

Only a,b,c,d, and e lie within the band of stability.

Q21.7.1

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

S21.7.1

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

\[_{27}^{60}\textrm{Co} \rightarrow _{0}^{0}\textrm{gamma} + _{27}^{60}\textrm{Co}\]

I would need to contain this isotope in a lead contain because cobalt-60 emits high energy photons (gamma rays), which can be stopped by lead.

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

\[_{42}^{99}\textrm{Mo} \rightarrow _{-1}^{0}\textrm{beta} + _{43}^{99}\textrm{Tc}\]

I would need to store this isotope in an aluminum container because it molybdenum releases a beta particle, which can be stopped by aluminum.

Q20.4.10

For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential (from the voltmeter) compare with the corresponding E° (the standard reduction potential from the chart)?

S20.4.10

a) measuring the potential of a Cl⁻/Cl₂ couple

b) measuring the pH of a solution

c) measuring the potential of a MnO₄⁻/Mn²⁺ couple

For b) I would use SHE, H⁺(aq, 1M)/H₂(g, 1 atm), as a reference electrode because I can use the determined concentration of H^{+} ions to find the pH value. For instance we can use the \[E_{cell} = E^{0}_{cell} - \frac{0.0591V}{e^{-} moles} \times logQ\] with Q containing \[\frac{P_{H^{+}}}{[H_{2}]}\]. For a) and c), I would also use SHE as a reference electrode because the the measured potentials for those cases would depend on the values on the voltmeter and these measured values would correspond with the values on the standard reduction potential chart only if the concentrations of both solutions (such as Cl^-(aq) and H⁺(aq) or MnO₄^-(aq) and H⁺(aq)) are the same .

Q20.4.11

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

\[H_{2}(g) + Cu^{2+}(aq) \rightarrow 2H^{+}(aq) + Cu(s)\]

S20.4.11

In this cell diagram, the left side of the cell diagram (containing H₂(g) and H⁺(aq)) undergoes oxidation whereas the right side of the diagram (containing Cu²⁺(aq) and Cu(s)) undergoes reduction. The Pt(s) is the inert anode that does not participate in the redox reactions and Cu(s) is a cathode that receives electrons. Each electrodes on placed at the ends of the cell diagram and each single, vertical line represents the phase boundary whereas the double vertical lines represent the two phase boundaries, or the salt bridge. More information can be found here.

Step 1: Identify the oxidation reaction

Oxidation Half Reaction: \(H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-}\)

Step 2: Identify the reduction reaction.

Reduction Half Reaction: \(Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s) \)

Solution: \(Pt(s) | H_{2}(g,1atm) | H^{+}(aq,1M)| | Cu^{2+}(aq,1M)| Cu(s)\)

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?

S20.8.2

The coating of a metal with a sacrificial layer is known as prophylactic protection. By applying a protective layer of another metal that is difficult to oxidize, the metal avoids corrosion. However, in galvanic protection, we use a more easily oxidized metal to protect another metal from corrosion. For example, in order to protect iron (Fe) from oxidizing, we have Zn(s) become the anode and this allows for Fe to be the cathode. The Zn(s) eventually dissolves, sacrificing itself to protect the iron underneath. Both the prophylactic protection and galvanic protection involve a metal being protected by a sacrificial layer of another metal. Additional information on corrosion can be found here.