Extra Credit 26

Q17.3.5

S17.3.5

3Cd(s)+2Al3+(aq)⟶3Cd2+(aq)+2Al(s)3Cd(s)+2Al3+(aq)⟶3Cd2+(aq)+2Al(s); −1.259 V; nonspontaneous

Phase II: This is partially correct because the person wrote the overall equation [ 3Cd(s) + 2Al³⁺(aq) --> 3Cd²⁺(aq) + 2Al(s) ] twice. This is correct because when you find the E° (subtracting the anode from the cathode), you would get -1.257 V. The slight difference is the result of using different reduction potentials for each species (Al and Cd). This is nonspontaneous because a negative overall cell potential translates to a positive ΔG.

• Cell Notation = Cd(s) | Cd²⁺(1M) || Al³⁺(1M) | Al(s)
  • We don't need to add the nitrate portion of "1M aluminum nitrate solution" because the Al is the one being reduced.
• Initial = Cd(s) + Al³⁺(aq) --> Cd²⁺(aq) + Al(s)
• Cathode (Reduction) = Al³⁺(aq) --> Al(s)
• Anode (Oxidation) = Cd(s) --> Cd²⁺(aq)
• E° = Cathode - Anode --> E° = -1.66 - (-0.403) --> E° = -1.257 V
• Overall = 3Cd(s) + 2Al³⁺(aq) --> 3Cd²⁺(aq) + 2Al(s)

Q19.1.24

S19.1.24

[CrO₄]^2- is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable.

Phase II: Partially Correct = Ionization energy measures the stability of the "outer shell" electrons of an atom, as it is the energy required to remove an electron from a gaseous atom or ion. The lower the ionization energy, the less energy it takes to remove an electron; therefore, more likely to lose an electron and become less stable. Since the ionization energy decreases as you go down a group, Tungsten is less stable than Chromium because it's lower in the group. Chromium is more stable than Tungsten. Chromium by itself is in the 3d orbital while Tungsten by itself is in the 5d orbital.

Q12.4.17

S12.4.17

252 days

for first order reaction: t_1/2 = 0.693 / k

k = 0.693 / 42

k = 0.0165

for first order reaction: [A] = [A]₀ e^-kt

1/64 initial means that: [A] = 1/64 [A]₀

therefore: 1/64 [A]₀ = [A]₀ e^-0.0165t

t = 252 days

Phase II: The answer of t = 252 days is correct (if we plug in our given values into the first order half life equation, we would get 252 days), however, the process of getting the answer is incorrect. \(\frac{1}{64}\) is supposed to represent \(\frac{[A]_{t}}{[A]_{0}}\). Therefore, [A]_t and [A]₀ shouldn't be canceled out through isolation.

Steps:

1st Order Equation:

\[ln[A]_{t}=-kt+ln[A]_{0}\]

We will now rearrange the equation to get it in terms of "t" (time), the variable we want.

\[\frac{ln\frac{[A]_{t}}{[A]_{0}}}{-k}=t\]

• Subtracting natural logs (ln) translates to dividing natural logs (ln)
• \(ln(\frac{x}{y})=ln(x)-ln(y)\)

We will now replace k with variables we know.

\[t_{1/2}=\frac{0.693}{k}\]

\[k=\frac{0.693}{t_{1/2}}\]

1st Order Equation Replaced with Variables we have:

\[\frac{ln\frac{[A]_{t}}{[A]_{0}}}{-\frac{0.693}{t_{1/2}}}=t\]

• \(\frac{[A]_{t}}{[A]_{0}}=\frac{1}{64}\)
• [A]_t does not equal 1 and [A]₀ does not equal 64, \(\frac{1}{64}\) is one total value to represent \(\frac{[A]_{t}}{[A]_{0}}\) as a whole. We're not given an initial or final amount, just the fraction of the initial dose left in the body in the specified time frame.

We will now plug in our known values into the equation.

\[t=\frac{ln\frac{1}{64}}{-\frac{0.693}{42}}\]

\[t=252.054=252\: days\]

Q21.3.1

S21.3.1

a. A nucleon is any particle contained in the nucleus of the atom, so it can refer to protons and neutrons.

• Phase II: Correct

b. An α particle is one product of natural radioactivity and is the nucleus of a helium atom.

• Phase II: Correct

c. A β particle is a product of natural radioactivity and is a high-speed electron.

• Phase II: Correct

d. A positron is a particle with the same mass as an electron but with a positive charge.

• Phase II: Correct

e. Gamma rays compose electromagnetic radiation of high energy and short wavelength.

• Phase II: Correct

f. Nuclide is a term used when referring to a single type of nucleus.

• Phase II: Incorrect = Nuclides are an atom or ion characterized by the contents of their nucleus (i.e. its number of protons, its number of neutrons, and its nuclear energy state).

g. The mass number is the sum of the number of protons and the number of neutrons in an element.

• Phase II: Partially Correct = The mass number is the total number of protons and neutrons in an atomic nucleus.

h. The atomic number is the number of protons in the nucleus of an element.

• Phase II: Correct

Phase II: The rest of the definitions (except the ones I fixed) are correct because they are the exact definitions from the dictionary and notes. I included 2 links to supplement these definitions.

• Atomic Structure:
  • https://chem.libretexts.org/Core/Phy...omic_Structure
• Radioactivity:
  • https://chem.libretexts.org/Under_Co..._Radioactivity

Q21.7.3

S21.7.3

all the radioactive reaction are first order reaction, t_1/2 = 0.693 / k

therefore, activity is inverse proportional with t_1/2.

Because uranium-232 (t1/2=68.9y) and uranium-233 (t1/2=159,200y), uranium-233's t_1/2 is larger than uranium-232, the activity of uranium-233 is smaller than uranium-232.

Phase II: This is correct. Uranium-232 has greater activity than uranium-233 because activity is inversely proportional to the half-life of a substance. This means that the longer the half-life of a substance, the lower its activity.

• Uranium-232 - t_1/2 = 68.9 years --> Higher activity than Uranium-233
• Uranium-233 - t_1/2 = 159,200 years --> Lower activity than Uranium-232

Q20.4.13

S20.4.13

1. Cl₂(g) + H₂(g) → 2Cl⁻(aq) + 2H⁺(aq); E° = 1.358 V
   • Phase II: This is correct because when you subtract the anode from the cathode, you would get E° = 1.358 V.
     • Cathode = Cl₂(g) → 2Cl⁻(aq) = +1.358 V
     • Anode = H₂(g) → 2H⁺(aq) = 0 V
     • E° = Cathode + Anode --> E° = 1.358 - 0 --> E° = 1.358 V
2. Br₂(l) + 2Fe²⁺(aq) → 2Br⁻(aq) + 2Fe³⁺(aq); E° = 0.316 V
   • Phase II: This is partially correct. The correct part is that when you subtract the anode from the cathode, you would get E° = 0.3163 V. However, the incorrect part is that the final phase of Br₂ in the reactants should be in aqueous form (aq), not liquid form (l).
     • Cathode = Br₂(aq) → 2Br⁻(aq) = +1.0873 V
     • Anode = 2Fe²⁺(aq) → 2Fe³⁺(aq) = +0.771
     • E° = Cathode + Anode --> E° = 1.0873 - 0.771 --> E° = 0.3163 V
3. 2Fe³⁺(aq) + Cd(s) → 2Fe²⁺(aq) + Cd²⁺(aq); E° = 1.174 V
   • Phase II: This is correct because when you subtract the anode from the cathode, you would get E° = 1.171 V.
     • Cathode = 2Fe³⁺(aq) → 2Fe²⁺(aq)​ = +0.771 V
     • Anode = Cd(s) → Cd²⁺(aq) = -0.403 V
     • E° = Cathode + Anode --> E° = 0.771 - (-0.40) --> E° = 1.174 V

Q20.4.15

S20.4.15

1. OH^-(aq) + 3H₂PO₂⁻(aq) + 2SbO₂⁻(aq) → 3HPO₃²⁻(aq) + 2Sb(s) + 2H₂O(l)
   • Phase II: Correct
   • Basic = Make sure to turn H⁺ molecules into H₂O by adding OH^- for every H⁺ molecule
2. 2H⁺(aq) + 2HNO₂(aq) + 2I⁻(aq) → NO(g) + I₂(s) + 2H₂O(l)
   • Phase II: Incorrect
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)
   • Equation = 2H+(aq) + 2HNO2(aq) + 2I-(aq) --> I2(s) + 2NO(g) + 2H2O(l)
3. 2OH^-(aq) + N₂O(g) + 2ClO⁻(aq) → 2Cl⁻(aq) + 2NO₂⁻(aq) + H₂O(l)
   • Phase II: Correct
   • Basic = Make sure to turn H⁺ molecules into H₂O by adding OH^- for every H⁺ molecule
4. 6OH^-(aq) + 3Br₂(l) → 5Br⁻(aq) + BrO₃⁻(aq)+ 3H₂O(l)
   • Phase II: Correct
   • Basic = Make sure to turn H⁺ molecules into H₂O by adding OH^- for every H⁺ molecule
5. 3Cl(CH₂)₂OH(aq) + 2K₂Cr₂O₇(aq) + 16H⁺(aq) → 3ClCH₂CO₂H(aq) + 4Cr³⁺(aq) + 4K⁺(aq) + 11H₂O(l)
   • Phase II: Correct
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)

Phase II: The rest of the redox reactions above are properly balanced in their respective solutions (except for the incorrect one). I know because the reactions are balanced. Also, the reactions in basic solutions don't have H+ in their final equations and the reactions in acidic solutions have H+ in their final equations. To supplement this section, I included a link to help figure out how to balance redox reactions.

• Balancing Redox Reactions:
  • https://chem.libretexts.org/Core/Ana...edox_reactions

Q20.4.16

S20.4.16

1. I⁻(aq) + 2HClO₂(aq) → IO₃⁻(aq) + Cl₂(g) + H₂O(l)
   • Phase II: Correct
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)
2. 4H⁺(aq) + 4Cr²⁺(aq) + O₂(g) → 4Cr³⁺(aq) + 2H₂O(l)
   • Phase II: Correct
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)
3. 2OH^-(aq) + 2CrO₂⁻(aq) + 3ClO⁻(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) +H₂O(l)
   • Phase II: Correct
   • Basic = Make sure to turn H⁺ molecules into H₂O by adding OH^- for every H⁺ molecule
4. S(s) + 2HNO₂(aq) → H₂SO₃(aq) + N₂O(g)
   • Phase II: Correct
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)
5. 3F(CH₂)₂OH(aq) + 2K₂Cr₂O₇(aq) + 16H⁺(aq) → 3FCH₂CO₂H(aq) + 4Cr³⁺(aq) + 4K⁺(aq) + 11H₂O(l)
   • Phase II: Correct
   • Acidic = Can leave H⁺ molecules by themselves (Don't need to add OH^-)

Phase II: The redox reactions above are balanced in their respective solutions (basic/acidic). Just like before in 20.4.15, the reactions are properly balanced because they are balanced. Plus, there are no H⁺ in the final equations of redox reactions in basic solutions, but in acidic solutions, there are H⁺. To supplement this section, I included a link to help figure out how to balance redox reactions.

• Balancing Redox Reactions:
  • https://chem.libretexts.org/Core/Ana...edox_reactions