Extra Credit 17

Q17.2.6

From the information provided, use cell notation to describe the following systems:

1. In one half-cell, a solution of Pt(NO₃)₂ forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO₃)₂ solution with all solute concentrations 1 M.
2. The cathode consists of a gold electrode in a 0.55 M Au(NO₃)₃ solution and the anode is a magnesium electrode in 0.75 M Mg(NO₃)₂ solution.
3. One half-cell consists of a silver electrode in a 1 M AgNO₃ solution, and in the other half-cell, a copper electrode in 1 M Cu(NO₃)₂ is oxidized.

Solutions:

Cell Diagram Set-up: X_{1 (phase)} | X_{2 (phase)} || Y_{1 (phase)} | Y_{2 (phase)}

Anode Cathode

1. i) Write the redox reactions for Pt and Cu metal.

Pt²⁺ (aq) +2e^- → Pt (s) and Cu (s) → Cu²⁺ (aq) +2e^-

ii) Identify which half-cell is anode and cathode.

Anode (oxidation): Cu (s) → Cu²⁺ (aq) + 2e^-

Cathode (reduction): Pt²⁺ (aq) +2e^- → Pt (s)

iii) Write out the cell notation.

Cu (s) | Cu²⁺ (aq) (1M) || Pt²⁺ (aq) (1M) | Pt (s)

2. i) Write out the redox reactions for Au and Mg metal.

Anode (oxidation): Mg (s) → Mg²⁺ (aq) +2e^-

Cathode (reduction): Au³⁺ (aq) + 3e^- → Au (s) and

ii) Write out the cell notation with the given electrodes.

Mg | Mg (s) | Mg²⁺ (aq) (.75 M) || Au³⁺ (aq) (.55 M) | Au (s) | Au

3. i) Write out the redox reactions for Au and Mg metal.

Ag⁺ (aq) + e^- → Ag (s) and Cu (s) → Cu²⁺ (aq) +2e^-

ii) Identify which half-cell is anode and cathode.

Anode (oxidation): Cu (s) → Cu²⁺ (aq) +2e^-

Cathode (reduction): Ag⁺ (aq) + e^- → Ag (s)

iii) Write out the cell notation with the given electrodes.

Cu | Cu²⁺ (aq) (1M) || Ag⁺ (aq) (1M) | Ag (s) | Ag

Q19.1.15

The standard reduction potential for the reaction [Co(H₂O)₆]³⁺ (aq) + e^- → [Co(H₂O)₆]²⁺ is about 1.8 V. The reduction potential for the reaction [Co(NH₃)₆]³⁺ (aq) + e^- → [Co(NH₃)₆]²⁺ is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H₂O)₆]²⁺ and/or [Co(NH₃)₆]²⁺, can be oxidized to the corresponding cobalt(III) complex by oxygen.

Given:

[Co(H₂O)₆]³⁺ (aq) + e^- → [Co(H₂O)₆]²⁺ E^o_cell = 1.8 V

[Co(NH₃)₆]³⁺ (aq) + e^- → [Co(NH₃)₆]²⁺ E_cell = 0.1 V

Solution:

i) Compare the standard reduction potentials of oxygen and [Co(H₂O)₆]²⁺.

- Since the standard reduction potential of oxygen is smaller than water (1.229 V < 1.8 V), oxygen is unable to oxidize the cobalt(III) complex because the reduction potential of the oxygen must be higher than that the other compound.

ii) Compare the standard reduction potential of oxygen and [Co(NH₃)₆]²⁺.

- Since the standard reduction potential of oxygen is greater than ammonium (1.229 V > 0.1V), the [Co(NH₃)₆]²⁺ is able to be reduced by oxygen.

Q19.3.7

Explain how the diphosphate ion, [O₃P−O−PO₃]⁴⁻, can function as a water softener that prevents the precipitation of Fe²⁺ as an insoluble iron salt.

Solution:

- In order to prove that [O₃P−O−PO₃]⁴⁻ is a sacrificial anode, preventing Fe²⁺ from precipitating, we must compare the standard reduction potential of the two ions.

- Since the standard reduction potential of [O₃P−O−PO₃]⁴⁻ is smaller than the value of Fe²⁺ (-.50 V < -0.44 V), disphosphate ion is more likely to react and be oxidized before Fe²⁺ precipitates as an insoluble iron salt. Therefore, diphosphate ion, [O₃P−O−PO₃]⁴⁻functions as a water softener in this situation.

Q12.4.7

What is the half-life for the first-order decay of carbon-14? (C-14⟶N-14+e^-) The rate constant for the decay is 1.21 × 10⁻⁴ year⁻¹.

Solution:

i) Identify the first-order reaction equation

ln[A]_o/[A]=kt

ln[A]_o/([A]_o/2)= ln2 = kt_1/2

t_1/2= 0.693/k

k = 0.693/t_1/2

ii) Note: rate constant for the decay = k = 1.21 x 10^-4 year ^-1

Plug in the decay constant into the half-life reaction equation to find half-life of Carbon-14

t_1/2 = 0.693/(1.21 x 10^-4 year ^-1)

= 5727.3 years or 5.73 x 10³ years

Q21.2.2

Write the following isotopes in nuclide notation (e.g., " C614C614 ")

1. oxygen-14
2. copper-70
3. tantalum-175
4. francium-217

Solutions:

Note: The nuclide notation is where A is the mass number and Z is the atomic number of an isotope.

1. i) Identify the atomic number and mass number of O from the periodic table.

Z = 8 A = 14

ii) Plug the given values into the formula of nuclide notation.

\[^{14}_{8}O\]

2. i) Identify the atomic number and mass number of Cu from the periodic table.

Z = 29 A = 70

ii) Plug the given values into the formula of nuclide notation.

\[^{70}_{29}Cu\]

3. i) Identify the atomic number and mass number of Ta from the periodic table.

Z = 73 A = 175

ii) Plug the given values into the formula of nuclide notation.

\[^{175}_{73}Ta\]

4. i) Identify the atomic number and mass number of Fr from the periodic table.

Z = 87 A = 217

ii) Plug the given values into the formula of nuclide notation.

\[^{217}_{87}Fr\]

Q21.5.5

Describe the components of a nuclear reactor.

- Fuel - nuclear reactor uses uranium-235 or uranium-238 as fuel to produce heat and then electricity during the process of fission (a nuclear reaction or radioactive decay process where nuclei splits into smaller parts)

- Reactor core - contains small pellets of uranium where uranium isotopes split and release heat during the chain reaction; the process depends on the moderator.

- Moderator - a substance, usually water or graphite, that slows down the neutrons that are being produced during fission of uranium nuclei so the nuclei can continue fission and produce more energy.

- Coolant - a substance (water, liquid sodium, helium, or others) that passes through the core, carries the heat produced from the reactor, and is transferred to the turbine.

- Turbine - a rotating device with blades that converts the heat or energy from the coolant and converts it into work or electricity

- Control rods - keeps the reactor running at a constant rate, increasing or decreasing the rate of fission.

Q20.4.5

Write a cell diagram representing a cell that contains the Ni/Ni²⁺ couple in one compartment and the SHE in the other compartment. What are the values of E°_cathode, E°_anode, and E°_cell?

Solution:

i) Identify the standard reduction potential of Ni/Ni²⁺

Ni ⇌ Ni²⁺ + 2e⁻ E° = -0.257 V

ii) Identify the standard reduction potential of SHE

2H⁺ + 2e^- ⇌ H₂ (g) E° = 0.0 V

iii) Identify which redox reaction is anode and cathode

Anode (: Ni ⇌ Ni²⁺ + 2e⁻ E° = -0.257 V because Ni is losing electrons, which increases its oxidation state

Cathode: 2H⁺ + 2e^- ⇌ H₂ (g) E^o = 0.0 V because H is gaining electrons, which reduces its oxidation state

iv) Find E^o_cell

E°_cell = E°_{cathode -} E°_anode

E°_cell = (-0.257 V) - 0 V = -0.257 V

v) Write the cell diagram representing the half-reaction cells.

Ni | Ni²⁺ (aq) (1M) || H⁺ (aq) (1M) | H₂ (g)

Q20.7.1

What advantage is there to using an alkaline battery rather than a Leclanché dry cell?

- Both alkaline battery and Lechanché dry cell are primary batteries that consists of a zinc anode and a magnesium oxide cathode. The center of the dry cell is a carbon electrode which has a electrolyte paste of ammonium chloride, graphite, zinc chloride, and water. Inside of the electrode, zinc and ammonium react together that causes polarization and leads to decreased amount of current due to the build up of ammonium molecules around the cathode. A Leclanché dry cell is inefficient in its production of electrical energy, has the tendency to leak more, performs poorly at low temperatures, and also has a low shelf life (spontaneous reactions between Zn and NH₄Cl in the cell cause corrosion). On the other hand, alkaline batteries have a higher energy density that allows the battery to last longer while more consistently providing the same amount of energy. Also, alkaline batteries are more stable at various temperatures and are thus able to function well at different temperatures. But although alkaline batteries are more expensive to reproduce, they are ultimately more cost efficient because of their longevity and performance.