Extra Credit 17

Q17.2.6

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

a)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.
b)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.
c)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.

a) The half reaction of platinum is required to determine which species is being oxidized and which species is being reduced.

Pt²+(aq)→Pt(s)Pt²+(aq)→Pt(s)

Through balancing the charges of NO₃ we can determine the charge of Platinum because there are 2 NO₃ ions, Platinum needs +2 to balance it out. The next step is to balance the electrons to balance the overall equation. This is the balanced half reaction:

Pt²⁺+(aq)+2e^-→Pt(s)

To determine if Pt is getting reduced or oxidized we can observe its change in charge of +2 charge to a 0 charge so it's being reduced. The next half reaction is regarding copper.

Cu(s)→Cu²⁺(aq)

To determine the charge of copper, we have to balance out the electrons and the product side is 2+ and the reactant side is 0. The electrons cancel out with the platinum half reaction.

Cu(s)→Cu²⁺(aq)+2e^-

Cu(s)+Pt²⁺(aq)→Cu²⁺(aq)+Pt(s)

Cell notation is written from anode to cathode, with the electrodes at the edges and they are merely for the transfer of electrons. Double lines represent the salt bridge and a single line represents a change in phase.

The cell notation for this is:

Cu(s)|Cu²⁺(aq,1M)||Pt²⁺(aq,1M)|Pt(s)

b) The anode is oxidized and the cathode is reduced, usually the anode is on the left side. The anode in this reaction is pre-determined to be magnesium.

Mg(s)→Mg²⁺(aq)+2e^-

Two electrons are added to the product side since it's +2.

Au is the cathode, so it is being reduced.

Au³⁺(aq)+3e^-→Au(s)

Three electrons are added to the reactant side because the 3+ needs to be balanced out.

To balance out the overall reaction, the top reaction is multiplied by 3 and the bottom reaction is multiplied by 2 to have a least common denominator of 6. The overall reaction is:

3Mg(s)+2Au³⁺(aq)→2Au(s)+3Mg²⁺(aq)

The cell notation for this reaction is as follows:

Mg(s)|Mg²⁺(aq,0.75M)||Au³⁺(aq,0.55M)|Au(s)

c) Copper is oxidized and is the anode.

Cu(s)→Cu²⁺(aq)+2e^-

Two electrons are added to the product side to balance the 2+ copper.

For silver, it is the opposite and being reduced as the cathode.

Ag⁺(aq)+e^-→Ag(s)

One electron is added to the reactant side to balance the + from Ag.

To balance the overall equation, the silver equation has to be multiplied by 2 to balance the 2e^- from copper.

2Ag⁺(aq)+Cu(s)→Cu²⁺(aq)+2Ag(s)

The cell notation is:

Cu(s)|Cu²⁺(aq,1M)||Ag⁺(aq,1M)|Ag(s)

Q19.1.15

The standard reduction potential for the reaction [Co(H₂O)₆]³⁺(aq)+e^-→[Co(H₂O)₆]²⁺(aq) is about 1.8 V. The reduction potential for the reaction [Co(NH3)₆]³⁺(aq)+e^-→[Co(NH₃)₆]²⁺(aq) 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.

To calculate the cell potential, we need to first determine the potentials for each half reaction. This in turn lets us determine which is being oxidized and which is being reduced. To determine the cell potential, the following equation is given:

E°_cell= E°_cathode - E°_anode

[Co(H2O)₆]³⁺ will be oxidized so it is the anode

O₂(g)+4H⁺(aq)+4e^-→2H₂O +1.229 V

O₂ is being reduced, so it is the cathode.

1.229V - 1.8V= -.571 V This reaction is not spontaneous because ΔE is negative.

For [Co(NH₃)₆]³⁺ it is again being oxidized, meaning it’s the anode.

1.229-.1= 1.129 V

This reaction is spontaneous because ΔE is positive.

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.

Iron usually reacts with water molecules and forms iron oxides, rust. The oxide layer will eventually allow more metal oxides to form as well, so diphosphate ion acts as a water softener by reacting with iron and keeping the iron from forming iron oxides.

Q12.4.7

What is the half-life for the first-order decay of carbon-14?

The rate constant for the decay is 1.21×10^-4year^-1

The half-life equation for first order is

t_1/2=ln2/k

with k being the rate constant. The rate constant for carbon-14 was given as 1.21×10−4year−11.21×10−4year−1.

Plug it in the equation.

t1/2=ln2/(1.21×10−4year−1)t1/2=ln2/(1.21×10−4year−1)

and solve for t1/2t1/2.

When you calculate it, the half life for carbon-14 is 5728.5 years.

To find the half life, we need to determine which equation to use depending on order reaction. The question states that this reaction is a first order reaction, so we use the half life equation for the first order.

The equation for a first order reaction is:

t_1/2=ln2/k

k is the rate constant, and the rate constant for carbon-14 was given as 1.21 x 10^-4year^-1

t_1/2=ln2/(1.21 x 10^-4year^-1)

and the half life for carbon-14 is 5728.5 years.

Q22.2.2

Write the following isotopes in nuclide notation

1. a) oxygen-14
2. b) copper-70
3. c) tantalum-175
4. d) francium-217

1. Oxygen with a mass number of 14, and the atomic number is 8 so the notation is:

2. Copper with a mass number of 70, and the atomic number is 29 so the notation is:

3. Tantalum with a mass number of 175, and the atomic number is 73 so the notation is:

4. Francium with a mass number of 217, and the atomic number is 87 so the notation is:

Q22.5.5

Describe the components of a nuclear reactor.

A nuclear reactor is where heat is produced through nuclear fission chain reactions and generates electricity. The energy released from the fission of the atoms is used as heat in a gas or water form, which produces steam-- this in turn moves the turbines to make electricity and power cities etc.

The components of a nuclear reactor are:

• nuclear fuel: requires an unstable isotope in a substantial quantity to start the reaction.
• moderator: the moderator slows down the neutrons produced in the nuclear reaction.
• coolant: Carries heat to an external boiler where it is transformed to electricity.
• control system: It controls the fission rate and is placed between fuel rods to absorb neutrons and adjusts the number of slow neutrons and keeps the rate of the chain reaction at a safe level.
• A shield and containment system. Protects the operator from radiation and high pressure and protects any serious malfunctions within the reactor.

Q22.4.3

What is the relationship between electron flow and the potential energy of valence electrons? If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?

Electrons flow from anode to cathode and the differences in potential energy makes the electrons flow from the higher potential energy to the lower potential energy. This is from the potential in the anode to become oxidized and the potential for the cathode to become reduced. The electrons are from the anode, which has a higher potential to become oxidized to the cathode. And the electrons favor a lower potential energy because the lowest energy form is more preferable to be stable. So the electrons would move from substance A to substance B moving from a more potential energy to a less potential energy.

Q20.7.1

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

An alkaline battery has double the energy density of a Leclanché cell. Which means that it produces twice the amount of energy while lasting even longer than a Leclanche battery. It's also less likely to leak and cause corroding, meaning it has a longer shelf life. This is largely due to the protective casing of the Leclache cell battery being made of zinc, and it serves as the anode for the cell which can have holes over time that allows leakage. Leclache dry cell batteries do not function well in low temperatures, and alkaline batteries perform better and work well in cold temperatures.