Extra Credit 17

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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.

Q17.2.6

a.\(Pt(NO_3)_2\rightarrow\) \(Pt+2NO_3^{-}\) : \(Pt^{2+}+2e^{-}\rightarrow\) \(Pt\) (oxidation state from 2+ to 0)

reduction reaction: Cathode

\(Cu+2NO_3^{-}\rightarrow\) \(Cu(NO_3)_2\) : \(Cu\rightarrow\) \(Cu^{2+}+2e^{-}\) (oxidation state from 0 to 2+):

oxidation reaction: Anode

Cu(s)|Cu²⁺(aq)||Pt²⁺(aq)|Pt(s)

b. Cathode: \(Au(NO_3)_3\rightarrow\) \(Au+3NO_3^{-}\): reduction: \(Au^{3+}+3e^{-}\rightarrow\) \(Au\) (oxidation state from 3+ to 0)

Anode: \(Mg+3NO_3^{-}\rightarrow\) \(Mg(NO_3)_3\): oxidation: \(Mg\rightarrow\) \(Mg^{3+}+3e^{-}\) (oxidation state from 0 to 3+)

Mg(s)|Mg³⁺(aq, 0.75M)||Au³⁺(aq, 0.55M)|Au(s)

C. \(AgNO_3\rightarrow\) \(Ag+NO_3^{-}\): the standard reduction potential of \(Ag^{+}+e^{-}\rightarrow\) \(Ag\) is 0.8V

\(Cu(NO_3)_2+2e^{-}\rightarrow\) \(Cu+2NO_3^{-}\): standard reduction potential of \(Cu^{2+}+2e^{-}\rightarrow\) \(Cu\) is 0.34V

The E°_cell value is found in the table of standard reduction potentials. The more negative the E°_cell value, the better the reducing agent, meaning the reaction will be more likely to be oxidized. Since silver has an E°_cell value of 0.8 V and copper has an E°_cell value of 0.34 V, we can see that copper is more negative. Therefore, copper will be the reducing agent and will be the anode. Silver has a more positive E°_cell value, meaning it is more likely to accept electrons, and will be the cathode.

Thus \(Ag^{+}+e^{-}\rightarrow\) \(Ag\) is the cathode, \(Cu^{2+}+2e^{-}\rightarrow\) \(Cu\) is the anode

Cu(s)|Cu²⁺(aq, 1M)||Ag⁺(aq, 1M)|Ag(s)

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Q19.1.15

The standard reduction potential for the reaction [Co(H₂O)₆]³⁺_(aq)+e⁻⟶[Co(H₂O)₆]²⁺_(aq)[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)[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.

Q19.1.15

given: the E° of [Co(H₂O)₆]³⁺(aq)+e⁻⟶[Co(H₂O)₆]²⁺(aq) is 1.8V

the E° of [Co(NH₃)₆]³⁺(aq)+e⁻⟶[Co(NH₃)₆]²⁺(aq) is 0.1V

asked: the E° to oxidize [Co(H₂O)₆]²⁺ and/or [Co(NH₃)₆]²⁺ to the corresponding cobalt(III) complex by oxygen

solution: to oxidize [Co(H₂O)₆]²⁺ and/or [Co(NH₃)₆]²⁺ by oxygen, oxygen is needed in the reaction.

\(O_2+2e^{-}+4H^{+}\rightarrow\) \(2H_2O\) E°=1.2V

\(E^{\circ}=E^{\circ}cathode-E^{\circ}anode\)

reduction reactions happen at cathodes, thus E° of [Co(H₂O)₆]³⁺(aq)+e⁻⟶[Co(H₂O)₆]²⁺(aq) is 1.8V-1.2V=0.6V

E° of [Co(NH₃)₆]³⁺(aq)+e⁻⟶[Co(NH₃)₆]²⁺(aq) is 0.1-1.2V=-1.1V

A reaction is spontaneous when the E° is positive and when the ΔG is negative.

In this equation, the reaction [Co(NH₃)₆]³⁺(aq)+e⁻⟶[Co(NH₃)₆]²⁺(aq) is non-spontaneous because \[E^{\circ}< 0\]. The reaction [Co(H₂O)₆]³⁺(aq)+e⁻⟶[Co(H₂O)₆]²⁺(aq) is spontaneous because \(E^{\circ}>0\)

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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.

Q19.3.7

The diphosphate ion binds very strongly to Fe²⁺ metal because it has multiple sites to bind with. It also has many oxygen molecules to bind to, making the a polar reaction between diphosphate and oxygen.

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Q12.4.7

What is the half-life for the first-order decay of carbon-14? \(^{14}_{6}C\rightarrow^{71}_{4}N+e^{-}\)

The rate constant for the decay is 1.21 × 10−4 year⁻¹.

Q12.4.7

given: rate constant = \(1.21x10^{-4}\) year^-1

solution:

Since the rate constant is given in years^-1, the reaction is a first order. The equation of first order half-life is

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

\(t_{1/2}\)=\((\frac{0.693}{1.21*10^-4})\) \(year^{-1}\)

thus \[t_{1/2}=5727.27273years\]

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Q21.2.2

Write the following isotopes in nuclide notation (e.g., " \(^{14}_{6}C\) ")

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

Q21.2.2

the formula used is \[_{Z}^{A}\textrm{M}\]

a. known: Z=8 for oxygen

given: A=14

\[\ce{_8^{14}O}\]

b.known: Z=29 for CU

given: A=70

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

c. known: Z=73 for Ta

given : A=175

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

d. known Z=87 for Fr

given A=217

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

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Q21.5.5

Describe the components of a nuclear reactor.

Q21.5.5

A nuclear reactor operates through the controlled fission of \(^{235}U\) at a slow, steady state.

the nuclear reactor include 5 parts

1. nuclear fuel: fissionable isotope in sufficient quantity to start the reaction. The continuous operation of the nuclear rector depends on a large enough fission to produce enough neutrons for the next fission in the chain reaction.
2. moderator: substance slow down neutrons produced in nuclear reaction (the neutrons are moving very fast, so they very hard to be absorbed by the nuclear fuel to cause any other additional reactions). It contains light nuclei that don't absorb the neutrons, but break apart some of the energy when they collide with neutrons.
3. coolant: carry heat to external boiler where it can be transferred to electricity
4. control system: Control fission rate by adjusting the number of slow neutrons through absorption. Control rods that readily absorb neutrons to help keep the reactor running at a constant rate.
5. shield and containment system: protect operator from radiation and high pressure resulting from high temperature produced by the nuclear reaction.

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Q20.4.3

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?

Q20.4.3

Potential energy difference causes the electrons flow through the circuit from the anode to the cathode. The anode is oxidized through the loss of the electrons and the cathode is reduced through the gain of electrons. Thus, as A has a higher potential energy, electrons will flow from substance A to substance B. Reactions tend to run in the direction from high potential energy to low potential energy, causing the reaction to react in a spontaneous manner.

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Q20.7.1

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

Q20.7.1

Alkaline batteries are a type of dry cell batteries that react in a basic environment. They do not have NH₄⁺ and therefore, have a more constant output of voltage as the cell is discharged. This battery employs the chemical reaction of zinc and magnesium oxide and use an alkaline electrolyte of potassium hydroxide to generate electric current. The higher energy density of the Alkaline battery allows for the same voltage to be produced, longer life and better performance. The content of Leclanché dry cell react in acidic environments. The Leclanché dry cell has an anode of Zn that is easily oxidized, meaning that it corrodes easier compared to the Alkaline battery.