Extra Credit 12

Q19.3C

For the electrochemical cells shown below, write the cell reactions and find E^∘_Cell using the Standard Reduction Potential table.

1. Cu(s)|Cu²⁺(aq)||Ag⁺(aq)|Ag(s)
2. Cu(s)|Cu²⁺(aq)||Fe²⁺(aq),Fe³⁺(aq)|Pt(s)
3. PbO₂(s)|Pb²⁺(aq)||Mn²⁺(aq)|MnO₂(s)
4. Cu(s)|Cu²⁺(aq)||O₂(g)|H₂O₂(aq)|Pt(s)

S19.3C

1. When writing the cell notation, the anode (oxidation equation) is always written first, then followed by the cathode (reduction equation). As a result, the following equation can be derived.

2Ag⁺(aq) + Cu (s) → 2Ag (s) + Cu²⁺ (aq)

Cu(s) is the anode, meaning the oxidized element, since its oxidation number increases from 0 to +2, whereas Ag⁺(aq) has a oxidation number which decreases from +1 to 0, making it the cathode in the cell.

The reduction potential of Ag⁺(s) = cathode = 0.8

The oxidation potential of Cu(s) = anode = -0.34

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 0.8 + (-0.34) = 0.46V

2. By looking at the cell notation, this problem can easily be solved just like the first one, the only difference in the cell in the inert metal, Pt(s) at the cathode. This inert metal only exists to provide a metal for the cell reaction to be complete since both elements in the cathode are aqueous solutions. As a result, it does not affect the chemical reaction equation in any way. Therefore the equation can be written as:

2Fe³⁺(aq) + Cu(s) → 2Fe²⁺(aq) + Cu²⁺(aq)

The anode of this reaction is Cu(s) since its oxidation number increases from 0 to +2, while Fe³⁺(aq) decreases from +3 to +2.

The reduction potential of 2Fe³⁺(aq) = cathode = 0.77

The oxidation potential of Cu(s) = anode = -0.34

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 0.77 + (-0.34) = 0.43V

3. Just like the first two, in order to solve for the cell potential, the anode and cathode must first be identified.

PbO₂(s) + Mn²⁺(aq) → Pb²⁺(aq) + MnO₂(s)

Based on the reaction equation, the anode is identified as Mn²⁺(aq) because its oxidation number increases from +2 to +4. The cathode is identified as PbO₂(s), since the oxidation number decreases from +4 to +2.

The reduction potential of PbO₂(s) = cathode = 1.455

The oxidation potential of Mn²⁺(aq) = anode = -1.229

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 1.455 + (-1.229) = 0.43V

4. This equation mimics that of the second question, the tricky part is the O₂(g) and H₂O₂(aq) when calculating the oxidation number.

Cu(s) + O₂(g) + 2H₃O(aq) → H₂O₂(aq) + Cu²⁺(aq)

In this reaction, the anode is Cu(s) since its oxidation number increases from 0 to +2. The cathode, on the other hand, is O₂(g) because its oxidation number goes from 0 to -1.

The reduction potential of O₂(g) = cathode = 0.695

The oxidation potential of Cu(s) = anode = -0.34

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 0.695 + (-0.34) = 0.355V

Q19.6A

Given the following voltaic cell diagram and using the standard gold electrode as the reduction half reaction, and metal MM as an oxidizing half reaction, find the E^∘ reduction half reaction:

Au³⁺(1M) + 3e⁻ → Au(s) E^∘ = 1.52

M³⁺(1M) + 3e⁻ E^∘ = ?

1. In E^∘ = 1.858V
2. Al E^∘ = 3.20V
3. La E^∘ = 3.90V
4. U E^∘ = 3.18V

S19.6A

1. First, in order to calculate the E^∘ with the given values, it is important to identify the equation and plug in the known values to solve for E^∘.

E^∘_Cell = E^∘_Cathode - E^∘_Anode

For the first question, we are given that the overall cell diagram has a E^∘_Cell of 1.858V, and that the reduction potential of Au³⁺(aq) is 1.52. So we can simply plug the values in and solve for E^∘_Anode.

E^∘_Cell = E^∘_Cathode - E^∘_Anode

_1.858 = 1.52 - E^∘_Anode

E^∘_Anode = 1.52 - 1.858

E^∘_Anode = -0.338V

2. In this question, the overall cell diagram is depicted as Al with a E^∘_Cell value of 3.2V. To solve for E^∘_Anode, simply plug in the values similarly to the previous question.

E^∘_Cell = E^∘_Cathode - E^∘_Anode

3.2 = 1.52 - E^∘_Anode

E^∘_Anode = 1.52 - 3.2

E^∘_Anode = -1.68V

3. Just like the previous question, determine the overall cell potential through the question and plug the individual values into the equation to solve for E^∘_Anode. Based on the question provided, the overall cell diagram is La with a cell potential of 3.9V

E^∘_Cell = E^∘_Cathode - E^∘_Anode

3.9 = 1.52 - E^∘_Anode

E^∘_Anode = 1.52 - 3.9

E^∘_Anode = -2.38V

4. The information given for question 4 suggests that the overall cell diagram is 3.18V. Plug this into the E^∘_Cell equation, along with the reduction potential of the gold cathode to solve for the E^∘_Anode value.

E^∘_Cell = E^∘_Cathode - E^∘_Anode

3.18 = 1.52 - E^∘_Anode

E^∘_Anode = 1.52 - 3.18

E^∘_Anode = -1.66V

Q19.6B

In each of the following examples, write the cell diagram and cathode. Indicate the direction of the electron flow; write a balanced equation.

1. Al(s) + Cu²⁺ → Cu(s) + Al³⁺
2. Fe³⁺ + Ag(s) → Ag⁺ + Fe²⁺
3. Cu(s) + 2Fe³⁺ → Cu²⁺ + 2Fe²⁺

S19.6B

1. A cell diagram includes multiple aspects, however, to accurately write a cell notation, the reduction (cathode) and oxidation (anode) reactions must be identified.

Reduction: Al³⁺(aq) + 3e^- → Al(s) E^∘_Cathode = -1.676V

Oxidation: Cu(s) → Cu²⁺(aq) + 2e^- E^∘_Anode = -0.340V

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = -1.676 + (-0.340) = -2.016

A cell diagram consists of the elements and lines which indicate the change of state or the separation of electrolytes, or usually referred to as a salt bridge. In most cases, the anode is usually written first, then followed by the cathode. Following these guidlines, the cell diagram for the above reaction can be written as:

Cu(s)|Cu²⁺(aq)||Al³⁺(aq)|Al(s)

In cells, the electron flow is always regarded as from the anode towards the cathode direction. As a result, the electron flow of this cell is from the copper electrode to the aluminum electrode.

2. In order to write a cell diagram for a cell, the reduction and oxidation reactions should be stated.

Reduction: Fe³⁺(aq) + e^- → Fe²⁺(s) E = 0.771V

Oxidation: Ag(s) → Ag⁺(aq) + e^- E = -0.800V

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 0.771+ (-0.800) = -0.029V

Through this, the cathode and anode can easily be identified. As a result, the cell diagram can be expressed in this way:

Ag(s)|Ag⁺(aq)||Fe³⁺(aq)|Fe²⁺(s)

As electrons move from the anode to the cathode, they flow from the silver electrode to the iron one.

3. Initially, the reduction and oxidation reactions can help determine the order in which the cell diagram is written, based on the oxidation numbers of the parents.

Reduction: Fe³⁺(aq) + e^- → Fe²⁺(s) E = 0.771V

Oxidation: Cu(s) → Cu²⁺(aq) + 2e^- E^∘_Anode = -0.340V

E^∘_Cell = E^∘_Cathode - E^∘_Anode

E^∘_Cell = 0.771+ (-0.340) = 0.431V

Based on the redox reactions, Fe³⁺(aq) undergoes reduction, categorizing it as the cathode of the reaction. This signifies that the anode of the reaction is the copper electrode. However, an important aspect to pay attention to is that in cases where the reaction is made up entirely of aqueous solutions, an inert electrode may be placed in the cell to facilitate the reaction process. As a result, the cell diagram for the given reaction is as follows:

Cu(s)|Cu²⁺(aq)||Fe³⁺(aq)|Fe²⁺(s)

Lastly, by reading the cell diagram, the electron flow can be determined to go from the copper electrode to the Fe²⁺ electrode.

Q19.37C

Using the Nernst equation and a list of E^∘_Cell values, calculate the E_Cell for the following cells:

1. Al(s)|Al3+(0.36M)||Sn4+(0.086M),Sn2+(0.54M)|Pt
2. Sn(s)|Sn2+(0.01M)||Pb2+(0.700M)|Pb(s)

S19.37C

1. Based on the cell notation given, the following equation can be derived.

Al(s) + Sn⁴⁺(aq) → Al³⁺(aq) + Sn²⁺(aq)

The anode from the equation is Al(s) since the oxidation goes from 0 to +3, whereas Sn⁴⁺(aq) is the cathode because the oxidation is reduced from +4 to +2.

The reduction potential of Sn⁴⁺(aq) = cathode = 0.154

The oxidation potential of Al(s) = anode = 1.676

E^∘_Cell = 0.154 + 1.676 = 1.83V

With the calculated E^∘_Cell value, we can plug in the value to the Nernst equation to calculate E_Cell. The variable n in the equation represents the number of moles transferred in the reaction. In this case, the number of moles is 2. Lastly, to calculate the value of Q, the formula is just like the equilibrium constant, K. Simply plug in the concentrations of the products over the reactants.

E_Cell = E^∘_Cell - (0.0592/n) x logQ

Q = ((0.54)(0.36)²)/(0.086) = 0.8138

E_Cell = 1.83 - (0.0592/(2)) x log(0.8138)

E_Cell = 1.833V

2. Firstly, to determine the anode and cathode, the reaction equation can be written.

Pb²⁺(aq) + Sn(s) → Sn²⁺(aq) + Pb(s)

The anode can be identified as Sn(s) since the oxidation number increases from 0 to +2, while the cathode is Pb²⁺(aq) since it reduces from +2 to 0.

The reduction potential of Pb²⁺(aq) = cathode = -0.125

The oxidation potential of Sn(s) = anode = 0.137

E^∘_Cell = -0.125 + 0.137 = 0.012V

Through plugging the E^∘_Cell value in the Nernst equation, we can determine the E_Cell value. In this reaction, the number of moles transferred is 2, while the value of Q is calculated based on the concentrations of the products over reactants.

Q = (0.01)/(0.700) = 0.0143

E_Cell = E^∘_Cell - (0.0592/n) x logQ

E_Cell = 0.012 - (0.0592/(2)) x log(0.0143)

E_Cell = 0.067V

Q20.57A

Predict the molecular formulas for the following metal carbonyls and explain how many electrons are contributed by the metal and how many are contributed by the carbon monoxide and which noble gas electron configuration does the molecule exhibit.

1. Titanium (Ti)
2. Manganese (Mn)
3. Tungsten (W)

S20.57A

1. In the question, it asks for the noble gas configuration the metal carbonyls exhibit. As a result, first we can determine the electron configuration of the nearest noble gas. By looking at the periodic table, the nearest noble gas to Titanium is [Kr]. It can be derived from the periodic table that the electron configuration of [Kr] consists of 36 electrons. On the other hand, the transition metal in the question, Titanium [Ti], has 22 electrons. The subtraction between the two suggests the idea of how many more electrons are needed to achieve the configuration of [Kr] for every one of transition metal.

Electrons needed = 36 - 22 = 14

The question refers to specifically metal carbonyls, simply meaning after disregarding the metal, how many electrons are contributed by the carbon monoxide, indicating the molecular formula of the metal carbonyls. Since every carbon monoxide complex ion donates 2 electrons, the remaining electrons have to be divided by two, this will indicate how many carbon monoxides will yield the correct number of electrons.

Number of CO in metal carbonyls = 14/2 = 7

Based on the calculations, the molecular formula can be seen as:

Ti(CO)₇

2. Manganese, an element on in the d block of transition metals has 25 electrons. The closest noble gas on the periodic table is [Kr] with 36 electrons.

Electrons needed = 36 - 25 = 11

Then, to divide the remaining electrons by 2 for the determination of the number of carbon monoxides in the molecular formula.

Number of CO in metal carbonyls = 11/2 = 5.5

It can be concluded that for every one manganese, there will be 5 metal carbonyls. This leaves the final molecular formula to be:

Mn(CO)₅

3. Based on the periodic table, tungsten has 74 electrons while the nearest noble gas is [Rn], with an electronic configuration of 86 electrons.

Electrons needed = 86 - 74 = 12

By dividing this amount by 2, we can determine how many carbon monoxide you need for one mole of tungsten.

Number of CO in metal carbonyls = 12/2 = 6

Concluding that:

W(CO)₆

Q21.3.5

Write a balanced equation for each of the following nuclear reactions:

1. the production of ¹⁷O from ¹⁴N by α particle bombardment
2. the production of ¹⁴C from ¹⁴N by neutron bombardment
3. the production of ²³³Th from ²³²Th by neutron bombardment
4. the production of ²³⁹U from ²³⁸U by ²₁H bombardment

S21.3.5

1. An "α particle" can be defined in a chemical reaction as the addition of ⁴₂He or ⁴₂α. The most important thing to check in writing a balanced equation is that the protons and neurons add up to the same amount on both sides of the equation.

¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H

2. A "neutron" bombardment can be represented as ¹₀n. In other words, the question asks for the balanced equation in which a ¹₀n is added to ¹⁴N, forming ¹⁴C. As a result, to balance the equation, one more proton and neuron is needed on the right side to equal it out.

¹⁴₇N + ¹₀n → ¹⁴₆C + ¹₁H

3. A "neutron" bombardment can be represented as ¹₀n. Just like the previous equation, ¹₀n is added to the existing element to form the new atomic mass of the element. In this example, the reaction equation only consists of 3 components because the addition of the neutron bombardment to the existing ²³²Th would form the new ²³³Th.

²³²₉₀Th + ¹₀n → ²³³₉₀Th

4. A "²₁H bombardment" basically means the addition of ²₁H to form ²³⁹U, however, the tricky aspect is that the existing elements is ²³⁸U. Since the original has an atomic mass of 238, and ²₁H adds 2 to the atomic mass, whereas the final product is 239 of the same element, another product must be formed to balance the reaction equation by 1 proton and 1 atomic mass.

²³⁸₉₂U + ²₁H → ²³⁹₉₂U + ¹₁H

Q24.3A

In the reaction A → B, [A] is found to be 0.750M at t=61.2s and 0.704 M at t=73.5s. Determine the average rate of the reaction during this time interval.

S24.3A

First, to solve for the average rate, the formula used can be written as:

Average rate = -Δ[A]/Δt

Δ[A] = change in concentration

Δt = change in time

Plug in the values provided in the question to solve for the average rate.

Average rate = -Δ[A]/Δt

Average rate = -((0.704M - 0.75M)/(61.2s - 73.5s))

Average rate = 3.7 x 10^-3 M/s

Q24.45C

Explain why:

1. A reaction rate cannot be calculated from the solely collision frequency.

2. The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency rises a lot more slowly.

3. Introducing a catalyst to a reaction mixture can have such a significant impact on the rate of the reaction, even if the temperature is held constant.

S24.45C

1. Collision frequency is the average speed in which 2 molecules collide with each other. This determines the average number of collisions for a unit of time for the molecules. Whereas to calculate the reaction rate, the many components taken into account is the temperature, concentration, and activation energy required to remove the electrons. As a result, with just the value of the collision frequency, the reaction rate cannot be determined.

2. Temperature and the rate of a chemical reaction is directly related to each other, if the temperature is increased, the rate of a reaction will go up as well. This happens due to the rise in kinetic energies along with the temperature, causing the molecules to move faster, increasing the reaction of collision frequency between molecules. On the other hand, having a steady collision rate with an increased kinetic energy around it would lead to high and low rates, as depicted like a sine curve.

3. A catalyst is a solution that speeds up the rate of a reaction by lowering its activation energy. This increased rate is based solely on the addition of a catalyst to the system, thus no additional change is required in the reaction. This means that without a change in the enthalpy exerted on or released by the system, the added catalyst will still have the same effect on the reaction, increasing its rate.

Q25.21D

The disintegration rate of a sample with ²⁰⁵₈₃Bi as the only radioactive nuclide is 3000 dis/h. The half life of ²⁰⁵₈₃Bi is 15.31 days. Estimate the number of atoms in the sample.

A = λN λ = 0.693/(t_1/2)

S25.21D

In order to solve for N, we can first rearrange the given equation to find our unknown.

A = λN

N = A/λ

We are given activity rate (A), so we only have to calculate λ in order to plug in all the variables to estimate the number of atoms in the sample.

λ = 0.693/(t_1/2)

λ = 0.693/(15.31 days) x (1 day/24 hours)

λ = 0.00189 h^-1

Lastly, just plug it into the newly written equation and solve for N.

N = A/λ

N = 3000 dis h^-1/0.00189 h^-1

N = 1587301.59 atoms ²⁰⁵₈₃Bi in the sample