Extra Credit 28

Q17.4.1

For the standard cell potentials given here, determine the ΔG° for the cell in kJ.

1. 0.000 V, n = 2
2. +0.434 V, n = 2
3. −2.439 V, n = 1

S17.4.1
This Question provides 'n' and standard cell potentials as givens, while asking for ΔG° in kj.
ΔG° can be related to 'n' and standard cell potentials (E°cell) with the equation:

\[ΔG°=−nFE°cell\]

'n' refers to the amount of moles transfered by the reaction.

'F' refers to Faraday's Constant, which is equal to 96,500 Joules/(Volt)(mol \(e^-\)) (C/Mol)

'E°cell' refers to standard cell potential, which is measured in volts. (1 Volt= 1 Joule /Coulomb)

Since we have all of our variables given to us, we can apply them to find ΔG°.

1) \[\Delta G°= - (2 \ mol \ of \ e^-* 96,500 \ \frac{C}{\ mol \ of \ e^-} * 0 \ V)\]

Since the standard potential is equal to 0, the ΔG°=0 KJ

2) \[\Delta G°= - (2 \ mol \ of \ e^-* 96,500 \ \frac{C}{\ mol \ of \ e^-} * 0.434 \ V)\]

After the units cancel out, ΔG°=-83762 J
Since the questions asks for KJ, we must convert the answer from J to KJ.

\[-83762 \ J * \frac{1 \ KJ}{1000 J}= -83.762 \ KJ\]

Therefore, ΔG°=-83.8 KJ

3)

\[\Delta G°= - (1 \ mol \ of \ e^- * 96,500 \ \frac{C}{\ mol \ of \ e^- } * -2.439 \ V)\]

After the units cancel, ΔG°=235363.5

Since the question asks for KJ, we must convert the answer from J to KJ

\[235263.5 \ J *\frac{1 \ KJ}{1000 J}= 235.3 \ KJ\]

Therefore, ΔG°=-235.3 KJ (typo: ΔG° is positive not negative. ΔG° = 235.3 KJ)

Q12.1.1

What is the difference between average rate, initial rate, and instantaneous rate?

S12.1.1

Instantaneous rate is the rate of the reaction at a specific time (Example: x=3 mins) that is so small that it is essentially instantaneous. It can be found by finding the slope of the tangent line at the corresponding point on the graph at that time, which is known as the derivative. This can be represented as

\[\lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t}\]

The Initial Rate is the instantaneous rate at the beginning of the reaction. It is essentially the starting derivative when time=0. The initial rate can also be described as the instantaneous rate before the initial concentrations of reactants have changed significantly.

The Average Rate is the average of instantaneous rates over a period of time. As the average rate, it is just an approximate value for a given area and is not as accurate as the instantaneous rate.

Q12.4.19

Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:

S12.4.19

When we are given this problem, we can instantly note that this is a first-order decomposition problem. We are asked to determine the average rate constant (K)for each of the experiments. By examining the given information, we have the amount of time (t) it took for each reaction and the percent of decomposed material. The formula that relates our given to what we are tasked to find is the First-order integrated rate law: \[ln\frac{A}{A_o}=-kt\]

With some manipulation, we can solve for k.

\[-(\frac{ln\frac{A}{A_o}}{t})=k\]

'A' refers to the concentration at a given time
'Ao' refers to the initial concentration.

'k' refers to the rate constant

't' refers to time

We can use the percent decomposition to find out our 'A' and 'Ao'. Since the initial amount (Ao) is 100% of the sample present, we can use the percent decomposed to find our concentration at the present time (A). For example, if 30% of the substance decomposed by time x, there would be 70% of the substance at time x. This would make the ratio of A to Ao 70/100.

Since we know what our variables are, we can plug them in and solve for k. The initial concentrations are not needed in the equation since we are given the percent decomposed. It is easier to treat 'A' and 'Ao' as percents rather than exact concentrations since it takes a little bit of extra work to use exact concentrations. However, treating 'A' and 'Ao' as percents or exact concentrations will yield the correct answer either way.

When [Ao] is 4.88, t(s)= 300 sec, and percent decomposed is 52%
\[k=-(\frac{ln\frac{48}{100}}{300 sec})= 0.00245 \frac{1}{sec} \]
When [Ao] is 3.52, t(s)= 300 sec, and percent decomposed is 52.9%
\[k=-(\frac{ln\frac{47.1}{100}}{300 sec})= 0.00251\frac{1}{sec} \]
When [Ao] is 2.29, t(s)= 300 sec, and percent decomposed is 53.2%
\[k=-(\frac{ln\frac{46.8}{100}}{300 sec})= 0.00253 \frac{1}{sec} \]
When [Ao] is 1.81, t(s)= 300 sec, and percent decomposed is 53.9%
\[k=-(\frac{ln\frac{46.1}{100}}{300 sec})= 0.00258 \frac{1}{sec} \]
When [Ao] is 5.33, t(s)= 180 sec, and percent decomposed is 34.6%
\[k=-(\frac{ln\frac{65.4}{100}}{180 sec})= 0.00236 \frac{1}{sec} \]
When [Ao] is 4.05, t(s)= 180 sec, and percent decomposed is 35.9%
\[k=-(\frac{ln\frac{64.1}{100}}{180 sec})= 0.00244 \frac{1}{sec} \]
When [Ao] is 2.95, t(s)= 180 sec, and percent decomposed is 36%
\[k=-(\frac{ln\frac{64}{100}}{180 sec})= 0.00247 \frac{1}{sec} \]
When [Ao] is 1.72, t(s)= 180 sec, and percent decomposed is 35.4%
\[k=-(\frac{ln\frac{64.6}{100}}{180 sec})= 0.00243 \frac{1}{sec} \]

Q21.3.3

Complete each of the following equations by adding the missing species:

a) \[_{13}^{27}Al+_{2}^{4}He \rightarrow ? + _{0}^{1} \]

b) \[_{94}^{239}Pu+ \ ? \rightarrow _{96}^{242}Cm + _{0}^{1}n\]

c) \[_{7}^{14}N+_{2}^{4}He \rightarrow ? + _{1}^{1} H\]

d) \[_{92}^{235}U \rightarrow ?+ \ _{55}^{135}Cs + 4_{0}^{1}n\]

S21.3.3

Nuclear accounting is a pertinent aspect of nuclear chemistry. The Mass numbers and the proton numbers must equal on both sides. Since we know this, we can deduce via algebra what the missing species is. The number in the superscript (above) refers to the mass number, while the number subscript (below) refers to the proton number. X will stand in for the missing Proton number, while Y will refer to the Mass number. The proton number will determine what element that the missing species, as protons determine what an element is.

a) \[_{13}^{27}Al+_{2}^{4}He \rightarrow ? + _{0}^{1} \]

From the proton numbers, we can set up an equation, \(13+2=X+0\). From this, we can deduce that the proton number is 15 if we solve for X. The element this corresponds to is Phosphorus.

For the mass numbers, we can set up the equation \(27+4=Y+1\). We can see from this that the mass number is 30.

Combined together, we can deduce that the missing species is \(_{15}^{30}P\).

b) \[_{94}^{239}Pu+ \ ? \rightarrow _{96}^{242}Cm + _{0}^{1}n\]

From the proton numbers, we can see that we should use the equation \(94+X=+96+0\). If we solve for X, we can determine that the proton number is 2. The element that this corresponds to is Helium.

For the mass numbers, we can set up the equation, \(239+Y=242+1\). If we solve this, we can see that the mass number Y for the missing species is 4.

Combined together, we can deduce that the missing species is \(_{2}^{4}He\).

c) \[_{7}^{14}N+_{2}^{4}He \rightarrow ? + _{1}^{1} H\]

From the proton numbers, we can set up an equation, \(7+2=X+1\). If we solve for X, we can deduce that the proton number is 8. The element this corresponds to is Oxygen.

For the mass numbers, we can set up the equation \(14+4=Y+1\). We can see from this that the mass number Y for the missing species is 17 .

Combined together, we can determine that the missing species is \(_{8}^{17}O\).

d) \[_{92}^{235}U \rightarrow ?+ \ _{55}^{135}Cs + 4_{0}^{1}n\]

From the proton numbers, we can set up an equation, \(92=X+55\). If we solve for X, we can deduce that the proton number is 37. The element that this corresponds to is Rubidium.

For the mass numbers, we can set up the equation \(235=Y+135+4\). It's important that we note that there are 4 neutrons. This means that it adds a total of 4 to the total. We can see from this that the mass number Y for the missing species is 96 .

Combined together, we can determine that the missing species is \(_{37}^{135}Rb\). (typo: mass number of missing species is 96 instead of 135 \(_{37}^{96}Rb\)

Q21.7.5

Given specimens neon-24 (t1/2=3.38min) and bismuth-211 (t1/2=2.14min) of equal mass, which one would have greater activity and why?

S21.7.5

Activity(A) is defined as the decrease in the number of the radioisotope's nuclei over time. Activity is synonymous with the 'rate of decay', which is a first-order system. It can be calculated with the formula: \[A=kN\]

Where

A= Activity
k= Decay Constant

N= # of atoms

In this question, we are given the half lives of both specimens. Since it is a first order process, we can use the half life formula for first order reactions: \[t_{\frac{1}{2}}= \frac{ln2}{k}= \frac{0.693}{k}\]

to calculate the decay constant, k.

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

Since we are given the half-lives, we can plug them in to find k


k for Neon-24: \[k=\frac{0.693}{3.38 min}=0.205\frac{1}{min}\]

k for Bismuth-211:\[k=\frac{0.693}{2.14 min}=0.324\frac{1}{min}\]

Now that we have k, we must calculate for N, which is the amount of atoms. Let's assume that we have 10 grams of each substance, since the masses are equivalent.

N for Neon-24: \[10 \ g \ Ne * \frac{1 \ mol}{24 \ g} * \frac{6.02*10^{23} \ atoms}{1 \ mol}= 2.508*10^{23} \ atoms\]

N for Bismuth-211: \[10 \ g \ Bi * \frac{1 \ mol \ }{211 \ g \ } * \frac{6.02*10^{23} \ atoms}{1 \ mol}=2.853*10^{22} \ atoms\]

Now that we have both k and N, we can calculate the Activity (A)

A for Neon-24: \[A=kN=0.205 \frac{1}{min}\ *(2.508*10^{23} atoms)= 5.14*10^{22} \frac{atoms}{min}\]

A for Bismuth-211:\[A=kN=0.324 \frac{1}{min}\ *(2.853*10^{22} atoms)= 9.244*10^{21} \frac{atoms}{min}\]

As we can see, the Activity for Neon-24 is higher than that of Bismuth-211. This is because activity is directly related to the amount of atoms. Since Neon-24 has much more atoms than Bismuth-211 due to the smaller molar mass, we can see that Neon-24 has much more activity than Bismuth-211. The more atoms present in a given amount of a substance the higher the activity of the particular specimen.

Q20.4.18

You have built a galvanic cell using an iron nail, a solution of FeCl₂, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is E_cell?

S20.4.18

The following is image is a visual representation of the galvanic cell that the question is describing.

Which is represented by this notation: \(Fe(s)|Fe^{2+}(aq)||2H^+(aq)|H_2(g)| Pt(s)\)

From the question, we can gleam that the iron nail is being corroded. Corrosion is defined as the 'unwanted' oxidation of a metal, thus affirming that the Iron nail is undergoing oxidized and that this process is a redox reaction. oxidation means that the Iron nail is losing electrons.

The half reaction for this situation is \(Fe(s) \rightarrow Fe^{2+}(aq)+ 2e^-\).

If something is being oxidized, then something else must be reduced. The Hydrogen, in this case, acts as the oxidizing agent and gets reduced.

This is shown by the half reaction \(2H^+(aq)+2e^- \rightarrow H_2(g) \).

In addition to the iron's corrosion, one should be able to observe the evolution of \(H_2\) gas.

Since we are under standard conditions, \(E°cell=E°(cathode)-E°(anode)\) can be used to find the E°cell. To find the E°(cathode) and E°(anode), we need to refer to a chart in order to determine their Standard Reduction Potentials (SRP).

The SRP of Hydrogen is equal to 0 Volts. The reduction of hydrogen is zero volts because it is used as a reference to measure the standard reduction potentials of other substances.
The SRP of Iron is -0.44 Volts.

Since Hydrogen is reduced, it is the cathode and Iron the anode.

Using the previously mentioned formula for E°cell:\[E°cell=0-(-0.045)\]

This results in E°cell=0.045. This positive number means that this reaction is spontaneous.

Q20.9.1

Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?

S20.9.1

An overvoltage is the value that occurs when a reaction needs more voltage than calculated in order to begin. It is the difference between the observed experimental potential value and the calculated value. This is because some reactions that are thermodynamically favored may require a high activation energy. Thus, an overvoltage may be required in order to start the reaction. (The following may or may not be similar to the previous explanation. Its simply how I would answer the question. In a thermodynamically favored electrochemical reaction that is taking place in a galvanic cell, electrons are flowing through a wire from the anode to the cathode producing a current. The frictional heating that occurs between the current and wire wastes some of the energy of the cell. Therefore, an increased voltage or overvoltage is required to ensure that the driving force behind the flow of electrons is sufficient enough for the reaction to occur.)

An analogy to this could be driving to a location. If you input a destination into a GPS, it will give you a time that tells you how far away you are from your destination. However, the actual amount of time that it would take to get to said location could be a bit longer than what the GPS calculated due to events like traffic and having to get gas. The 'overvoltage' in this case would be the difference between the calculated GPS time (the calculated voltage required) and the time it actually took (experimentally observed voltage).

Q20.9.3

Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?

S20.9.3

A mixture of molten salts is generally used, rather than a pure salt, due to the fact that mixtures of molten salts lowers the melting point of the entire solution. More specifically, the melting point of the pure salt is so high that practical electrolysis is not an option. This means other methods of electrolysis are needed and these methods are often very costly.

For example, the Downs Cell is a cell which is used to produce sodium metal. In it, molten \(NaCl\) undergoes electrolysis. However, the melting point of \(NaCl \)is 801 °C. When mixed with another salt, namely \(CaCl_2\) (which has a melting point of 772°C), the combined melting point is lowered to about 600°C.

The lowering of the temperature is integral to the industrialization of electrolysis, which is why the mixture of molten salts is much more generally used than a pure salt. This reduction means that less energy is needed to preform the electrolysis, which means less money spent on production.