Extra Credit 28
- Page ID
- 82888
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For the standard cell potentials given here, determine the ΔG° for the cell in kJ.
- 0.000 V, n = 2
- +0.434 V, n = 2
- −2.439 V, n = 1
S17.4.1
GIVEN:
Standard Cell Potentials: 0.000V, +0.434V, -2.439V
Number of Electrons: n=2, n=2, n=1
UNKNOWN:
ΔG°: ?
Equation:
ΔG° = -nFE°cell
F = 96 485 C mol−1
a. Determine the equation to use
ΔG° = -nFE°cell
b. Plug in known values to each corresponding equation
(Make sure to divide the last step by 1000 kJ in order to convert from Joules to Kilojoules)
1. ΔG° = -(2 e-)(96 485 C mol-1)(0.000V)
ΔG° = 0.000 kJ K‑1 mol-1
2. ΔG° = -(2 e-)(96 485 C mol-1)(0.434V)
ΔG° = -83749 J K‑1 mol-1
( -83749 J K‑1 mol-1)/1000 kJ = -83.7 kJ K‑1 mol-1
3. ΔG° = -(1 e-)(96 485 C mol-1)(-2.439V)
ΔG° = 235327 J K‑1 mol-1
( 235327 J K‑1 mol-1)/1000 kJ = 235.3 kJ K‑1 mol-1
Q12.1.1
What is the difference between average rate, initial rate, and instantaneous rate?
S12.1.1
Instantaneous rate is defined to be the rate of a reaction that is changing at any point in time. The times of these changes can be short, therefore the products and reactants of the reaction change by only a small amount.
Average rate is more of an approximation of the instantaneous rate within a given interval. As the name suggests, it is the average rate over a specific period of time.
Initial rate is the instantaneous rate of a reaction when the reaction starts from the beginning, more specifically, when the products start to form.
Example:
A⟶B
- We see here that A is our reactant and B is our product
- Δ[B]/Δt would the instantaneous rate of change in a graph of Δt in the x-axis and ΔB in the y axis.
- If we wanted to see the instantaneous rate of the change of B between 30-60 seconds, this would be known as average rate
- If we wanted to see what the instantaneous rate of B would be from the initial time that the reaction began, then this would be initial 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:
| Initial [C3H5N3O9] (M) | 4.88 | 3.52 | 2.29 | 1.81 | 5.33 | 4.05 | 2.95 | 1.72 |
|---|---|---|---|---|---|---|---|---|
| t (s) | 300 | 300 | 300 | 300 | 180 | 180 | 180 | 180 |
| % Decomposed | 52.0 | 52.9 | 53.2 | 53.9 | 34.6 | 35.9 | 36.0 | 35.4 |
S12.4.19
a.
GIVEN:
initial time: 300 seconds
initial concentration: 4.88 M
% decomposed: 52.0
UNKNOWN:
average rate that it decomposes
1st set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 4.88M
Solve for At --> (100 - percent decomposed)% of initial concentration = (100-52)%=48% of 4.88M=.48*4.88=2.34M (same for rest of the sets of data)
[At] = (100-52)% of 4.88M
=48% of 4.88M
=2.34M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(4.88M/2.34M)= k*300s
0.735= k*300s
k=2.45 × 10−3s−1
2nd set of data
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 3.52M
[At] = (100-52.9)% of 3.52M
=47.1% of 3.52M
=1.65792M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(3.52M/1.65792M)= k*300s
0.752= k*300s
k=2.51 × 10−3s−1
3rd set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 2.29M
[At] = (100-53.2)% of 2.29M
=46.8% of 2.29M
=1.072M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(2.29M/1.072M)= k*300s
0.759= k*300s
k=2.53 × 10−3s−1
4th set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 1.81M
[At] = (100-53.9)% of 1.81M
=46.1% of 1.81M
=0.834M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(1.81M/0.834M)= k*300s
0.77= k*300s
k=2.58 × 10−3s−1
5th set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 5.33M
[At] = (100-34.6)% of 5.33M
=65.4% of 5.33M
=3.48582M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(5.33M/3.48582M)= k*180s
0.42= k*180s
k=2.35 × 10−3s−1
6th set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 4.05M
[At] = (100-35.9)% of 4.05M
=64.1% of 4.05M
=2.596M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(4.05M/2.596M)= k*180s
0.445= k*180s
k=2.47 × 10−3s−1
7th set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 2.95M
[At] = (100-36)% of 2.95M
=64.0% of 2.95M
=1.888M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(2.95M/1.888M)= k*180s
0.446= k*180s
k=2.48 × 10−3s−1
8th set of data:
Equation to use:
ln(A initial/At)=kt
A=concentration of the reactant
k=reaction rate
t=time
a) find At
[A initial] = 1.72M
[At] = (100-35.4)% of 1.72M
=64.6% of 1.72M
=1.111M
b) substitute in values to equation to solve for k
ln(A initial/At)=kt
ln(1.72M/1.111M)= k*180s
0.437= k*180s
k=2.43 × 10−3s−1
Q21.3.3
Complete each of the following equations by adding the missing species:
-
1327Al+24He⟶?+01n
(typo) ( i tried to fix the end products) -
94239Pu+?⟶ 24296Cm+ 10n
-
714N+ 24He⟶ ?+ 11H
- 92235U⟶ ?+55135Cs+401n
S21.3.1
The main idea for nuclear reaction problems is to make sure that everything in the equation in balanced. This is targeted mainly for the atomic mass and Z (the atomic number). The student has to make sure that both sides of a reaction add up to the same numbers for both.
For problem 1, we have:
1327Al+24He⟶?+01n
From the notation, we know that the atomic mass of an element is the top number while the atomic number is the bottom one. It is crucial that we do not mix these up because it could create complications.
On the left side, if we add up the atomic masses of He and Al, we get 4+27= 31 amu. If we add up the atomic numbers for both we get 13+2=15. With this information, we have to make a judgment on the right side about with element to fill in the questions mark. We are given a neutron that has no change on atomic number, but an added number to the atomic mass. Thus, we would have to find an element that has an atomic number of 15 so that it is balanced on both sides.
The element that best represents this would be P because it has an atomic mass of 15. For the atomic mass, we have to make sure that it still adds up to 31 like on the left side, so we will give the number 30 for it because 30+1=1 which is the same as the left side.
So the end product would end up looking like this:
1327Al+24He⟶3015P+01n
1327Al+24He⟶3015P+01n
For problem 2, we can follow the same type of idea:
94239Pu+?⟶ 24296Cm+ 10n
On the right hand side, we know that we have an atomic mass of 242+1=243 amu and we are dealing with an atomic number that adds up to 96+0=96. On the left hand side, we are given the element Pu that has an atomic mass of 239 and an atomic number of 94. We need to make sure that both sides are balanced, so we must subtract the atomic masses so that we can see what element we need. 243-239=4. The same is applied for the atomic number. 96-94=2.
We need to find an element that has an atomic number of 2 and an atomic mass of 4. The answer is He.
94239Pu+24He⟶ 24296Cm+ 10n
94239Pu+24He2⟶ 24296Cm+ 10n
Problem 3 is nothing different from what we have already done:
714N+ 24He⟶ ?+ 11H
On the left side we are given an atomic mass of 14+4=18 amu and an added atomic number of 7+2=9. We have to make sure that this adds up to the same thing on the right hand side. Since H has an atomic mass of 1 and an atomic number of 1, we need to look for an element that has an atomic number of 8 so that it would add up to 9 like it does on the other side. Oxygen is the element to do this and its atomic mass would have to be 17 so that it would add up to 18 with the hydrogen.
As a result, you end up getting the answer:
714N+ 24He⟶ 817O+ 11H
714N+ 24He⟶ 817O+ 11H
Problem 4 should be a piece of cake now:
92235U⟶ ?+55135Cs+401n
Do not be fooled by the coefficient in front of the neutron, that only just adds 4 more amu to the right side of the reaction! On the left side, we have an atomic mass of 235 and an atomic number of 92. So, we have to make sure that the right side is added up to the same numbers. If we add up the amu from Cs and the neutron, we get 135+4=139 amu. The atomic number would be just 55.
We have to subtract the atomic number from the left to the right to find out which element we need. 92-55=37 means that we need the element Rb. The atomic mass would be 235-139= 96 amu.
92235U⟶ 3796Rb+55135Cs+401n
92235U⟶ 3796Rb+55135Cs+401n
Q21.7.5
Given specimens neon-24 (t1/2=3.38min) and bismuth-211 (t1/2=2.14) of equal mass, which one would have greater activity and why?
S21.7.5
From problem, we can tell that bismuth-211 would have the most activity because it has a faster half-life that neon-24. Half-life is defined to be the time it takes for something to fall to half of its original value, which usually applies for radioactive decay and isotopes. Bismuth-211 would have more activity since it reaches its half-life faster than it does for neon-24, which is over a minute longer than the other.
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 Ecell?
S20.4.18
SHE stands for the standard hydrogen electrode which has the potential of 0V. According to the list of potentials at SHE, the Fe2+ has the reduction potential of -0.440V, this means that it would be the one at the anode and the hydrogen at the cathode. We know this because the one that has the lowest voltage is deemed to be the one getting oxidized, therefore being the anode. It makes sense that this is the one to corrode because it get oxidized in the process. I begin to notice the hydrogen starts to plate out as a result of the corrosion and the hydrogen is being gassed off because it is not a solid. The standard potential of the cell would come out to:
0-(-0.440)= 0.440 V
Remember that you must subtract the anode voltage from the cathode voltage!
A SHE is a standard hydrogen electrode which has a potential of 0V. Fe2+ has a reduction potential of -0.440 which means it would be oxidized so it would be at the anode and hydrogen electrode is the cathode. In addition this makes sense because if Iron corrodes that means it gets oxidized. Ecell at standard condition would be reduction potential of the cathode-anode so it would be
0-(-0.440)= 0.440 V
Q20.9.3
Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?
S20.9.3
Molten salts are better for electrolysis because they prepare active metals to react more readily with water. Also, any metal that is does not readily react with water to make hydrogen could be helped with molten salts that are within aqueous solutions. There is also the chance for competition for different electrolytic reactions.
Q20.9.1
Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?
S20.9.1
Overvoltage is defined to be the additional driving force and a voltage that is 0.4-0.6V higher that the calculated needed value. It may be needed in order to overcome additional barriers such as large activation energy during the formation of a gas. Overvoltage is needed in all electrolytic processes, so many reactions needs more energy and thus are favored thermodynamically.