Extra Credit 13
- Page ID
- 83520
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Q19.4A
Given the reaction:
\(\mathrm{2MnO_4^-(aq) + PbO(s) + 6H^+(aq) \rightarrow 2MnO_2 (s) + 5PbO_2 (s) + 3H_2O (l)}\)
\(\mathrm{E^\circ_{cell}: +1.23\,V}\)
What does \(\mathrm{E^\circ}\) \(\ce{PbO2(s)}\) equal?
First, break the equation into two half reactions:
2MnO4- (aq) → 2MnO2 and
PbO(s) → 5PbO2
Normally, we would have to balance these half reactions, but just from these two half reactions, we can tell that Mn is reduced, as its charge goes from +7 to +4. This means than Mn is the cathode.
Since we know Eºcell = Eºcathode - Eºanode, and we know the value of MnO4- reduction thanks to the standard potential table to be +1.55 V, we can plug in the values, which gives us +1.23 = 1.55- Eºanode. This results in Eºanode = +0.32V
Q19.41A
If [Cu2+] is maintained at 1.0 M, what is the minimum [Ag+] for which the reaction from 19.2, given below, is spontaneous in the forward direction?
Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)
First, we split this into two separate half-reactions.
2Ag+ + 2e- → 2Ag
Cu → Cu2+ + 2e-
Since Ag decreases from 1 to O, we know Ag is being reduced, making it the cathode. Given the table of standard potentials, we know the reduction half reaction for both Ag+ and Cu2+ at .800V and .340V respectively. Since we know that Ag is the cathode, we subtract using the equation Eºcell = Eºcathode - Eºanode. In this case, its 0.800-0.340 = Eºcell, which is 0.460V.
Since we are striving to maintain the molarity, the Ecell would be 0. Then, we would use the equation Ecell = E°cell - \(\frac{0.0592V}{n}\)lnQ. n is the number of electrons, and F is Faradays' constant. Q is equal to the molarity of reactants over the molarity of the products. Plugging in our known values, and assuming standard values, our equation is
0 = 0.46- \(\frac{8.3145*298}{2*96485}\)ln(1/x2). We can rearrange this to get
.46 = ((8.3145*298)/(2*96485))ln(1/x2).
Next, you can use natural log properties to make ln(1/x2) equal to ln1-lnx2. Since the ln1 is 0, the equation becomes
.46 = -((8.3145*298)/(2*96485))ln(x2)
Simplify the equation and divide to isolate the ln(x2). Then make both sides powers of e. For example, making a 2 on one side e2. This eliminates the ln from one side. Then, take the square root of both sides to turn the x2 into an x, and the resulting value will be what x equals.
.46/(-.0128399285) = ln(x2)
e(-35.82574471) = x2
(2.76106585*10-16)(1/2) = x
1.661645524*10-8 = x
Q20.57B
How many electrons would be in the metals carbonyl?
- Chromium
- Nickel
1. Since carbonyl is CO, and we know it to have no charge, we tag it on to the element itself. Since we know Cr is in group 6, we know it to have 6 electrons. Since we must fulfill the rule of 18 electrons, the total must be 18. CO has an overall 2 electrons, as C has 4 and O has a tendency to remove 2 electrons. 2*6= 12 electrons, plus 6 more from the Cr makes 18.
2. The same applies here. However, since Ni is in group 10, we know there are 10 electrons. Since we must satisfy the rule of 18, we need 8 more electrons. Since CO has a charge of 2, we need 4 more. As a result, the resulting compound will be Ni(CO)4.
Q24.3B
What is the average rate of reaction over a time interval for [A] if it is 0.455 M at t=80.25s and 0.474 M at t=82.4s?
Since we are trying to find the rate of the reaction, we need to find the amount that changes during the interval. Since the problem tells us our amount is increasing, we know that the rate is positive. This can be expressed as:
rate = d[A]/dt
We simply plug in the values to the equation, and solve
rate = (.474M - .455M)/(82.4s - 80.25s)
rate = 8.8372093 * 10(-3)
Q24.46A
If even a small spark is introduced into a mixture of H2(g) and O2(g), a highly exothermic explosive reaction occurs. Without the spark, the mixture remains unreacted indefinitely.
- Explain this difference in behavior.
- Why is the nature of the reaction independent of the size of the spark?
1. The spark was the cause of the differing behavior between the two systems.
2. The reaction would remain the same regardless of the size of the spark. The size of the spark would affect the rate of the reaction, but it wouldn't change the reaction itself.
Q25.21E
Estimate the number of Fe57 atoms are in the sample given that the half-life of Fe57 is 4.8 years and the disintegration rate for a sample containing only Fe57 as the only radioactive nuclide is 6650 dis/h
Disintegration rates are first order reactions, meaning the equation to find the half life is
t1/2 = .693/k. I couldn't find lambda, so I'm using k.
Since we know the half life of Fe57, we can solve for k. However, because we have dis/h, we need to convert years into hours. The equation is rearranged as such, which gives us:
k = .693/4.8 years * (1 year/ 365.25 days) * (1 day/ 24 hours) = 1.646988364 * 10(-5)
Now, since we have k, we can now find the number of atoms, with the following equation.
N = A/k. Plug in values to get N = ((6650 dis/h)/ (1.646988364 * 10(-5)))
The final answer should be N = 4.037672727 * 108 Fe57 atoms
Q19.7
Why is nuclear fission considered a “chain reaction”? What is “critical” about critical mass? Why does nuclear fission produce radioactive waste?
Nuclear fission can be initiated by firing a neutron into the atom in question. This atom would then be energized, and would break off into other elements and other neutrons, which could repeat the process with other atoms of the same initial element.
Critical mass is the the smallest amount of material needed to sustain the chain reaction to release all of the atomic energy.
Nuclear fission produces radioactive products because the resulting products "relax" from the their excited state by emitting gamma rays
Q21.3.6
If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation.