Extra Credit 11
- Page ID
- 83404
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Edited by Emily Shane in bold
Q19.3B
Write cell reactions for the electrochemical cells diagrammed here, and use data from Table P2 to calculate \(\mathrm{E^\circ_{cell}}\) for each reaction.
- \(\mathrm{Al(s)| Al^{3+}|| Zn^{2+}(aq)| Zn(s)}\)
- \(\mathrm{Pt (s)| Fe^{2+}(aq), Fe^{3+}(aq)|| Cu^{2+}(aq)| Cu(s)}\)
- \(\mathrm{Pt (s)| Cr_2O_7, Cr^{3+}(aq)|| Ag^+(aq)| Ag(s)}\)
- \(\mathrm{O_2^-(aq)| O_2(g)|| H^+(aq)|H_2(g)| C(s)}\)
S19.3B
To calculate Ecell for each reaction, determine at which half reaction oxidation and reduction occurs. Then, subtract the tabulated Eanode, where reduction occurs, from the tabulated Ecathode, where oxidation occurs, to find the Ecell for the given reaction.
- Half reaction:
Reduction: \(\mathrm{Al(s) + 3e^- \rightarrow Al^{3+} \quad E=-2.310\,V}\) We know reduction occurs at Al(s) because electrons are gained.
Oxidation: \(\mathrm{Zn^{2+} \rightarrow Zn(s) + 2e^- \quad E=-0.763\,V}\) We know oxidation occurs at Zn(s) because electrons are lost.
\(\begin{align}
\mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
& = -2.310- -0.763\,\mathrm V \\
& = 1.447\,\mathrm{V}
\end{align}\)
- Reduction: \(\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+} \quad E=0.771\,V}\) We know reduction occurs at Fe because electrons are gained.
Oxidation: \(\mathrm{Cu(s)\rightarrow Cu^{2+} + 2e^- \quad E= 0.520\,V}\) We know oxidation occurs at Cu(s) because electrons are lost.
\(\begin{align}
\mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
& = 0.520- 0.771\,\mathrm V \\
& = 0.149\,\mathrm{V}
\end{align}\)
- Reduction: \(\mathrm{Cr^{6+} + 3e^- \rightarrow Cr^{3+} \quad E=1.33\,V}\) We know reduction occurs at Cr because electrons are gained.
Oxidation: \(\mathrm{Ag(s) \rightarrow Ag^+ + e^- \quad E=0.800\,V}\) We know oxidation occurs at Ag(s) because electrons are lost.
\(\begin{align}
\mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
& = \mathrm{0.800\,V-1.33\,V} \\
& = -0.53\,\mathrm{V}
\end{align}\)
- Reduction: \(\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O(l) \quad E = 1.229\,V}\) We know reduction occurs at O2 because electrons are gained.
Oxidation: \(\mathrm{H_2 \rightarrow 2H^+ + e^- \quad E=0\,V}\) We know oxidation occurs at H2 because electrons are lost.
\(\begin{align}
\mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
& = 0- 1.229\,\mathrm V \\
& = -1.229\,\mathrm{V}
\end{align}\)
Q19.37B
- \(\mathrm{Al(s) | Al^{3+}(aq) (0.20\:M) || Sn^{2+}(aq)(0.90\:M)| Sn(s)}\)
- \(\mathrm{Zn(s) | Zn^{2+}(aq) (0.42\:M) || Cl^- (0.02\:M), Cl_2 (g,\: 0.5\: atm) | Pt(s) }\)
S19.37B
Calculate \(\mathrm{E^\circ_{cell}}\) using the tabulated values, subtracting the \(\mathrm{E^\circ_{cell}}\) of the anode or oxidation equation from the \(\mathrm{E^\circ_{cell}}\) of the cathode or reduction equation. Then use Nernst equation, \(\mathrm{E_{cell} = E^\circ_{cell} - \dfrac{0.0592}{n} \log \{Q}\) to calculate \(\mathrm{E_{cell}}\). Make sure that units for all values match others and concel to give you the proper answer units.
- Oxidation: \(\mathrm{Al(s) \rightarrow Al^{3+}(aq) (0.20\:M) + 3e^- \quad E^\circ_{cell} = -1.676\:V}\)
- Oxidation: \(\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) (0.42\:M) + 2e^- \hspace{33 pt} \quad E^\circ_{cell} = -0.763\:V}\)
Q20.27B
Create a standard electrode potential diagram for the reduction of Fe3+ to Fe2+ in acidic solution.
S20.27B
\(\mathrm{|| Fe^{3+}(aq), Fe^{2+}(aq)| Pt(s)}\)
In the standard electrode potential diagram for this reduction, the salt bridge allows electrons from the acidic solution to the iron metal ion solution. This electron gain is what causes Fe3+ to be reduced to Fe2+. Electrons flow from anode to cathode.
Q24.1C
For the reaction 3A+2B→C+2D reactant A is found to disappear at the rate of 4.6 X 10-5 Ms-1
- What is the rate of reaction?
- What is the rate of disappearance of the reactant B?
- What is the rate of appearance for product D?
S24.1C
- \(\mathrm{-\dfrac{1}{3} \left(\dfrac{- \Delta[A]}{\Delta t}\right) = \dfrac{1}{3} (4.6 \times 10^{-5}\, Ms^{-1}) = 1.5 \times 10^{-5}\, Ms^{-1}}\)
- A is a reactant, so its rate of reaction will be negative and it has a coefficient of 3 which translates to (1/3) for the reaction rate. So in order to find the overall rate of reaction, you multiply A's disappearance rate by (-1/3).
- Rate disappearance of \(\mathrm{B = (rate\: of\: reaction) \times 2(coefficient\: of\: B)}\)
\(\mathrm{(1.5 \times 10^{-5}\,Ms^{-1})(2) = 3.1 \times 10^{-5}\, Ms^{-1}}\)- In order to find the rate of disppearance of B, you simply multiply the rate of reaction by the coefficient of B, which in this case is 2.
- Same process as B) (multiply rate of reaction by 2) OR:
Rate of appearance of \(\mathrm{D = -\dfrac{1}{3}(2)\left(\dfrac{\Delta[A]}{\Delta t}\right) = -\dfrac{2}{3}(4.6 \times 10^{-5}\, Ms^{-1}) = 3.1 \times 10^{-5}\, Ms^{-1}}\)
For reaction rates the rates of appearances and rates of disappearances are equal, however, reactants have negative equations (since their quantity is decreasing) and products have positive equations since their quantity is increasing.
Q24.45B
Answer the following:
- What two factors does the rate of a reaction depend on other than the frequency of collisions?
- Why does the rate of reaction increase dramatically with temperature?
- What is the net effect of the addition of a catalyst?
S24.45B
-
- 1) Whether or not the occurring collisions have enough energy to get over the activation energy and become products.
- 2) Whether or not the molecules involved in the collisions are properly orientated for the reaction to occur. A combination of these is needed for a specific reaction to occur at the proper speed.
- Collision frequency does not increase dramatically with temperature. However, the percentage of molecules with enough energy to get over the activation energy and become products increases. With higher energy molecules (due to the increased temperature) reactions are more likely to be successful.
- The addition of a catalyst would have the net effect of decreasing the energy barrier for the reactants to become products. The catalyst does so by enabling an alternative mechanism with a lower activation energy. Due to this lowered activation energy, reactions can occur more quickly and successfully.
Q25.21C
The rate of disintegration for a sample with \(\ce{^{210}_{85}At}\) is 9871 dis day-1. The half-life of \(\ce{^{210}_{85}At}\) is 4.3 years. Estimate the mass of \(\ce{^{210}_{85}At}\) in the sample.
S25.21C
\(\mathrm{\lambda = \dfrac{0.693}{4.3\,y}\times\dfrac{1\,y}{365.25\,d}=4.41\times10^{-4}\,d^{-1}}\)
\(\mathrm{mass = \dfrac{9871\,atoms/day}{4.41\times10^{-4}\,d^{-1}}\times\dfrac{1\,mole}{6.022\times10^{23}\,atoms}\times\dfrac{210\,grams}{1\,mole}=7.81\times10^{-15}\,g\,present}\)
Since the amount of disintegrating atoms is proportional to the current number of atoms, conversions are needed to plug into the corresponding formula. First, you must convert the half-life, given in years, to days-1. Then, find the number of atoms using your converted inverse half-life and the rate of disintegration. With the number of atoms, use Avogadro's number, 6.002x1023 atoms per mole, to convert to moles, and finally use the atomic mass, 210 grams per mole, to find the mass of the sample.
Q19.5
Write balanced nuclear equations for:
- positron emission by Sr-83
- the fusion of two C-12 nuclei to give another nucleus and a neutron.
- the fission of U-235 to give Ba-140, another nucleus and an excess of two neutrons.
S19.5
In a nuclear reaction, the nucleus of an atom is bombarded with energized particles. These reactions lead to fusion, fission, or radioactive decay. This leads to a different identity of the nucleus, which in this case is found algebraically using the atomic and mass numbers.
1. \(\ce{^{83}_{38}Sr}\) → \(\ce{^{83}_{37}Rb}\) + \(\ce{^{0}_{+1}β}\)
- The given nuclear structure, in this case Sr-83, is on the reactant side of the equation. Since a positron is being emitted, \(\ce{^{0}_{+1}β}\) it would be on the product side. Then, you simply find the missing nucleus to balance the equation. The mass number to begin is 83, and 83-0=83. This means the mass number of the unknown is 83. The number of protons is originally 38, and 38-(+1)=37. This leaves us with Rb-83.
2. 2\(\ce{^{12}_{6}C}\) → \(\ce{^{23}_{12}Mg}\) + \(\ce{^{1}_{0}n}\)
- To begin, we know there are two C-12 nuclei on the reactant side and the fusion of the two result in a neutron and an unknown nucleus. To find this unknown nucleus, we simply balance the mass number and number of protons for the overall equation. For the mass number, (12+12)-1=23 and for the number of protons, (6+6)-0=12. We then know that the resulting nucleus is Mg-23.
3. \(\ce{^{235}_{92}U}\) → \(\ce{^{140}_{56}Ba}\) + \(\ce{^{93}_{36}Kr}\) + 2\(\ce{^{1}_{0}n}\)
- When U-235 undergoes fission, the product needs to have an overall mass number of 235 and an overall number of protons, 92. We know that in the product, there is Ba-140, two neutons and an unknown nucleus. This unknown can be found by balancing the equation. 235-(140+1+1)=93 gives the mass number and 92-(56+0+0)=36 gives the number of protons. This means that the unknown nucleus is Kr-93.
Q21.3.4
During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.
S21.3.4
A stars energy comes from nuclear fusion processes, which is multiple nuclei joining together to form a heavier nucleus accompanied by the release or absorption of energy. For stars like the sun, which have internal temperatures less than fifteen million Kelvin, the dominant fusion process is proton-proton fusion (the fusion of four protons into an alpha particle). Stars are powered by the conversion of hydrogen into helium through nuclear fusion and paired with the spontaneous release of positrons, gamma rays, and energy. For more massive stars which can achieve higher temperatures, the carbon cycle fusion becomes the dominant mechanism. For older stars which are collapsing at the center, the temperature can exceed one hundred million Kelvin and initiate the helium fusion process called the triple-alpha process. The stars are formed through the aggregation of interstellar dust
The fusion of helium and hydrogen nuclei in stars is provides that energy and synthesizes new nuclei as a byproduct of that fusion process. Stars are formed by the aggregation of interstellar dust. In the beginning of its life, it is powered by a series of nuclear fusion reactions in which hydrogen is converted to helium. This overall reaction is paired with the release of two positrons, two gamma rays, and a large amount of energy. These reactions are responsible for its release of solar heat.