Extra Credit 7

Q19.1F

From the observations listed, estimate the value of E∘ for the imaginary half reaction M2++2e−→M(s), where M is an unknown metal to be determined under the following conditions:

1. The metal M reacts with HCl(aq).
2. The metal displaces Fe3+ but does not displace Sn4+.
3. The metal reacts with HNO3(aq).
4. The metal can displace K+(aq).

S19.1F

1. in order to estimate the value of E∘ for this question we must refer to Standard Reduction Potentials by Value. Only those metals whose standard reduction potentials are lower than that of hydrogen react with HCl. For this reaction to occur, the reduction potential of the reactant receiving the electrons must be lower than the reduction potential of the reactant. So the estimated E∘ has to be less than 0.

2. Referring back to the table, the E∘ has to be less than 0.771 (Fe3+) and bigger than 0.154 (Sn4+) for it to be able to displace Fe3+ but not Sn4+.

3. The E∘ value must be lower than the E∘ for NO3 which is equal to 0.956_.

4. Since metal M has to get oxidized in order rto displaceK+ ion into solution, the E∘ has to be lower than -2.924.

Q19.35A

A voltaic cell has an Ecell value of 1.536 V. What is the concentration of Ag+ in the cell?

Zn(s)|Zn2+(2.00M)||Ag+(xM)|Ag(s)

S19.35A

in order to answer this question we need to go through several steps.

1. Write the half-reactions for each process. and determine their standard potential reduction.

Zn(s)→Zn²⁺(aq)+2e^- -E⁰=0.763 V

2Ag⁺(aq)+2e^-→Ag(s) E⁰=0.800 V

2. Add the cell potentials to get the overall standard cell potential.

Zn(s)+2Ag⁺(aq)→Zn²⁺(aq)+Ag(s)

0.763+0.800=1.563 V overall standard cell potential

3. use Nernst equation to determine the concentration of [Ag⁺]

E=E⁰_cell−(0.0592/n)log([Zn²⁺]/[Ag⁺]²)
1.536=1.563−(0.0592/2)log(2/x²)
x²=0.2448
x=0.495=[Ag⁺]

Q20.21B

Write half-equation to represent each of the following in an acidic solution.

1. MnO₄^-(aq) as an oxidizing agent.
2. Cr²⁺(aq) as a reducing agent.

S20.21

1: Write the “skeleton” half-reactions

MnO₄⁻ → MnO₂

2: Balance half-reaction “atomically” balance O atoms by adding H2O • balance H atoms by adding H+

MnO₄⁻+4H⁺ → MnO₂+2H₂O

3: Balance the electric charges by adding electrons

MnO₄⁻+4H⁺+3e⁻→MnO₂+2H₂O

b. follow the same procedure as part a.

1: skeleton half reaction

Cr2⁺(aq)→ Cr³⁺(aq)

2: we do not need to add any H⁺ or H₂O.

3: balance e^-

Cr²⁺(aq)+e^-→ Cr³⁼⁺(aq)

Q21.27A

Which of the following complex ions would you expect to have the largest overall KfKf, and why? [Cr(CO)6]2+[Cr(CO)6]2+, [Cr(en)3]2+[Cr(en)3]2+, [Cr(NH3)6]3+[Cr(NH3)6]3+, [Cr(CO)2(NH3)2(en)][Cr(CO)2(NH3)2(en)]

S21.27A

Due to its chelating effect [Cr(en)3]2+[Cr(en)3]2+ would have the largest overall Kf.

Q24.36A

For the disproportionation of p-toluenesulfinic acid,

3ArSO2H→ArSO2SAr+ArSO3H+H2O

(where Ar=p-CH3C6H4), the following data were obtained:t=0min, [ArSO2H]=0.140M; 15min, 0⋅0965M; 30min, 0⋅0852M, 45min, 0⋅0740M; 60min, 0⋅0668M; 120min, 0⋅0493M; 180min, 0⋅0365M; 300min, 0⋅0196M

1. Show that this reaction is second order
2. What is the value of the rate constant, kk?
3. At what time would [ArSO2H]=0.0600M?
4. At what time would [ArSO2H]=0.0300M?
5. At what time would [ArSO2H]=0.0150M?

S24.36A

1. Make graphs of concentration vs. time (zeroth order), natural log of concentration vs. time (first order), and one over concentration vs. time (second order) Determine which graph results in a straight line. This graph reflects the order of the reaction. For this problem, the straight line should be in the 3rd graph, meaning the reaction is second order. with the following data.

2. the value of k can be determined by finding the slope; rise/run. In this problem we can use the two points (0,7.143) and (900,10.363).

k=(10.363-7.143)/(900-0)=3.57x10^-3 M^-2s^-1

3. using the equation 1/[A]=kt+1/[A]_o

1/0.06=(3.57x10^-3t)+1/0.14 and solving for t we get

t=2668 s/60 s =44.5 min

4. following the same procedure as question 3 we get t=122.3 min

5. following the same procedure as question 3 we get t=277.9 min

Q25.13C

Write a nuclear equation for the formation of 242-Curium from the bombardment of bismuth-209 with aluminum-27, then followed by an emission of five α-particles.

\[\ce{^{27}_{13}Al + ^{209}_{83}Bi \rightarrow ^{242}_{96}Cm + 6n}\]

\[\ce{^{242}_{96}Cm \rightarrow ^{222}_{86}Rn+5\alpha}\]

Q19.1

Balance each of the following nuclear equations and indicate the type of nuclear reaction (α-emission, β-emission, fission, fusion, or “other”).

1. \(\mathrm{\ce{^{239}_{94}Pu}+\ce{^1_0n}\rightarrow \ce{^{130}_{50}Sn}+{?}+3^1_0n}\)
2. \(\mathrm{{?} + \ce{^{6}_{3}Li} \rightarrow 2^4_2He}\)
3. \(\mathrm{\ce{^{210}_{84}Po}\rightarrow \ce{^4_2He}}+{?}\)
4. \(\mathrm{\ce{^{235}_{92}U}+{^1_0n}\rightarrow \ce{^{72}_{30}?}+{?}+4^1_0n}\)
5. \(\mathrm{\ce{^{125}_{53}I}\rightarrow \ce{^{125}_{53}I}+{?}}\)
6. \(\mathrm{\ce{^{238}_{92}U}\rightarrow {?}+\ce{^{234}_{?}Th}}\)
7. \(\mathrm{\ce{^{235}_{92}U}+{^1_0n}\rightarrow \ce{^{86}_{?}Br}+\ce{^{147}_{?}?}+{?^1_0n}}\)
8. \(\mathrm{\ce{^{234}_{90}Th}\rightarrow {?}+\ce{^{234}_{91}?}}\)

S19.1

to answer the following questions we need to make sure the total atomic number of reactants are equal to the total atomic number of products. Also the total atomic mass of the reactants and total atomic mass of products have to equal eachother.

1. fission

\(\mathrm{\ce{^{239}_{94}Pu}+\ce{^1_0n}\rightarrow \ce{^{130}_{50}Sn}+{^{107}_{44}Ru}+3^1_0n}\)

2. α-emission

\(\mathrm{{^{2}_{1}H} + \ce{^{6}_{3}Li} \rightarrow 2^4_2He}\)

3. α-emission

\(\mathrm{\ce{^{210}_{84}Po}\rightarrow \ce{^4_2He}}+{^{206}_{82}Pb}\)

4. fission

\(\mathrm{\ce{^{235}_{92}U}+{^1_0n}\rightarrow \ce{^{72}_{30}Zn}+{^{160}_{62}Sm}+4^1_0n}\)

5. gamma emission

\(\mathrm{\ce{^{125}_{53}I}\rightarrow \ce{^{125}_{53}I}+{^{0}_{0}gamma}}\)

6. α-emission

\(\mathrm{\ce{^{238}_{92}U}\rightarrow {^{4}_{2}He}+\ce{^{234}_{90}Th}}\)

7. Fusion

\(\mathrm{\ce{^{235}_{92}U}+{^1_0n}\rightarrow \ce{^{86}_{35}Br}+\ce{^{147}_{57}La}+{3^1_0n}}\)

8. β-emission

\(\mathrm{\ce{^{234}_{90}Th}\rightarrow {^{0}_{1}e}+\ce{^{234}_{91}Pa}}\)

Q21.2.35

How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture?

S21.2.35

E=mc² c²=931.5 MeV/u

substituting the mass by the change in the masses of deuterium, tritium, and nuclei, we can find the amount of energy released.
[(2∙2.014u)−(3.016u+1.007u]931.5 MeV/u
E=4.03MeV or 6.457e−16 kJ/mol
[(2.014u+3.016u)−(4.003u+1.007u)]931.5 MeV/u = 17.58 MeV converting it to KJ 2.817e−15 kJ/mol

The fusion of a deuterium nucleus and tritium nucleus would be a better choice.