Extra Credit 7

From the observations listed, estimate the value of \(E^{\circ}\) for the imaginary half reaction \(M^{2+}+2e^−→ M_{(s)}\), 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 \(Fe^{3+}\) but does not displace \(Sn^{4+}\).
3. The metal reacts with \(HN{O_3}_{(aq)}\).
4. The metal can displace \(K^+_{(aq)}\).

1. Metals that react with \(HCl_{(aq)}\) will have an E value that is less than 0 (between -3.040 and 0). This is because these metals must be reactive enough to displace hydrogen gas from the acidic solution, and metals above \(H^+\) in the activity series are unable to do so.

2. The E value for these reactions will be between 0.154 and 0.771. Metals that will displace \(Fe^{3+}\) but not \(Sn^{4+}\) in a solution are below \(Fe^{3+}\) in the activity series but above \(Sn^{4+}\). This means that the metal is a stronger reducing agent than \(Fe^{3+}\) and will take its electrons, but is not strong enough to do so with \(Sn^{4+}\).

3. The E value for metals that react with \({HN{O_3}_{(aq)}}\) must be lower than 0.956, which is the E value for \(HN{O_3}_{(aq)}\) in an aqueous solution. Any metal below \({HN{O_3}_{(aq)}}\) in the activity series will displace it because it is a stronger reducing agent and will take the electrons from \(NO_3^- {(aq)}\) in its acidic solution.

4. In order to displace \(K^+_{(aq)}\) from a solution, its E value must be lower than -2.924. It must be a stronger reducing agent than \(K^+\) in order to take its electrons.

First, we must calculate the \(E^{\circ}cell\) value for the reaction

\(E^{\circ}cell= E^{\circ}cat - E^{\circ}an\)

The \(E^{\circ}cell\) for \(Zn_{(aq)} → Zn^{2+} _{(aq)} +2e^-\) is -0.763 (this occurs at the anode because it is being oxidized)

We know that Zn is being oxidized because its reduction potential is less than that of Ag. For this reason, you reverse the sign of the reduction potential.

The \( E^{\circ}cell\) for \(Ag^{+} + e^{-}\rightarrow Ag_{(s)}\) is 0.800 (this occurs at the cathode because it is being reduced)

0.800-(-0.763)= 1.563= \(E^{\circ}cell\).

Then, we will use the Nernst equation to find Q for the reaction.

\(Ecell= E^{\circ}cell - \frac{0.0592 V}{n} logQ\)

\(1.536 = 1.563- \frac{0.0592 V}{2}logQ\)

0.9122 = logQ

\(Q = 8.169 = \frac{[Zn^{2+}]}{[Ag^+]^2} = \frac{2.00}{[Ag^+]^2} = 8.169\)

\([Ag^+]^2= 0.2448 Ag^+= 0.4948\)

Therefore, the concentration of Ag^+ in the cell is 0.4948 M.

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.

Write the possible half equations to represent each of the following

1. Cu²⁺ as an oxidizing agent
2. Mn²⁺ as a reducing agent

Since \(MnO{_4^-} _{(aq)}\) is acting as an oxidizing agent, we know that it is being reduced and therefore will gain electrons in its half reaction.

1. \(MnO{_4^-} _{(aq)} + 8H^+ _{(aq)} + 5e^- \rightarrow Mn^{2+}_ {(aq)} + H_2O_ {(l)}\)

Since \(Cr^{2+}_{(aq)}\) is acting as a reducing agent, we know that it is being oxidized and will therefore lose electrons in its half reaction.

2. \(Cr^{2+}_{(aq)} \rightarrow Cr^{3+}_{(aq)} + e^-\)

Since \(Cu^{2+}_{(aq)}\) is acting as an oxidizing agent, we know that it is being reduced and will gain electrons in its half reaction.

1. \(Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}\)

Since \(Mn^{2+}_{(aq)}\) is acting as a reducing agent, we know that it is being oxidized and will lose electrons in its half reaction.
2. \(Mn^{2+}_{(aq)} + 4H_2O_{(l)} \rightarrow MnO{_4^-}_{(aq)} + 8H^+_{(aq)} + 5e^-\)

Q21.27A

Which of the following complex ions would you expect to have the largest overall \(K_f\), and why? \([Cr(CO)_6]^{2+}[Cr(CO)_6]^{2+}, [Cr(en)_3]^{2+}[Cr(en)_3]^{2+}, [Cr(NH3)_6]^{3+}[Cr(NH3)_6]^{3+}, [Cr(CO)_2(NH3)_2(en)]\)

\([Cr(en)_3]^{2+}\) has the largest overall \(K_f\). This is due to the chelating effect; because en is a polydente ligand and therefore is more stable, it has a larger \(K_f\).

Q24.36A

For the disproportionation of p-toluenesulfinic acid,

\(3ArSO_2H→ArSO_2SAr+ArSO_3H+H_2O\)

(where \(Ar=p-CH_3C_6H_4\), the following data were obtained: t=0min, \([ArSO_2H]\)=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, k?
3. At what time would \([ArSO_2H]\)=0.0600M?
4. At what time would \([ArSO_2H]\)=0.0300M?
5. At what time would \([ArSO_2H]\)=0.0150M?

a. We know that this reaction is a second order reaction because when plotting \(\frac{1}{Reactant}\) vs. \(t\), the slope is a straight line.

\(k=\frac{1}{{t_{1/2}}[A]_0}\)

b.

Taking the slope of the line, we obtain an approximate k value of 0.0023. Using the equation \(k=\frac{1}{{t_{1/2}}[A]_0}\), we can solve for t at multiple concentrations.

c. \(\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}\)
\(\frac{1}{[0.0600]}= (0.0023)(t) + \frac{1}{[0.140]}\)
9.52= (0.0023)(t)
t= 4141 seconds= 69.1 minutes

d. \(\frac{1}{[0.0300]}= (0.0023)(t) + \frac{1}{[0.140]}\)
26.2= (0.0023)(t)
t= 11390 seconds= 189 minutes

e. \(\frac{1}{[0.0150]}= (0.0023)(t) + \frac{1}{[0.140]}\)
59.5 = (0.0023)(t)
t= 25880 seconds= 431 minutes

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.

Adding the total mass numbers of Bismuth and Aluminum, we get 242 for the 242-Curium, as expected. We can either add the amount of protons in Aluminum and Bismuth (13 and 83, respectively), to get 96 protons for Curium, or simply look at the periodic table. Therefore, the nuclear equation can be shown as:

\(^{209}_{83}Bi+{^{27}_{13}Al} \rightarrow {^{242}_{96}Cm}\)

An alpha particle has a total mass number of 4 and 2 protons. Since 5 alpha particles are being emitted, we multiply each of these numbers by 5 to get 20 and 10 for the total mass loss and proton loss. Subtracting 242 by 20, we get a total mass number of 222. Subtracting 96 by 10, we get a proton mass of 86, meaning that the element created by this emission is Radon. Thus, 222-Radon is created by this nuclear reaction, which can be shown below.
\({^{242}_{96}Cm} \rightarrow {^{222}_{86}Rn} +5\alpha\)

Q19.1

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

a.\({^{239}_{94}Pu} + {^1_0n} \rightarrow {^{130}_{50}Sn} + {3^1_0n} + ?\)

b.\(? + {^6_3Li} \rightarrow {2^4_2He}\)

c. \({^{210}_{84}Po} \rightarrow {^4_2He} + ?\)

d.\({^{235}_{92}U} + {^1_0n} \rightarrow {^{72}_{30}?} + {4^1_0n} + ?\)

e. \({^{125}_{53}I} \rightarrow {^{125}_{53}I} +?\)

f. \({^{238}_{92}U} \rightarrow ? + {^{234}_{?}Th}\)

g.\({^{235}_{92}U} + {^1_0n} \rightarrow {^{86}_{?}Br} + {^{147}_{?}?} + {?^1_0n}\)

h. \({^{234}_{90}Th} \rightarrow ? + {^{234}_{91}?}\)

a. The total mass number on the reactant side is 239+ 1 (240), and the number of protons is 94. We know that the total mass number and amount of protons on the products side must be equal. Since we know that the total mass of Sn and 3 neutrons is 130 + 3 (133) and the number of protons is 50, the missing element must have a total mass of 240-133 (107) and has 94-50 (44) protons. The missing element is therefore 107-Ruthenium. This is a fission reaction.

\({^{239}_{94}Pu} + {^1_0n} \rightarrow {^{130}_{50}Sn} + {3^1_0n} + {^{107}_{44}Ru}\)
b. The total mass number of the products is 2x4 (8) and the number of protons is 2x2 (4). Thus, we know that the element on the reactant side has a total mass number of 8-6 (2), and has 4-3 (1) proton. This element is 2-Hydrogen. This is a fusion reaction.

\({^2_1H} + {^6_3Li} \rightarrow {2^4_2He}\)
c. Since 210-Polonium undergoes alpha decay, we know that it must lose 2 protons and its total mass number will decrease by 4. Thus, the total mass number will be 206, and it will have 82 protons. This means that the missing element is 206-Lead. This is an alpha decay reaction.

\({^{210}_{84}Po} \rightarrow {^4_2He} + {^{206}_{82}Pb}\)
d. The total mass number on the reactant side is 235+1 (236), and the amount of protons is 92. Therefore, the products side must also have the same amount of both. We know that because the missing element has 30 protons, it is Zinc. To solve for the other missing element, we will subtract the total mass number and amount of protons on both sides. 236- (72+ 4)= 160, and 92-30 is 62. Therefore, the missing element is 160-Samarium. This is a fission reaction.

\({^{235}_{92}U} + {^1_0n} \rightarrow {^{72}_{30}Zn} + {4^1_0n} + {^{160}_{62}Sm}\)
e. Since the mass of the element does not change and thus the element does not change, it undergoes gamma decay (other).

\({^{125}_{53}I} \rightarrow {^{125}_{53}I} + \gamma\)
f. We know that the missing element must have a total mass of 238-234 (4) and has 92-90 (2) protons. Thus, the missing element is 4-Helium and this is an alpha-decay reaction.

\({^{238}_{92}U} \rightarrow {^4_2He} + {^{234}_{90}Th}\)
g. The total mass of the reactants is 235+1 (236), and the total number of protons is 92. Thus, we know that the products must also have the same amount. 236- (86+147) is 3, so there are three neutrons. We know that Bromine has 35 protons, so 92-35 is 57. Thus, the missing element is 147-Lanthanum. This is a fission reaction.

\({^{235}_{92}U} + {^1_0n} \rightarrow {^{86}_{35}Br} + {^{147}_{57}La} + {3^1_0n}\)
h. The total mass number on both of the reactants and products sides are equal, but the element to the right has one more proton. Thus, we know that it is a \(\beta\) reaction and it loses one electron. The element created is then 234-Protactinium.

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

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?

In order to solve for the amount of energy released by these fusion reactions, we will use the following equation

\(E=mc^2\) \(c^2= 931.5\) MeV/u

By using this and 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
\(E= 4.03 MeV\) or \(6.457e^{-16}\) kJ/mol
\([(2.014u+ 3.016u)-(4.003u+1.007u)]931.5\) MeV/u = 17.58 MeV or \(2.817e^{-15}\) kJ/mol

The fusion of a deuterium nucleus and tritium nucleus would be a better choice for a fusion reactor fuel mixture because it releases more energy per mole.