Extra Credit 8

Q17.1.7

Balance the following in basic solution:

1. SO₃^2-(aq)+Cu(OH)₂(s)⟶SO₄^2-(aq)+Cu(OH)(s)
2. O₂(g)+Mn(OH)₂(s)⟶MnO₂(s)
3. NO₃^-(aq)+H₂(g)⟶NO(g)
4. Al(s)+CrO₄^2-(aq)⟶Al(OH)₃(s)+Cr(OH)₄^-(aq)

S17.1.7

In order to balance the equation, it is important to list the reduction and oxidation reaction separately. And then, make sure the number of electrons of both half-reactions are the same. Then just sum them up into an overall reaction.

1. Reduction: 2e^-+2Cu(OH)₂(s)+H⁺⟶2Cu(OH)(s)+H₂O(l)
Oxidation:SO₃^2-(aq)+H₂O(l)⟶SO₄^2-(aq)+2e^-+2H⁺(aq)
Overall: SO₃^2-(aq)+2Cu(OH)₂(s)⟶2Cu(OH)(s)+SO₄^2-(aq)+H₂O(l)

2.Reduction:O₂(g)+2e^-⟶MnO₂(s)
Oxidation:Mn(OH)₂(s)⟶MnO₂(s)+2e^-+2H⁺(aq)
Overall:2Mn(OH)₂(s)+O₂(g)⟶2MnO₂(s)+2H₂O(l)

3.Reduction:NO₃^-(aq)+3e^-+4H⁺(aq)⟶NO(g)+2H₂O(l)
Oxidation:H₂(g)⟶2H⁺(aq)+2e^-
Overall: 2NO₃^-(aq)+3H₂(g)+2H⁺(aq)⟶2NO(g)+4H₂O(l)

Above solution is incorrect. In a basic solution, the overall reaction is: 2NO₃^-(aq) + 2H₂O(l) +3H₂(g) ⟶ 2NO(g) + 4H₂O(l) +2OH^-

4.Reduction:CrO₄^2-(aq)+2e^-+4H⁺(aq)⟶Cr(OH)₄^-(aq)
Oxidation:Al(s)+3H₂O(l)⟶Al(OH)₃(s)+3e^-+3H⁺(aq)
Overall:2Al(s)+6H₂O(l)+3CrO₄^2-(aq)​​​​​​​+6H⁺(aq)​​​​​​​⟶3Cr(OH)₄^-(aq)​​​​​​​+2Al(OH)₃(s)​​​​​​​

Above solution is incorrect.

The reduction half reaction is: CrO₄^2- (aq) + 4H₂O(l) + 3e^- ⟶ Cr(OH)₄^- (aq) +4OH^-

The oxidation half reaction is: Al (s)+3H₂O(l) + 3OH^- ⟶ Al(OH)₃ (s) +3e^-+ 3H₂O(l)

The overall reaction is: Al (s) + CrO₄^2- (aq) + 4H₂O(l) ⟶Al(OH)₃ (s)+Cr(OH)₄^- (aq)+OH^{​​​​​​​-}

Q19.1.6

Which of the following is the strongest oxidizing agent: VO₄^3- , CrO ₄^2- , or MnO₄^-

S19.1.6

From the periodic table, it is clear to see that Mn has the most valence electrons among the three. Therefore, Mn is most likely to gain electrons. Thus, MnO₄^- will be most likely to undergo reduction. Therefore, it is the strongest oxidizing agent.

Vanadium has 5 valence electrons in VO₄^3- , Chromium has 6 valence electrons in CrO₄^2- , and Manganese has 7 valence electrons in MnO₄^-. Because Mn has the greatest number of valence electrons, it is most electronegative. The more an element is electronegative, the higher will be is oxidizing power, therefore Mn is the strongest oxidizing agent.

Q19.2.8

Specify whether the following complexes have isomers.

1. tetrahedral [Ni(CO)₂(Cl)₂]
2. trigonal bipyramidal [Mn(CO)₄NO]
3. [Pt(en)₂Cl₂]Cl₂

S19.2.8

1.none; the Tetrahedral structure has no geometric isomers. It has two pairs of same ligands connected to the central atom, Ni. So it won't have any optimal isomers either. And there are no structural isomers. Recall that it is not possible for geometric isomers to display geometric isomers.

2. none; it is not possible for [Mn(CO)₄NO] to display isomers.

3.The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer. Refer below for an image of what the [Pt(en)₂Cl₂]Cl₂ isomers would look like.

Q12.3.21

The annual production of HNO₃ in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.

1. 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)
2. 2NO(g)+O2(g)⟶2NO2(g)
3. 3NO2(g)+H2O(l)⟶2HNO3(aq)+NO(g)

The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O₂, what is the rate of formation of NO₂ when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 × 10⁻⁶ L²/mol²/s.

S12.3.21

Because the second reaction limits the rate of the entire process, the rate law for the reaction is:

rate=k[NO]²[O₂]

Given the K value and initial concentrations of NO and O₂, substitute the values for the desired rate.

rate = k[NO]²[O₂]

rate = 5.8 × 10⁻⁶ L²/mol²/s × [0.75]²× [0.5]

rate =1.63 × 10⁻⁶

Q12.7.1

Account for the increase in reaction rate brought about by a catalyst.

S12.7.1

The general mode of action for a catalyst is to provide a mechanism by which the reactants can unite more readily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium. Catalysts alter mechanisms, but are not consumed in reactions.

Q21.4.24

A ⁷₄Be atom (mass = 7.0169 amu) decays into a ⁷₃Li atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?

S21.4.24

reaction: ⁷₄Be ⟶⁷₃Li + ⁰_-1e

First, amu must be converted to kg. The conversion unit of amu to kg is 1 amu/1.66054*10^-27 kg. The constants become:

Mass of electrons=9.10938356 × 10^-31kg, ⁷₄Be=1.1651836× 10^-26kg, ⁷₃Li=1.165034× 10^-26kg.

To calculate the energy, we need to know the mass deficit between the experimental value and the calculated value.

Mass deficit = (mass of ⁷₃Li + ⁰_-1e) - (mass of ⁷₄Be)

Mass deficit = (9.10938356 × 10^-31 + 1.165034× 10^-26 ) - (1.1651836 × 10^-26)

Mass deficit = -5.8506164× 10^-31kg

To calculate the energy, the formula is: E=mc² . The constant, c, is equal to 2.998× 10⁸.

E = -5.8506164 × 10^-31 × (2.998× 10⁸)

E =-5.2585364 × 10^-14 J

The conversion unit of J to MeV is 1 J/6.242*10¹² MeV.

E =-0.328238 MeV

Q20.5.23

For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?

S20.5.23

Reduction reaction of oxygen to water: O₂(g) +4H⁺(aq)+4e^-⟶ 2H₂O(l). Assume this reaction happens in bar=1. Because pH=7, [H]=10^-7 M

According to the Nernst equation, \[E = E^o - \dfrac{0.0592\, V}{n} \log Q \tag{Nernst Equation @ 298 K}\]

E=1.23V-(0.059V/2)× log(1/pO₂[H⁺]⁴).

Therefore, the potential for O₂ reduction at pH = 7 and p(O2) = 1 bar is:

E=1.23V-0.826V=0.404V

In solution of NaOH:

Na⁺(aq)+e^-⟶ Na(s)

E=(RT/NF)lnQ=0.0592× log(1/0.85)=0.00417V