Extra Credit 37

Q17.5.5

An inventor proposes using a SHE (standard hydrogen electrode) in a new battery for smartphones that also removes toxic carbon monoxide from the air:

• Anode:

  CO(g)+H2O(l)⟶CO2(g)+2H+(aq)+2e−CO(g)+H2O(l)⟶CO2(g)+2H+(aq)+2e−

  with E^o(anode)=−0.53V
• Cathode:

  2H+(aq)+2e−⟶H2(g)2H+(aq)+2e−⟶H2(g)

  with E^o(cathode)=0V
• Overall reaction:

  CO(g)+H2O(l)⟶CO2(g)+H2(g)CO(g)+H2O(l)⟶CO2(g)+H2(g)

  with E^ocell=+0.53V

Would this make a good battery for smartphones? Why or why not?

Solution:

This question relates to batteries and cell potentials. A good smartphone battery is one that can be recharged thousands of times, and the reaction must be spontaneous. Since we are given the overall cell potential, we can see that it is a positive value. A positive E^ocell means that the forward reaction is spontaneous, and because of this we can conclude that this battery is probably a good one to use in smartphones.

Q12.2.4

In the PhET Reactions & Rates interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.

1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
2. Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.

Solution:

1. The rate of the reaction seems slow.

2. As the temperature increases, so will the kinetic energy. This means that the average energy is higher than the potential energy. This means that the reactants will decrease and the products will increase. There will be about equal amounts of all the molecules; however, A will be slightly in excess.

Q12.5.9

The rate constant at 325 °C for the decomposition reaction C₄H₈⟶2C₂H₄C₄H₈⟶2C₂H₄ is 6.1 × 10⁻⁸ s⁻¹, and the activation energy is 261 kJ per mole of C₄H₈. Determine the frequency factor for the reaction.

Solution:

Using the Arrhenius equation allows me to find the frequency factor, A.

k=Ae^-Ea/RT​

k, Ea, R, and T are all known values. k, Ea, and T are given in the problem as 6.1x10^-8, 261 kJ, and 598 K, respectively.

So, plugging them into the equation gives:

6.1x10^-8 s^-1=Ae^{(-261000 J)/(8.3145 J/mol)(598 K)}

Take e^{(-261000 J)/(8.3145 J/mol)(598)} and you'll get 1.59 x 10^-23. Divide k, 6.1 x 10^-8, by 1.59 x 10^-23 and you get A=3.9 x 10¹⁵s^-1

^Q21.4.4

Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles.

Solution:

^{This question relates to radioactive decay. Because elements with an atomic number greater than 83 are unstable, they release alpha particles because it produces a more stable daughter nucleus quicker. Because alpha decay reduces Z by 2, it is able to lower the repulsion between the protons in the nucleus quicker. The graph above depicts the belt of stability, the stable isotopes are those that fall on the greenish line, so the goal of any unstable isotope is to decay enough to fall on that line. Elements with an atomic number greater than 83 tend to emit alpha particles because it reduces the repulsion between protons in the nucleus faster than any other sort of decay would.}

Q20.2.8

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

1. Mg(s) + Cu²⁺(aq) →
2. Au(s) + Ag⁺(aq) →
3. Cr(s) + Pb²⁺(aq) →
4. K(s) + H₂O(l) →
5. Hg(l) + Pb²⁺(aq) →

Solution:

This question relates to half reactions, so we need to use the activity series, which is shown below.

1. Let's take a look at A. We are given Mg(s) + Cu²⁺(aq) →, and our goal is to predict what happens next. In order to do this, let's take a look at the half reactions of Mg and Cu:

Mg(s) + 2e^- → Mg²⁺(aq)

Cu²⁺(aq) → Cu(s) + 2e^-

By adding these two reactions together and cancelling the electrons, we get:

Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu (s).

As you can see in this equation, the Mg was oxidized, meaning it lost electrons, and the Cu was reduced, meaning it gained electrons.

2. The half reactions we have from Au and Ag are:

Au(s) + e^- → Au⁺ (aq)

Ag⁺(aq) → Ag(s) + e^-

By adding the reactions together and cancelling the electrons, we get:

Au(s) + Ag⁺(aq) → Au(s) + Ag⁺(aq)

Since nothing changed in this reaction, this reactions is very unlikely to occur.

3. The half reactions we get from Cr and Pb are:

Cr(s) + 2e^- → Cr²⁺(aq)

Pb²⁺(aq) → Pb(s) + 2e^-

By adding these reactions together and cancelling the electrons, we get:

Cr(s) + Pb²⁺(aq) → Cr²⁺(aq) + Pb(s)

Here, you can see the Cr was oxidized because it lost electrons and the Pb was reduced because it gained electrons.

4. This reaction is actually a replacement reaction, so we don't have to use the activity series. We simply balance the equation. It becomes:

2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)

5. Let's look at the half reactions of Hg and Pb:

Hg(l) → Hg²⁺(aq) + 2e^-

Pb²⁺(aq) → Pb(s) + 2e^-

This reaction is incredibly unlikely to occur, as the reactants are both metals and have a positive charge, so they would repel each other.

Q20.5.3

In the equation w_max = −nFE°_cell, which quantities are extensive properties and which are intensive properties?

Solution:

Before being able to answer this question, the difference between extensive and intensive properties must be understood. An extensive property is a property that varies as the amount of material changes (i.e. mass, volume, enthalpies, and current); whereas, an intensive property is a property that does not vary as the amount of material changes (i.e. tempertaure, pressure, average speed of molecules, and voltage). By understanding this, one can see that w_max and n are both extensive properties because they are proportional to each other. E°_cell is an intensive property because the cell potential doesn't change with how much work is being done to the surroundings, and it is independent of the system size or the materials in the system.

Q20.9.12

Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.

1. MgBr₂
2. Hg(CH₃CO₂)₂
3. Al₂(SO₄)₃

Solution:

Recall where oxidation and reduction take place: oxidation occurs at the anode, and reduction occurs at the cathode. This question also pertains to the activity series in order to determine what is being reduced and what is being oxidized. Since these are all in aqueous solutions, they will involve the half reaction of water, which is: 2H₂O(aq) + 2e^- → H₂(g) + 2OH^- when is is being reduced, or 2H₂O (aq) → O₂(g) + 4H⁺(aq) + 4e^- when it is being oxidized.

1. MgBr₂

Anode: 2Br^-(aq)→ Br₂(l) + 2e^-

Cathode: 2H₂O(aq) + 2e^- → H₂(g) + 2OH^-

2. Hg(CH₃CO₂)₂

Anode: 2H₂O (aq) → O₂(g) + 4H⁺(aq) + 4e^-

Cathode: Hg²⁺(aq) + 2e^- → Hg(l)

3. Al₂(SO₄)₃

Anode: 2H₂O (aq) → O₂(g) + 4H⁺(aq) + 4e^-

Cathode: 2H₂O(aq) + 2e^- → H₂(g) + 2OH^-

Q14.7.6

An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred.

Solution:

A catalyst is a compound that is added to reactions to lower the activation energy and increase the reaction rate. Essentially, it is used in order to speed up a reaction. With that said, a homogenous catalyst is a catalyst that is in the same phase as the reaction mixture, gas or liquid. For example, if the reaction mixture is all gases, then the catalyst is also a gas, or if the reaction mixture is all liquids, then the catalyst is also liquid. One reason why homogenous catalysts may be preferred is because it allows for greater interaction with the reaction mixture because they are all in the same phase, which means a quicker reaction. In short, this means that a homogenous catalyst mixes better with the reactants, which allows for a faster reaction.