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) ⟶ 2 H⁺(aq) + 2e^- with E^o(anode)=-.53 V

Cathode:

2H⁺(aq) + 2e^-⟶ H₂(g) with E^o(cathode)=0V

Overall reaction:

CO(g) + H₂O(l) ->CO₂(g) + H₂(g) with E^ocell=+.53 V

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

S17.5.5

Standards for a good battery include energy requirements, cost efficiency, and toxicity of components. First, we look at the components in the equations. The gases produced, CO2 and H2, are not harmful to the environment. Furthermore toxicity is even removed from the environment in that the reactant, CO, is turned into a less toxic gas. Next, we look at the energy requirements and spontaniety. The battery is good since Ecell standard is positive, making the reaction spontaneous. No outside power supply, which often causes undesired reactions that often damage the battery, needs to be added. However, there are a lot of smartphone batteries that are recharged, being possible through a secondary battery. Although a secondary battery can be recharged, it is not the most preferred method since external power supply cause outside reactions that ultimately stack up and kill the battery. When the battery is killed, a new battery is needed, costing money from both manufacturing and buying. When no outside power supply is needed, there are no outside reactions that damage the battery, making the battery more long-lived. Therefore, the proposed overall reaction makes a good battery for 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.

S12.2.4

a. On the simulation, we select the default setting and the reaction A+BC. In the default setting, we see frequent collisions, a low initial temperature, and a total average energy lower than the energy of activation. The collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Although we see moving and frequently colliding reactants, the rate of the forward reaction is actually slow because it takes a long time for the products, AB and C, to start appearing. This is apparent mainly because the fractions of collisions with required energy is low, coming from the average energy of the molecules being lower than the energy of activation.

b. The reaction proceeds at an even faster rate. Again, the collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Because molecules have a higher amount of energy, they have more kinetic energy. With an increased kinetic energy, the molecules not only collide more but also increase in the fraction of collision. However, the forward reaction and the backward reaction both proceed at a fast rate, so both happen almost simultaneously. It takes a shorter time for both reactions to happen. With both of the reactions adding up together overall, there is eventually a state of equilibrium. The process at which equilibrium is reached, however, is faster. Therefore, the amount of products of A+BC stays the same after a while (actually having the equilibrium favor toward reactants since the backward reaction has a smaller activation energy).

Q12.5.9

The rate constant at 325^oC for the decomposition reaction C₄H₈ ⟶ 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.

S12.5.9

We look at the Arrhenius equation k=Ae^(-Ea/(RT)). We use this equation since this equation provides a pathway for us to find A, the frequency factor and we have the other variables and values that allow us to solve for A. We have k= 6.1E-8 s^-1, Ea= 261 kJ/mol, R=8.3145 J/(mol*K), and T=325^oC.

First, we look at the term -Ea/(RT). Because this term is an exponent, the term cannot have a unit since exponents are whole numbers, which don't have units. Let's make R have a kJ unit. This is done by using the unit relation 1 kJ= 1000 J. By dimensional analysis, we do [(8.3145 J)/(mol*K)]*[1 kJ/1000J], which is equal to 8.3145E-3 kJ/(mol*K). Let's also let look at T. We need to convert Celcius into Kelvin since R has a unit of K, not ^oC. We convert Celcius to Kelvin by adding 273 to the celcius value. So, 325+273=598 K. Now, lets find the value of the overall exponent by plugging the values in.

We do -261 kJ/mol divided by the product of 8.3145E-3 kJ/(mol*K) and 598 K. The product of R and T using dimensional analysis is [8.3145E-3 kJ/(mol*K)]*[598 K], which is equal to 4.97 kJ/mol. The K unit is canceled out since K/K is equal to 1. We then divide -Ea by RT. Plugging in, we have (-261 kJ/mol)/(4.97 kJ/mol), which is equal to -52.5. Likewise as before, we have kJ/mol unit canceled out because (kJ/mol)/(kJ/mol) is equal to 1. Therefore, the exponent term is equal to -52.5.

Now, let's find A. Plugging in the Arrhenius equation, we have 6.1E-8 s^-1= A*(e^-52.9). All we need to do to find A is to divide k by Ae^(-Ea/(RT)). By doing so, we have 6.1E-8 s^-1/e^-52.5, we have 3.82E15 s^-1. But if we consider significant figures, which is 2 since 6.1E-8 has the least amount of significant figures, the answer is 3.8E15 s^-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.

S21.4.4

In nuclear chemistry, there are three main types of decays. Alpha decay is when a nuclei releases a helium-4 particle along with a daughter nucleus. Beta decay is when a nuclei releases an electron along with an electron. Gamma decay is when a nuclei just releases a gamma ray without any change in the composition of the nucleus.

Explaining the observation assigned, too many nucleons (protons and neutrons) lead to an unstable nucleus. Nuclear force is significant only at a certain small radius from the center. When nucleons stack up to nuclei of nuclides of atomic numbers greater than 83, the resulting nucleus has a radius greater than the radius at which nuclear force is significant. Because many of the outer nucleons are outside the range where nuclear force is significant, they tend to be "unbound" and be "loose." These nuclides are superheavy elements, and the most common type of decay includes alpha decay. The other types of decay, beta and gamma, do not change the atomic mass. Alpha decay is the only emission that helps unstable nuclei stabilize in mass. Therefore, these unstable nuclides normally emit alpha particles to achieve a mass that will make them stable.

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) →

S20.2.8

For activity series, we are using the oxidation activity series chart. The higher up the chart, the more likely oxidation will take place.

a. Cu(s) and Mg²⁺(aq) will be formed as products. Let's take a look at the reactants. We see that Mg has no charge and Cu²⁺ has a charge. If we look at the activity series in terms of oxidation, we see that Mg(s) is more likely to be oxidized than Cu(s). So, Cu²⁺ will gain electrons (be reduced) and Mg will lose electrons (be oxidized). We have the half reactions from the activity series chart:

Mg(s) → Mg²⁺(aq) + 2e^-

Cu²⁺(aq) + 2e^- → Cu(s)

Adding the half reactions together, we get the electrons canceled out and having the net ionic equation Mg(s) + Cu²⁺(aq) → Cu(s) + Mg²⁺(aq)

b. Nothing will happen. Let's take a look at the reactants. We see that Au has no charge and Ag⁺ has a charge. If we look at the activity series in terms of oxidation, we see that Ag(s) is more likely to be oxidized than Au(s). Au being oxidized and Ag⁺ being reduced is less likely. The reaction is not spontaneous, and there needs to be an external power supply for a reaction to even happen.

c. Cr³⁺(aq) and Pb(s) will be formed as products. Let's take a look at the reactants. We see that Mg has no charge and Cu²⁺ has a charge. If we look at the activity series in terms of oxidation, we see that Cr(s) is more likely to be oxidized than Pb(s). Pb²⁺ will gain electrons (be reduced) and Cr will lose electrons (be oxidized). We have the half reactions from the activity series chart:

Cr(s) → Cr³⁺(aq) + 3e^-

Pb²⁺(aq) + 2e^- → Pb(s).

Adding the half reactions together after the oxidation half reaction is multiplied by a coefficient of 2 and the reduction half reaction is multiplied by a coefficient of 3, we get the electrons canceled out and having the net ionic equation 2 Cr(s) + 3 Pb²⁺(aq) → 3 Pb(s) + 2 Cr³⁺(aq).

d. KOH and H₂ will be formed as products. Let's take a look at the reactants, K and H₂O. We see that an alkali metal, which would be K, is reacting with water. When an alkali metal reacts with water, a strong base and hydrogen gas is formed. K is highly likely to be oxidized, as alkali metals are good reducing agents, and the H₂O molecule is reduced.

We get the unbalanced equation: K(s) + H₂O(l)→ KOH(aq) + H₂(g)

We add the coefficient of 2 on K, H₂O, and KOH to get the net ionic equation 2 K(s) + 2 H₂O(l) → 2 KOH(aq) + H₂(g)

e. Nothing will happen. Let's take a look at the reactants. We see that Hg has no charge and Pb²⁺ has a charge. If we look at the activity series in terms of oxidation, we see that Pb(s) is more likely to be oxidized than Hg(l). The reaction is not spontaneous, and there needs to be an external power supply for a reaction to even happen.

Q20.5.3

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

S20.5.3

Extensive properties are values that change when the amount of material changes. Intensive properties are constant values independent of the amount of material.

Let's take a look at the quantities given above. One quantity we see here is E^ocell. No matter what the coefficients of the half/net reactions are, meaning that the amount of material is taken out/added, the E^ocell will always be constant. Since E^ocell stays the same no matter how much material there is, it is an intensive property. Another quantity we see here is n, which is the number of electrons. The number of electrons depend on how much chemicals are added. If we look at half reactions, the number of electrons change when a half reaction is multiplied/divided by a coefficient. When a half reaction is multiplied/divided, the amount of reactants and products (types of material) changes. Since n changes as the amount of material changes, it is an extensive property. A third quantity we can look at is w_max. w_max has a direct relation with n, F, and E^ocell. F and E^ocell are constant values, but n changes depending on the amount of electrons, also a type of material. When n changes, w_max has a direct relation of change as well. Also, the amount of reactants and products affect the numerical value of n, so the amount of reactants and products also have an effect on the value of w_max. Because w_max changes as the amount of material changes, it is an extensive property. The final quantity left to look at is F. It seems that it is an intensive property since it is a constant value. However, F is the same no matter what the given condition is. It does not take regards of the amount of material there is, its constant value is just there. However, F also disqualifies from being classified as an extensive property since it does not change in value. Therefore, F is neither an extensive nor an intensive property.

Therefore, the extensive properties are n (number of electrons) and w_max. The intensive property is E^ocell.

Q20.9.12

8. 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

S20.9.12

Here, electrolysis of aqueous solutions are mentioned, and electrodes are not mentioned. We need to note four important things for electrolysis of aqueous solutions:
1. Cations of less active metals are reduced on the cathode.

2. Anions of less active nonmetals are oxidized on the anode.

3. Cations of more active metals are not reduced and cannot take place by electrolysis of aqueous solution. The cathode half reaction will go with the reduction of water.

4. Anions of more active nonmetals and oxoanions of elements in their highest oxidation state are not oxidized. The anode half reaction will go with the oxidation of water.

Referring to the pictures below, active metals and active metals, in the case of electrolysis, are metals and nonmetals located below the half reaction of 2 H₂O(l) + 2e^- ⟶ H₂(g)+ 2 OH^-(aq) and above the half reaction 2 H₂O(l) ⟶ O2(g) + 4 H⁺(aq) + 4e^-.

a. The aqueous solution of MgBr₂ consists of Mg²⁺ and Br^- ions. Br^- is an anion of a less active nonmetal, so it is oxidized. However, Mg²⁺ is a cation of a more active metal, so the reduction of this cation cannot take place by electrolysis. H₂O is reduced to H₂ instead.

The half reaction for the reduction of H₂O is:

2 H₂O(l) + 2e^- ⟶ H₂(g)+ 2 OH^-(aq)

The half reaction for the oxidation of Br^- is:

2 Br^-(g) ⟶ Br₂(g) + 2e^-

Adding the half reactions together(we can just do so since each half reaction has the same number of electrons), we get the net equation:

2H₂O(l) + 2 Br^-(aq) ⟶ H₂(g) + Br₂(g) + 2 OH^-(aq)
H₂(g) is formed at the cathode, where reduction takes place, while Br₂(g) is formed at the anode, where oxidation takes place.

b. The aqueous solution of Hg(CH₃CO₂)₂ consists of Hg²⁺ and CH₃CO₂^- ions. Hg²⁺ is a cation of a less active element, so it is reduced. CH₃CO₂^- is a special case. CH₃CO₂^- is an anion that is in its most oxidized organic state. However, CO₂ is the most oxidized form of carbon. The acetate ion undergoes a combustion reaction to form CO₂ and H₂O, being done by acetate reacting with O₂.

The half reaction for the reduction of Hg²⁺ is:

2 Hg²⁺(aq) + 2e- ⟶ 2 Hg(l)

The reaction for the combustion of acetate ion is:

4 CH₃CO₂^-(aq) + 7 O2(g) ⟶ 8 CO₂(g) + 6 H₂O (l) + e^-

Adding the half reactions together after we multiply the combustion reaction by 2, we get the net equation:

2 Hg²⁺(aq) + 8 CH₃CO₂^- (aq) + 14 O₂ (g) ⟶ 2 Hg(l) + 16 CO2(g) + 12 H₂O(l).

Hg(l) is formed at the cathode, where reduction takes place, while CO₂(g) is formed at the anode, where oxidation takes place.

c. The aqueous solution of Al₂(SO₄)₃ consists of Al³⁺(aq) and SO₄^2-(aq) ions. The Al³⁺ ion does not react since it is the cation of a more active metal. SO₄^2- also stays the same since it is the oxoanion of an element in its highest oxidation state (S has an oxidation state of 6, which is the highest possible). Therefore, oxidation and reduction of water will take place.

The half reaction for the reduction of water is:

2 H₂O(l) + 2e^- ⟶ H₂(g)+ 2 OH^-(aq)

The half reaction for the oxidation of water is:

2 H₂O(l) ⟶ O2(g) + 4 H⁺(aq) + 4e^-

The incomplete net reaction after we multiply the reduction half reaction by a coefficient by 2 is 4H₂O(l) + 4e^- + 2H₂O(l) → 2H₂(g) + 4OH^-(aq) + O₂(g) + 4H⁺(aq) + 4e^-. However, if we take into the account that H⁺(aq) + OH^-(aq) → H₂O(l), and that there are four moles of H⁺ ions and four moles OH^- ions to form four moles of H₂O, we then need to cancel out the extra water moles on both sides of the equation. By doing so, we get: 2H₂O(l) → 2H₂(g) + O₂(g). H₂(g) is formed on the cathode, where reduction takes place, and O₂(g) is formed on the anode, where oxidation takes place.

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.

S14.7.6

A catalyst speeds up a reaction by lowering the energy of activation. There are two types of catalysts: homogenous and heterogenous. A homogenous catalyst is a catalyst with the same type of phase as the reactants. A heterogenous catalyst is a catalyst with a different type of phase as the reactants. The problem lies in that although both types of catalysts ultimately speed up the reaction, each type has certain advantages and disadvantages. Here, we are looking at why a homogenous catalyst may be preferred. Although homogenous catalysts have a number of operational difficulties, it can still be preferred because of safety and a faster reaction rate. Unlike a heterogenous catalyst at which reactions only take place at the exposed active surface, a homogenous catalyst does not have to rely on the surface and has a better contact with the reactants since all of the catalyst is exposed. Additonally, in a homogenous catalyst, poisoning is not an issue while in heterogenous catalysts, residue on surfaces can also cause poisoning. The transfer of heat can be very problematic when using a heterogeneous catalysts as opposed to a homogeneous catalyst.