Extra Credit 45

Q17.6.4

Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?

SOLUTION:

In order to predict this, we need to be familiar with the concept of standard reduction potential table (example of a table: http://www.chemeddl.org/services/moo...ndardTable.htm). The metals with the most negative standard reduction potential are good reducing agents and, thus, best electron donors (have least tendency to accept electrons), and the metals with the most positive standard reduction potential are the best oxidizing agents and, therefore, the best electron acceptors (have bigger tendency to accept electrons).

The electrons flow from the most the metals with the most negative standard reduction potential to the metals with the most positive standard reduction potential. Therefore, the metal with the most negative standard reduction potential will be donating electrons and the metals with the most positive standard reduction potential will be accepting electrons.

Corrosion means oxidation: so if B is more likely to oxidize (give up electrons) than A, then metal B has lower standard potential than the metal A. Same logic works when A corrodes and C does not. If A (an anode) is more likely to oxidize than C (cathode), then metal A has lower standard potential than the metal C.

We can rank of metals in terms of standard reduction potential: B<A<C. C is the least likely to oxidize and B is the most like to oxidize.

Then, we can determine what will happen if B and C come in contact. Because B has lower standard reduction potential than metal C, metal B will oxidize (corrode) and metal C will not.

*NOTE: Also, although the answer (I am not sure if it's official or a student just accidently posted it there) posted below a question is correct, the explanation is wrong. The answer argues that metals with higher standard potential will be the donors, when higher positive E values means higher tendency to accept an electron. The metals with lower E values are the ones that donate an electron)

Q12.3.8

(I think in the prompt notation was supposed to be a little different: the proton number of nitrogen should have been 7 instead of 6 and the subscripts notation should be flipped. Right now, in the actual prompt it says \({^{6}_{14}}C\), which is incorrect because proton number 14 indicates Si). I corrected it to my best understanding.

The rate constant for the radioactive decay of ¹⁴C is 1.21 × 10⁻⁴ year⁻¹. The products of the decay are nitrogen atoms and electrons (beta particles):

\({^{14}_{6}}C\)⟶\({^{14}_{7}}N\)+e^–

rate=k[\({^{14}_{6}}C\)]

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10⁻⁹ M?

SOLUTION:

We can compare the rate of formation of reactants and products for any equation like:

aA ⟶ bB + cC

Where rate = -(1/a)*(Δ[A]/Δt) = (1/b)*(Δ[B]/Δt) = (1/c)*(Δ[C]/Δt)

In the similar fashion, we can construct an equation to compare rates of disappearance of \({^{14}_{6}}C\) and formation of \({^{14}_{7}}N\):

rate=k[\({^{14}_{6}}C\)] = (1/n)* Δ[\({^{14}_{7}}N\)]/Δt = (1/1)* Δ[\({^{14}_{7}}N\)]/Δt, where n is the coefficient in front of \({^{14}_{7}}N\). Δ[\({^{14}_{7}}N\)]/Δt should be equal to the rate of production of N atoms.

We are given k and [\({^{14}_{6}}C\)]. Plug the values in: Δ[\({^{14}_{7}}N\)]/Δt = k[\({^{14}_{6}}C\)] = (1.21 × 10⁻⁴ year⁻¹)*(6.5 × 10⁻⁹ M) = 7.865*10^–13M*year^–1 = 7.9*10^–13M*year^–1 (check sig figs and units)

Q12.5.17

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A+BC⟶AB+C reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?

SOLUTION:

The molecule A has to hit BC molecule with a certain orientation, otherwise the reaction will not happen. Orientation is one of key points of The Collision Theory: A certain orientation is needed to make a reaction successful. When the atoms align as needed for reaction (180 angle between A and B) bond between B and C will break and a bond between A and B will form. Even if the molecules moving with enough energy, the improper orientation might make the reaction impossible.

Q21.4.12

Write a nuclear reaction for each step in the formation of Pb from Th, which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.

SOLUTION:

While writing nuclear reactions, keep two rules in mind:

• Total (sum it up) mass number (number at the upper left corner of an element) on the left side should equal to the total (sum it up) mass number on the right side.
• Total (sum it up) number of protons (number on the bottom left corner of an element) on the left side should be equal to the total (sum it up) number of protons on the right side.

Use this example for help:

\({^{a}_{x}}A ⟶ {^b_{y}}B +{^{c}_{z}}C\),

where a = b+c and x = y+z

α-particle emission means that \({^4_{2}}He\) is released

β-particle emission means that \({^0_{–1}}e\) is released

There are seven steps:

1) α-particle emission: \({^{228}_{90}}Th ⟶ {^4_{2}}He +{^{224}_{88}}Ra\)

2) α-particle emission: \({^{224}_{88}}Ra ⟶ {^4_{2}}He +{^{220}_{86}}Rn\)

3) α-particle emission:\({^{220}_{86}}Rn ⟶ {^4_{2}}He +{^{216}_{84}}Po\)

4) α-particle emission:\({^{216}_{84}}Po ⟶ {^4_{2}}He +{^{212}_{82}}Pb\)

5) β-particle emission (or electron emission):\({^{212}_{82}}Pb ⟶ {^0_{–1}}e +{^{212}_{83}}Bi\)

6) β-particle emission (or electron emission):\({^{212}_{83}}Bi ⟶ {^0_{–1}}e +{^{212}_{84}}Po\)

7) α-particle emission:\({^{212}_{84}}Po ⟶ {^4_{2}}He +{^{208}_{82}}Pb\)

Q20.2.16

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

SOLUTION:

First, let's consider definitions of:

• an acid–base reaction: when an acid and base react, a salt and water is produced. This is also known as a neutralization reaction.
• a precipitation reaction: an insoluble compound appears as one of the products.
• redox reaction: this reaction consists of the transfer of electrons from one compound to another. During redox reaction, one of the reactants is oxidized (loses electrons) and another one is reduced (gains electrons). Oxidation and reduction always happen in pair: if one substance is oxidized, then another one is reduced. Redox reactions are easy to stop if you compare the oxidation states before and after redox reaction.

Then, let's balance the reactions and classify the reactions:

1. Zn(s) + 2HCl(aq) →ZnCl₂(s) + H₂(g) This is a redox reaction because the initial oxidation state of Zn is 0 and the final is 2+. This means that Zn has been oxidized (lost electrons). We can also see that initial oxidation state of H is +1, while the final oxidation state of H is 0. This means that H has been reduced (gained electrons). NOTE: Although the answer (I don't now if it is the official answer or not) below says this is a precipitation reaction because ZnCl₂ is insoluble, I believe that this is a mistake. Many sources says that all chlorides are soluble with exception of (Hg₂)²⁺, Pb²⁺, and Ag⁺. Source: https://chem.libretexts.org/Core/Phy...lubility_Rules
2. 3HNO₃(aq) + AlCl₃(aq) → no reaction will be observed because both reactants are acids.
3. K₂CrO₄(aq) + Ba(NO₃)₂(aq) → BaCrO4(s) + 2KNO3(aq) This is a precipitation reaction because one of the products is insoluble, and no change in oxidation states is observed. Also, the reaction is happening between two salts, so it is not acid/base reaction.
4. Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s) This is a redox reaction because we can see the transfer of electrons from one compound to each other. The initial oxidation state of Zn is 0 and the final is 2+. This means that Zn has been oxidized (lost electrons). We can also see that initial oxidation state of Ni is +2, while the final oxidation state of Ni is 0. This means that Ni has been reduced (gained electrons).

Q20.5.11

The chemical equation for the combustion of butane is as follows:

C₄H₁₀(g)+(13/2)O₂(g)→4CO₂(g)+5H₂O(g)

This reaction has ΔH° = −2877 kJ/mol. Calculate E°_cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?

SOLUTION:

The solution consists of the three steps:

First step: calculate ΔS° from the equation: ΔS°_rxn = sumS°products – sumΔS°_reactants

The values were taken from http://www.ars-chemia.net/Permanent_...ThermoData.pdf

C₄H₁₀ S^o = 310.1 J/(K*mol) (reactant)

H₂O S^o = 188.83 J/(K*mol) (reactant)

O₂ S^o = 205.14 J/(K*mol) (product)

CO₂ S^o = 213.74 J/(K*mol) (product)

ΔS°_rxn = (4*213.74+5*188.83) – (310.1 + 6.5*205.14) = 155.60 J/(K*mol) This is change in entropy.

Second step: Use ΔG°= ΔH° - TΔS° to find ΔG°. We are given ΔH° (convert: −2877000 J/mol) and T (in Kelvin), and already found ΔS°.

ΔG°= ΔH° - TΔS° = −2877000 J/mol – 298K*(155.60 J/(K*mol) ) = –2923368.80 J/mol

Since ΔG° is negative, the reaction is highly spontaneous.

Third step: find E°_cell using ΔG° = –nFE°_cell. E°_{cell =} ΔG°/(–n*F), where n (number of electrons transferred is 10) and F (Faraday's constant is equal to 96485.33 C*mol⁻¹) are constants, and we have calculated ΔG° in a previous step.

E°_{cell =} ΔG°/(–n*F) = (–2923368.8 J/mol) /(–10* 96485.33 C*mol⁻¹) = 3.03 (J/C) or V. (J/C is Volts)

Note: I am really not sure about this solution, this needs to be rechecked, maybe there is a better way.

Q24.6.7

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

SOLUTION:

a. [TiCl₆]³⁻

First, we will determine the structure. Since Ti has 6 ligands, the structure is octahedral.

Then, we check whether Cl^– is a strong or weak ligand. On the list of Spectrochemical Series, we see that Cl^– is a weak ligand (it is one of the first ones, and the ligands are listed in the order of increasing strength), meaning that it will result in high spin if the number of unpaired electrons is greater than 3.

Now, we will determine the number of unpaired electrons present. First, there are 6 Cl^– ligands and 3– overall charge of the compound, meaning that Ti has charge of 3+ (use equation to solve x + 6*(-1) = -3). The electron configuration of Ti³⁺ is [Ar]d¹ (first remove all (2) electrons from s-orbitals, and then one more form d-orbital in order to reflect the charge). Therefore, there is only one electron and it is, of course, unpaired.

We see that we only have 1 unpaired electron, so the spin is low by definition (it is impossible for a spin to be high when the number of unpaired electrons is lower than 3).

b. [CoCl₄]²⁻

First, we will determine the structure. Since Co has 4 ligands, the structure is tetrahedral.

Then, we check whether Cl^– is a strong or weak ligand. On the list of Spectrochemical Series, we see that Cl^– is a weak ligand (it is one of the first ones, and the ligands are listed in the order of increasing strength), meaning that it will result in high spin if the number of unpaired electrons is greater than 3.

Now, we will determine the number of unpaired electrons present. First, there are 4 Cl^– ligands and 2– overall charge of the compound, meaning that Co has charge of 2+ (use equation to solve x + 4*(-1) = -2). The electron configuration of Co²⁺ is [Ar]d⁷ (remove all (2) electrons from s-orbitals to reflect the charge). Therefore, there are 7 electron, 3 of which are unpaired. We have 7 electrons, so it is high spin.