Extra Credit 39

Q 17.5.7

Explain what happens to battery voltage as a battery is used, in terms of the Nernst equation.

Let us clarify some definitions before answering this question. What is a battery? A battery is denoted as an electrochemical cell that uses chemical energy and converts it into electricity; because it is an electrochemical cell, it is often contained in some kind of container--imagine your typical alkaline battery. Some types of batteries can only be used once (primary cells) while others can be recharged again as needed (secondary cells). What is a voltage? Voltage is often termed potential; it is measured in joules per coulombs and describes the energy contained in a unit of charge. It is easier to think of it in terms of work; how much work needs to occur to move an electric current from one point to another. This unit for voltage is termed volts. The Nernst equation can be used to calculate the cell potential under nonstandard conditions. It relates thermodynamics and the equilibrium constant. Under nonstandard conditions this is the equation:

E = E^º– (kT/nF)ln([Red]/[Ox]

n is the number of electrons in the reaction, F is denoted as Faraday's constant (96485 C), that is the charge of each mole of electron, and E is the potential difference. Looking at the equation we know that if Q is 1, then the ln1 will give us 0; as such, when the reaction proceeds, the two chemical species under nonstandard conditions begin to approach equilibrium—that is the ratio of their concentration begin to equalize, indicating the system is now in equilibrium. A system under equilibrium indicates a dead battery and a voltage of zero. A zero voltage indicates there is no potential difference between the two terminals, hence the battery does not work.

Q 12.3.2

Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

1. What is the order of the reaction with respect to that reactant?

2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

To solve this problem, let us consider a hypothetical scenario, in which we give arbitrary values in order to understand what is going on.

Working with easy numbers, we can get this table for part a:

2^x= 4, thus we can conclude x = 2. The order of reaction with respect to reactant A is 2. In general using this knowledge, we make the assumption that if doubling the concentration of a reactant quadruples reaction rate, it is in second order and follows the following rate law: \[rate = [A]^2\]

For part b, we can make a similar table:

3^x= 3, thus we can conclude x = 1. The order of reaction with respect to reactant B is 1. This first order rate law depends linearly on only one reactant concentration, and follows the following rate law: \[rate = [A]\]

Q 12.5.11

An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate

Let us begin by making a what is given, and what is asked table:

The equation that encompasses what is being, and what is being given is the Arrhenius Equation at different temperature:

ln(k₂/k₁) = -E_a/R (1/T₂ - 1/T₁)

we do not know the k, but we do not it increases by 1.47. We can use this information to cancel the k's, by setting k₁=1 and k₂= 1.47(1) = 1.47

ln(1.47k2/k₁) = -E_a/R (1/T₂ - 1/T₁)

we cancel out k₁ (and convert temperature from C^o to K by adding 273) and get:

ln(1.47) = -Ea/8.314 J/mol·k (1/310 K - 1/303 K)

E_a = 4.3·10⁴ J/mol or 43 kJ/mol

Q 21.4.6

Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.

To approach this problem, it is necessary to understand the meaning of positron emission, beta decay, and alpha decay. Alpha decays happens with heavy nuclei (Z> 83). The daughter cell produced has a larger n:p ratio than the parent nuclide. If a parent nuclide is below the belt of stability, it can come closer to the belt of stability through an emission of alpha particles. A positron emission is a beta + decay in which a proton is converted into a neutron, positron (e+) and an antineutrino. This occurs with a parent nuclides that have a low n:p (neutron to proton) ratio and desire to approach the belt of stability by increasing n:p ratio of their daughter nuclide. Whereas beta decay occurs when the n:p ratio is large, and so the daughter nuclide has a lower n:p ratio.

a) Because the problem concerns heavy, unstable nuclides; The parent nuclide lying under the belt of stability will go through alpha decay to come closer to the belt of stability. Alpha decay typically occurs with very heavy elements, and is a form of radioactive decay in which the nucleus emits an alpha particle (helium-4 nucleus)

b) The parent nuclide lying above the belt of stability will go through beta decay to come closer to the belt of stability. Beta decay typically occurs above the belt of stability and consists of the nucleus emitting beta particles (electrons), resulting in a daughter nucleus that has a higher atomic number.

This section is covered in greater detail: here

Q 20.2.10

Balance each redox reaction under the conditions indicated.

1. MnO₄⁻(aq) + S₂O₃²⁻(aq) → Mn²⁺(aq) + SO₄²⁻(aq); acidic solution
2. Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq); acidic solution
3. Fe(s) + CrO42−(aq) → Fe2O3(s) + Cr2O3(s); basic solution
4. Cl₂(aq) → ClO₃⁻(aq) + Cl⁻(aq); acidic solution
5. CO₃²⁻(aq) + N₂H₄(aq) → CO(g) + N₂(g); basic solution

Problem # 1

1. Begin by giving oxidation numbers to each element:

MnO₄⁻(aq) + S₂O₃²⁻(aq) → Mn²⁺(aq) + SO₄²⁻(aq); acidic solution +7 -2 +2 -2 +2 +6 -2

2. Split into corresponding half reactions:

Oxidation: S₂O₃^2-→ SO₄^2-

Reduction: MnO₄^-→ Mn²⁺

3. Balance any element that is not oxygen or hydrogen first.

S₂O₃^2-→ 2SO₄^2-

4. Balance oxygen using H₂O

5 H₂O + S₂O₃^2-→ 2SO₄²

MnO₄^-→ Mn²⁺ + 4 H₂O

5. Balance hydrogen using H⁺ ions

5 H₂O + S₂O₃^2-→ 2SO₄² + 10 H⁺

8H⁺+ MnO₄^-→ Mn²⁺ + 4 H₂O

6. Balance change using e^-

5 H₂O + S₂O₃^2-→ 2SO₄² + 10 H⁺ + 8e^-

5e^-+ 8H⁺+ MnO₄^-→ Mn²⁺ + 4 H₂O

7. Make both equations have equal electrons

25 H₂O + 5S₂O₃^2-→ 10 SO₄² + 50 H⁺ + 40e^-

40e^-+ 64 H⁺+ 8 MnO₄^-→ 8 Mn²⁺ + 32 H₂O

8. Overall equation;

5 S₂O₃^2- + 8 MnO₄^- + 14H⁺→ 10 SO₄^2- + 8 Mn²⁺ + 7 H₂O

Check if your mass and charge are balanced.

*original solution was different, changes made in red.

Problem #2

1. Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq); acidic solution +2 +6 -2 +3 +3

2. Oxidation: Fe²⁺→ Fe³⁺

Reduction: Cr₂O₇^2- → Cr³⁺

3. Fe²⁺→ Fe³⁺

Cr₂O₇^2- → 2 Cr³⁺

4. Fe²⁺→ Fe³⁺

Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

5. Fe²⁺→ Fe³⁺

14 H⁺ + Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

6. Fe²⁺→ Fe³⁺ + e^-

6e^- + 14 H⁺ + Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

7. 6 Fe²⁺→ 6 Fe³⁺ + 6e^-

6e^- + 14 H⁺ + Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

8. 6 Fe²⁺ + 14 H⁺ + Cr₂O₇^2- → 6 Fe³⁺ + 2 Cr³⁺ + 7 H₂O

Problem # 3

1. Fe(s) + CrO₄²⁻(aq) → Fe₂O₃(s) + Cr₂O₃(s) 0 +6 -2 +3 -2 +3 -2

2. Oxidation: Fe→ Fe₂O₃

Reduction: CrO₄^2- → Cr₂O₃

3. 2 Fe→ Fe₂O₃

2 CrO₄^2- → Cr₂O₃

4. 3 H₂O + 2 Fe→ Fe₂O₃

2 CrO₄^2- → Cr₂O₃ + 5 H₂O

5. 3 H₂O + 2 Fe→ Fe₂O₃ + 6 H⁺

10 H⁺+ 2 CrO₄^2- → Cr₂O₃ + 5 H₂O

6. 3 H₂O + 2 Fe→ Fe₂O₃ + 6 H⁺ + 6e^-

6e^- + 10 H⁺+ 2 CrO₄^2- → Cr₂O₃ + 5 H₂O

7. 3 H₂O + 2 Fe + 4 H⁺+ 2 CrO₄^2-→Fe₂O₃ + Cr₂O₃ + 2 H₂O

This is where acidic acid solutions differ from basic solution; in acidic solution you would stop here. But in basic solutions, you must add OH

8. Add 4 OH^- on each side to make H₂O, and cancel out like terms.

2 Fe + 2H₂O + 2 CrO₄^2-→Fe₂O₃ + 4 OH- + Cr₂O₃

Problem # 4

1. Cl₂(aq) → ClO₃⁻(aq) + Cl⁻(aq) 0 +5 -2 -1

2. Oxidation: Cl₂→ ClO₃^-

Reduction: Cl₂→ Cl^-

3. Cl₂→ 2 ClO₃^-

Cl₂→ 2 Cl^-

4. 6 H₂O + Cl₂→ 2 ClO₃^-

Cl₂→ 2 Cl^-

5. 6 H₂O + Cl₂→ 2 ClO₃^- + 12 H⁺

Cl₂→ 2 Cl^-

6. 6 H₂O + Cl₂→ 2 ClO₃^- + 12 H⁺ + 10e^-

2e^-+ Cl₂→ 2 Cl^-

7. 6 H₂O + Cl₂→ 2 ClO₃^- + 12 H⁺ + 10e^-

10e^-+ 5 Cl₂→ 10 Cl^-

8. 6 H₂O + 6 Cl₂ → 2 ClO₃^- + 12 H⁺ + 10 Cl^-

9. Divide to get a reduced form of coefficents

3 H₂O + 3 Cl₂ → ClO₃^- + 6 H⁺ + 5 Cl^-

Problem # 5

1. CO₃²⁻(aq) + N₂H₄(aq) → CO(g) + N₂(g) +3 -2 -2 +1 +2 -2 0

2. Reduced: CO₃²⁻→ CO

Oxidized: N₂H₄→ N₂

3. CO₃²⁻→ CO + 2 H₂O

N₂H₄→ N₂

4. 4 H⁺ + CO₃²⁻→ CO + 2 H₂O

N₂H₄→ N₂ + 4 H⁺

5. 2e^- + 4 H⁺ + CO₃²⁻→ CO + 2 H₂O

N₂H₄→ N₂ + 4 H⁺ + 4e^-

6. 4e^- + 8 H⁺ + 2 CO₃²⁻→ 2 CO + 4 H₂O

N₂H₄→ N₂ + 4 H⁺ + 4e^-

7. 4 H⁺ + 2 CO₃²⁻+ N₂H₄→ 2 CO + 4 H₂O + N₂

8. Add 4 OH^-

2 CO₃²⁻+ N₂H₄→ 2 CO + N₂ + 4 OH^-

Q 20.5.5

State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.

This statement is true. Under equilibrium conditions, it is possible to determine the equilibrium constant. A numerical value of K will give you an indication if the reaction will proceed to the right or to the left. Moreover, you can use electrochemical methods to determine if a certain reaction is spontaneous or nonspontaneous; that is if your delta G is negative or positive. This triangle connects thermodynamics and equilibrium, as each is related to electrochemistry:

Note: This triangle concerns equilibrium, and nonstandard condition.

In regard to non-standard conditions, the Nerst equation is used. This follows the equation: \[E_{cell}=E^o_{cell}-\frac{0.0592}{n}logQ\] If you can conclude the cell potential, you can conclude the sign of the delta G as you do not need a numerical value to infer whether the considered reaction is spontaneous or nonspontaneous. If the reaction is spontaneous as it written, then you can deduce the reaction will eventually proceed to the right. If it is not, then the reversible reaction is favored.

Q 24.6.1

Describe crystal field theory in terms of its assumptions

a. regarding metal–ligand interactions

b. weaknesses and strengths compared with valence bond theory.

CFT, or crystal field theory explains the multiple arrangements that have the same energy (often termed degeneracy) in the d and f orbitals, due to the presence of metal ligands. According to the theory, the arrangement or symmetry of a ligand around the central metal can determine the magnetic and color properties of a ligand complex. Here is what the CFT theory assumes:

• The interaction between the central and ligand are considered electrostatic, meaning it considers the interactions (that is attraction or repulsion) between electrical charges
• For simplification, the ligands are considered point charges and so we do not consider them to be moving in space, nor do we consider them having volume or area.
• The electrons on the ligand are repulsed by those on the central atom

These assumptions are simple when considering a low charge metal, such as an alkali metal as they do not form many coordination complexes—transition metals, however, do. When you consider a non-spherical field with a transition metal, all the orbitals are no longer of the same energy; Due to the repulsion between the ligands and transition metal cations, the degenerate orbitals split; the dx²-y², z² (note: these lie on the axes, so they experience a greater repulsion) will be destabilized from the xy, xz, yz (as these come between the axes). The corresponding splitting is known as crystal field splitting. Here is an example of an octahedral complex:

For more practice, regarding this topic please visit this link

Where the electrons go is predicted by Hund’s rule. But for d⁴-d⁷, you could have low spin complex or high spin complex; which one it really is, is dependent on the strength of the ligand (see spectrochemical series)

b) The valence bond theory explains bonding; It states that when an atomic orbital overlap with another orbital, to form pi or sigma bonds, there is a formation of a covalent bond. By knowing the strength and geometry of this overlap, it is plausible to determine geometry and strength of the corresponding covalent bond.

While the VB theory describes the formation of the covalent bond, it does not cover magnetism or the spectrochemical series. Moreover it fails to explain the bond formation of CH4; according to the VB theory there should only be two C-H bonds. It also assumes that electrons are localized, meaning the electrons always stay close to the one atom.

CFT can explain the different colors exhibited by coordination complexes, and magnetism. But with CFT we assume that the ligands are point charges, and so it should follow that anionic ligands should have the greatest splitting effect—they are actually found to be on the low end of our spectrochemical series. It assumes the interaction between the ligand metal is ionic, and does not consider the covalent character.

Q 14.7.9

Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.)

a) An enzyme is a type of protein, which themselves are made of amino acids. The tertiary structure of the protein is made of ionic and hydrogen interactions. These interactions keep the active site of the enzyme in place. The binding site is what helps determines substrate specificity. Changing the concentration of H⁺ will change the interactions of the protein which will affect the specific binding site; consequently the substrate cannot bind, catalysis cannot take place, and the reaction will not continue.

b) Basic and Acidic amino acids have different functional groups; basic amino acids have an ammine function group, whereas acidic amino acids have a carboxyl functional group. The interaction these different functional groups have with an increased or decreased concentration of H⁺ ions would be different; as such even though the active site itself isn’t directly affected, there is a possibility that the enzyme changes its 3D shape and subsequently the interaction it has with the substrate