Extra Credit 39

Q17.5.7

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

S12.3.2

The definition of a battery is a device consisting of two or more electrochemical cells, which convert chemical energy into electrical energy to produce a current. There are two types: primary and secondary.

The Nernst Equation: E_cell = E°_cell - RT/nF ln Q

Where E_cell is the cell potential, E^°_cell is the standard cell potential, R is the gas constant, T is the temperature, F is Faraday's constant, n is the number of moles, and Q is the ratio of molar concentrations of product ions to reactant ions raised to the powers of their balancing coefficients.

Under standard conditions and at equilibrium, Q is equal to K and E^°_cell is equal to 0. Because E_cell = -nFE^°_cell this results in the following equation:

-nFE^°_cell = -RT ln K

E^°_cell = RT/nF ln K

Then, substitute RT/nF lnK for E^°_cell in the Nernst Equation and simplify to calculate E_cell:

E_cell = RT/nF ln K - RT/nF ln Q

E_cell = RT/nF (ln K - ln Q)

E_cell = RT/nF (ln K/Q)

It can be observed that RT/nF (ln K/Q) will approach zero, leading the cell potential to reach zero and resulting in a dead battery.

Q12.3.2

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?

S12.3.2

1. 2

If doubling a particular reactant concentration, makes the reaction rate quadruple, then the rate law is second-order dependent in that species.

2^x = 4

x = 2

2. 1

If tripling a particular reactant concentration, makes the reaction rate triple, then the rate law is first-order dependent in that species.

3^x = 3

x = 1

Q12.5.11

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?

S12.5.11

Use the equation: ln (K₂/K₁) = -E_a/R (1/T₂ - 1/T₁)

The rate of the reaction (K) increases 1.47 times, so we can assign the values for K₂ and K₁. For this example, let's assign K₂ the value of 1.47 and assign K₁ the value of 1. Given in the problem, T₁ is 30^°C and T₂ is 37^°C. R is the constant 8.3145 J/mol K.

Plug in the given values into the equation, making sure to convert Celsius to Kelvin and Joules to kilojoules.

ln(1.47/1) = -E_a/8.3145J/molK (1/310K - 1/303K)

E_a = 43.0 kj/mol

Q21.4.6

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.

S21.4.6

(a) If they are below the band of stability, they decompose to emit a positron in order to gain stability as they have more protons to the number of neutrons.

(b) If they are above the band of stability, they decompose to emit a beta particle as they have an excess of neutrons.

Q20.2.10

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

S20.2.10

For each redox reaction, separate it into the reduction half-reaction and oxidation half-reaction. Then balance it based off whether it's an acidic or basic solution.

1. MnO₄⁻(aq) + S₂O₃²⁻(aq) → Mn²⁺(aq) + SO₄²⁻(aq)

Step 1: Write the half-reactions.

reduction half-reaction: MnO₄⁻(aq) → Mn²⁺(aq)

oxidation half-reaction: S₂O₃²⁻(aq) → SO₄²⁻(aq)

Step 2: Balance the half-reactions by adding H₂O and H⁺

8H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4H₂O(l)

5H₂O(l) + S₂O₃²⁻(aq) → 2SO₄²⁻(aq) + 10H⁺(aq)

Step 3: Balance the charges using electrons.

5e^- + 8H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4H₂O(l)

5H₂O(l) + S₂O₃²⁻(aq) → 2SO₄²⁻(aq) + 10H⁺(aq) + 8e^-

Step 4: Balance the number of electrons in each half-reaction in order to cancel them out and balance the rest of the reaction.

8(5e^- + 8H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4H₂O(l))

5(5H₂O(l) + S₂O₃²⁻(aq) → 2SO₄²⁻(aq) + 10H⁺(aq) + 8e^- )

Step 5: Cancel out the common factors in H₂O and H⁺ and add the two half-reactions to get the final balanced reaction.

14H⁺ + 8MnO₄^-(aq) + 5S₂O₃²⁻(aq) → 10SO₄²⁻(aq) + 8Mn²⁺(aq) + 7H₂O(l)

2. Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq); acidic solution

6(Fe²⁺(aq) → Fe³⁺(aq) + e^-)

6e^- + 14H⁺(aq) + Cr₂O₇²⁻(aq) → 2Cr³⁺(aq) + 7H₂O(l)

14H⁺(aq) + 6Fe²⁺(aq) + CrO₇^2-(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

3. Fe(s) + CrO₄²⁻(aq) → Fe₂O₃(s) + Cr₂O₃(s); basic solution

3H₂O(l) + 2Fe(s) → Fe₂O₃(s) + 6H⁺(aq) +6e^-

6e^- + 10H⁺(aq)+ CrO₄²⁻(aq) → Cr₂O₃(s) + 5H₂O(l)

In a basic solution, H⁺ protons are balanced using OH^-

4OH^- + 4H⁺ + 3H₂O + 2CrO₄²⁺ + 2Fe(s) → Fe₂O₃(aq) + Cr₂O₃(s) + 2H₂O(l) + 6H⁺(aq) + 4OH^-

Combine the hydroxide ions and protons to form H₂O.

2H₂O(l) + 2CrO₄²⁺ (aq)+ 2Fe(s) → Fe₂O₃(aq) + Cr₂O₃(s) + 2H₂O(l) + 4OH^-(aq)

4. Cl₂(aq) → ClO₃⁻(aq) + Cl⁻(aq); acidic solution

6H₂O(l) + Cl₂(aq) → 2ClO₃⁻(aq) +12H⁺(aq) + 10e^-

5(2e^- + Cl₂(aq) → 2Cl⁻(aq))

6H₂O(l) + 6Cl₂(aq) → 2ClO₃⁻(aq) + 10Cl⁻(aq) + 12H⁺(aq)

5. CO₃²⁻(aq) + N₂H₄(aq) → CO(g) + N₂(g); basic solution

2e^- + 4H⁺ + CO₃²⁻(aq) → CO(g) +2H₂O(l)

N₂H₄(aq) → N₂(g) + 4H⁺ + 4e^-

4OH^- + 4H⁺ + 2CO₃²⁻(aq) + N₂H₄(aq) → 2CO(g) +4H₂O(l) + N₂ + 4OH^-

2CO₃²⁻(aq) + N₂H₄(aq) → 2CO(g) + N₂ + 4OH^-

Q20.5.5

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.

S20.5.5

I agree with this statement because for example, when looking at a reduction potential half-reaction chart, this tells us which reactions take place at an anode and a cathode in an electrochemical cell. Electrons will flow from anode to cathode because oxidation takes place at the anode and reduction takes place at the cathodes. The voltage as a result of this reaction determines what direction the reactions will go in. Sometimes, if we apply a little bit of work, it will be easy for the reverse reaction to occur.

Q24.6.1

Describe crystal field theory in terms of its

1. assumptions regarding metal–ligand interactions.
2. weaknesses and strengths compared with valence bond theory.

S24.6.1

1. D-subshell electrons determine how many electrons are in the outer shell in order to give to a higher energy level. If you have under three electrons, crystal field theory is an estimate of how many electrons transition metals have. We study tendencies because we don't know if the electrons are going to go high energy or low energy and because of that there are two possibilities so we study the tendencies to find which possibility will occur.

2. Valence bond theory is a weaker theory because it has no ability to gage where the electrons are and find them but crystal field theory is superior and has the ability to find the electrons.

Q14.7.9

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.)

S14.7.9

(a) A change in pH could affect the catalytic activity due effects at the catalytic center - reactants won't be able to bind as well to the catalyst and vice versa. The reactants - a change in pH, reactants bind at a catalyst because there are lone pairs in the catalyst. For example, in an acidic solution, a hydrogen atom will want an electron from the catalyst to become neutral, and inhibit the catalyst's ability to bind to other reactants and change the catalytic center.

(b) A catalyst is a protein meaning it is made up of amino acids which are held together by intermolecular forces. If the pH is changed enough and enough electrons are taken away from the catalyst, it will denature; however, this would take a large pH change.