Extra Credit 31

Q17.4.4

Determine ΔG and ΔG° for each of the reactions in the previous problem.

Answer 17.4.4

ΔG (Gibbs free energy) is the amount of energy that is associated with a chemical reaction that can be used to do work under non-standard conditions. ΔG^o means the same as ΔG except the Knott (^o) means under standard conditions (1M, 25^o C, 1 atm). When ΔG or ΔG^o is negative, this means that the reaction occurs spontaneously in the forward reaction. If the answer is positive, this means the reaction occurs non-spontaneously in the forward reaction.

The two equations we are looking at are:

\[{\Delta G}= {-nFE_{cell}} \]

\[{\Delta G^{o}}= {-nFE^{o}_{cell}}\]

Based on what was previously said, you can see that you will use the first equation if the reaction is NOT under standard conditions (1M, 25^o C, 1 atm). However, if the reaction is under standard conditions, we will use the second equation. The n in the equations represents the number of electrons transferred between the to reactions. F stands for Faradays constant (96,485 C mol^-1). E_cell is the potential difference between two half reactions in a cell. It measures the ability of electrons to flow from one cell to another. E^o_cell is defined the same but, once again, the Knott (^o) represents under standard conditions. In relation to Gibbs free energy, if E_cell or E^o_cell is positive, this means that the Gibbs free energy will be negative and, therefore, spontaneous.

In order to determine ΔG and ΔG^o we must first determine the number of electrons transferred between the two half reactions. To do this we must balance each half reaction and balance the number of electrons transferred to find n.

a)

\[Hg(l)+S^{2-}(aq,\,0.10M)+2Ag^+(aq,\, 0.25M) ⟶ 2Ag(s) + HgS(s)\]

Following the rules of oxidation, we find that the oxidation reaction includes Hg(l)⟶ HgS(s) and the reduction reaction includes 2Ag (aq)⟶2Ag(s). After balancing the half reactions we find that 2 electrons are transferred amongst the reactions, therefore n=2. We are given the E^o_cell and E_cell potentials and F stands for Faradays constant. Now all we need to do is plug the numbers into the equation:

\[{\Delta G^{o}}= ({-2})({96,485C{mol^{-1}})(1.50V})\] = -289,455 J K^-1 mol^-1

\[{\Delta G}= ({-2})({96,485C{mol^-1}})({1.43V})\] = -275,947.1 J K^-1 mol^-1

Once again, a negative ΔG indicates a spontaneous reaction. Therefore, both occur spontaneously under standard and non-standard conditions. The larger the value of ΔG, the more energy is required to get from non-standard/ standard conditions to equilibrium.

b) You can determine which elements/compounds are apart of the oxidation and reduction reactions by looking at their standard reduction potentials. Whichever standard reduction potential is the most negative, that is the more favorable one which means it is included in the oxidation half reaction. We find that the aluminum electrode in 0.015 M aluminum nitrate solution is the oxidation reaction and the half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution is the reduction reaction. Through balancing the half reactions we find that the number of electrons transferred is 6, so n=6. Now we can plug in all the given numbers.

\[{\Delta G^{o}}= ({-6})({96,485C{mol^{-1}})(1.405V})\] = -813,368.55 J K^-1 mol^-1

\[{\Delta G}= ({-6})({96,485C{mol^{-1}})(1.423V})\] = -823,788.93 J K^-1 mol^-1

The answers tell us that the reaction will occur spontaneously under non-standard and standard conditions. Also, the reactions will require a lot of energy in order to reach equilibrium.

c) The question tells us what the oxidation and reduction reactions are and so our job now is to balance each half reaction and figure our how many electrons are transferred. Once you have balanced both reactions, the electrons transferred should be equal to n=6.

\[{\Delta G^{o}}= ({-6})({96,485C{mol^{-1}})(−2.749 V})\]

= 1,591,423.59 J K^-1 mol^-1

\[{\Delta G}= ({-6})({96,485C{mol{^-1}})(−2.757 V})\]

= 1,596,054.87 J K^-1 mol^-1

The amount of energy is very large which means it will take a lot of energy to reach equilibrium. However, this time the answers are positive which means the reactions will occur non-spontaneously.

Q12.1.4

A study of the rate of dimerization of C₄H₆ gave the data shown in the table:

\[{2C_{4}H_6}\rightarrow{C_{8}H_{12}}\]

1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
2. Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C₄H₆]. What are the units of this rate?
3. Determine the average rate of formation of C₈H₁₂ at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).

Answer 12.1.4

a) The average rate looks at the difference in concentration final and concentration initial divided by the difference of the final time and the initial time. The equation looks like:

\[rate=\dfrac{\Delta Concentration}{\Delta t} \]

\[rate=\dfrac{\Delta [C_4H_6]}{\Delta t}\]

Next, plug the given numbers into your equation.

\[rate=\dfrac{\Delta [5.04e{-3} M] - [1.00e{-2} M]}{\ 1600s - 0s}\]

\[rate=\dfrac{[-.00496M]}{\ 1600s}\]

= −3.1e−6 M/s

\[rate=\dfrac{\Delta [3.37e{-3} M] - [5.04e{-3} M]}{\ 3200s - 1600s}\]

\[rate=\dfrac{[-.00167 M]}{\ 1600s}\]

= −1.04375e−6M/s

The negative rates represent the slope of the line which means it can be negative indicating a downward slope.

b) The instantaneous rate of change of a certain concentration at a single moment so we need to find the slope of the tangent line at the point. In order to do this, we can take one point on the left side of the given point (3200s, 3.37e-3 M) and one point from the right side of the given point. To find the slope of the tangent, we must average the two other slopes to get the estimate instantaneous rate.

1. Plot the graph *I could not figure out how to paste the graph onto the edit sheet, errors kept popping up.

2. Find the slope of (1600s, 5.04e-3 M) and (3200s, 3.37e-3 M)

\[{[3.37e{-3} M]- [5.04e{-3} M]\over 1600s}\]

Calculate,

\[{-.00167M\over 1600s}\] = -1.04375e-6 M/s

3. Find the slope of (3200s, 3.37e-3M) and (4800s, 2.53e-7M)

\[{[2.53e{-3}M] - [3.37e{-3}M]\over 4800s - 3200s}\]

Calculate,

\[{[-8.4e{-4}M]\over 1600s}\] = -5.25e-7 M/s

4. Find the average of the slopes to find the instantaneous rate

\[{-1.04375e{-6} M/s + -5.25e{-7} M/s\over 2}\]

= -7.84e-7 M/s

The negative instantaneous rate means the function is decreasing at that point.

c) When looking at the rate of a reactant, we must put a negative sign in front of the equation so that the rate will always come out positive. Also, we see that 2 moles of C₄H₆ are used for every one mole of C₈H₁₂ is two times greater as the change in concentration of C₄H₆ is two times greater as the change in concentration of C₈H₁₂ over the same time and so we must show this in both of our equations:

average rate of formation:

\[-{1\over 2}\dfrac{\Delta [C_4H_6]}{\Delta t} = \dfrac{\Delta [C_8H_{12}]}{\Delta t}\]

Calculate,

\[-{1\over 2}({-3.1e-6 M/s}) = \dfrac{\Delta [C_8H_{12}]}{\Delta t}\]

= 1.55e-6 M/s

The formation of C₈H₁₂ occurs at a rate of 1.55e-6 M/s

instantaneous rate of formation:

\[-{1\over 2}\dfrac{d[C_4H_6]}{d t} = \dfrac{d[C_8H_{12}]}{d t}\]

Calculate,

\[-{1\over 2}({-7.84e{-7} M/s}) = \dfrac{d[C_8H_{12}]}{d t}\]

= -3.92e-7 M/s

The instantaneous rate of formation for C₈H₁₂ is -3.92e-7 M/s

Q12.5.2

When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?

Answer 12.5.2

The following factors determine the rate at which the reaction occurs: temperature, activation energy (E_a), concentration, surface area, and chemical nature.

Temperature: If you increase temperature, the particles will move faster causing more collisions to occur. This would then increase the rate of a reaction. On the other hand, if you decrease temperature, the particles will move slower/ less freely causing less collisions to occur, then the rate of the reaction decreases.

• Activation energy (E_a): Activation energy is the minimum amount of energy required for a reaction to occur. Therefore, if there is a large activation energy, the reaction needs more energy to overcome the barrier in order for the reaction to occur and so the reaction will be slower. If there is a small activation energy, the reaction needs less energy for the reaction to occur and so the reaction rate will be faster.
• Concentration: If there is a big concentration, then the distance between particles decreases which leads to more collisions, therefore, the rate of reaction increases. If there is a small concentration, the distance between particles increases which leads to less collisions, therefore, the rate of reaction increases.
• Surface area: Increasing the surface area of a reaction would expose more of the particles to the reactant which would cause more collisions to occur and then the rate of reaction increases. Decreasing the surface area of a reaction would expose less of the particles to the reactant which would cause less collisions to occur and then result in a slower reaction rate.
• Chemical nature: Gases tend to have faster reaction rates than liquids or solids because their particles can move more freely which means they can collide more often. Ionic compounds will react faster than molecular compounds. Bond strength affects the reaction rate because breaking weaker bonds require less energy which means the reaction rate increases. Also, breaking few number of bonds require less energy than breaking more bonds, therefore the reaction rate increases.

For the reaction to occur, there must be contacts between molecules. Concentration plays a key role in how often the reactants will collide and therefore form product. Other factors such as temperature and catalysts can make a large impact on the speed of the reaction and how often the molecules will collide. An increase in temperature would cause the molecules to move faster and therefore collide more often (increasing reaction rate.) A catalyst would similarly increase the rate of the reaction.

Q21.3.6

Technetium-99 is prepared from ⁹⁸Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as ⁹⁹Tc^*. This excited nucleus relaxes to the ground state, represented as ⁹⁹Tc, by emitting a γ ray. The ground state of ⁹⁹Tc then emits a β particle. Write the equations for each of these nuclear reactions.

Answer 21.3.6

1) \[{^{98}_{42}Mo} + {^{1}_{0}n}\rightarrow {^{99}_{42}Mo} \]

A neutron changes the mass by 1 because mass number includes the number of protons and the number of neutrons. However, a neutron does not change the number of protons. Therefore, we get the isotope Mo-99. The number of protons and the mass number should be equal on both sides of the reaction.

2) \[{^{99}_{42}Mo} \rightarrow {^{99}_{43}Tc^{*}} + {^{0}_{-1}\beta}\]

Since the isotope Mo-99 emits a beta particle, it is added to the right side of the reaction. A beta particle does not change the mass number because mass number does not include electrons, but it does change the number of protons by -1. The number of protons and the mass number should be equal on both sides.

3) \[{^{99}_{43}Tc^{*}} \rightarrow {^{0}_{0}\gamma} + {^{99}_{43}Tc}\]

The gamma particle does not change the mass number or the number of protons. Gamma particles attempt to return the element from an excited state (^*) to the ground state.

4) \[{^{99}_{43}Tc} \rightarrow {^{0}_{-1}\beta} + {^{99}_{44}Ru}\]

Again, the mass number and number of protons should be equal on both sides. Since it emits a beta particle, it goes on the right side.

Q20.2.2

If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?

Answer 20.2.2

If there is a poor reductant and a poor oxidant, the redox reaction is not likely to occur. This is because the poor reductant will not be strong enough to oxidize and the poor oxidant will not be strong enough to reduce. However, if there is a strong oxidant, that oxidant will be strong enough to cause a reduction. If there is a reduction, then there must be an oxidation as well. You cannot have one without the other. On the standard reduction potential table, species that have standard reduction potentials below H₂ are stronger reductants than H₂. Similarly, species above H₂ will be stronger oxidants. This will aid in determinig the strength of oxidants and reductants by their proximity to H₂.

Q20.4.21

Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.

1. Se(s) + Br₂(l) → H₂SeO₃(aq) + Br⁻(aq)
2. NO₃⁻(aq) + S(s) → HNO₂(aq) + H₂SO₃(aq)
3. Fe³⁺(aq) + Cr³⁺(aq) → Fe²⁺(aq) + Cr₂O₇²⁻(aq)

Answer 20.4.21

In order to find out which element is being oxidized and which element is being reduced, you must determine each element's oxidation number. Whichever element's oxidation number increases, that is the element involved in oxidation. If the an element's oxidation number is reduced, that element is involved in reduction. There are some rules for assigning oxidation numbers and some of those rules are: 1) Any element in its free state will have an oxidation number of zero. 2) If you are looking at a neutral compound, the oxidation numbers of all the elements in the compound must add up to zero. 3) A monoatomic ion's oxidation number will be the same as its charge. 4) The sum of oxidation numbers in a polyatomic ion must equal the ion's charge. Once you have determined which elements are involved in the oxidation and reduction reactions, you can begin to balance them. In order to balance an acidic solution, you must first balance the equation. To do this, check if all the elements are balanced. Then check and balance oxygen (if there is oxygen) by adding the correct amount of H₂O to the other side. Once you have balanced your oxygen and other elements, you must now balance hydrogen since you just added some through the addition of H₂O. Next, you want to determine the charge of all the reactants and the charge of all of the products, you want the charges on both sides to equal each other. If they are not already equal, you need to add the correct amount of electrons to the product side (for oxidation reaction) or to the left side (for reduction reaction). Then, if the electrons for the oxidation and reduction reaction are not the same, you must multiply or divide one or both equations so that the electrons are the same. Once you have done that you can add the two reactions together and balance the equation in order to get the overall reaction. Tah-Dah!!!

a)

\[Se(s) + Br_2(l) → H_2SeO_3(aq) + Br^−(aq)\]

We find that the oxidation number of Se and Br₂ are each zero, H₂ has an oxidation number of +1, Se has an oxidation number of +4, O₃ has an oxidation number of -2, and Br^- has an oxidation number of -1. We can see now that Se(s) is oxidized to H₂SeO₃(aq) and Br₂(l) is being reduced to Br⁻(aq). Now, we can follow the steps given above to balance each reaction and figure out the overall reaction.

Oxidation:

\[Se(s) → H_2SeO_3(aq)\]

Balance oxygen:

\[3H_2O(l) + Se(s) → H_2SeO_3(aq)\]

Balance hydrogen:

\[3H_2O(l) + Se(s) → H_2SeO_3(aq) + 6H^+\]

Balance charges/add electrons:

\[3H_2O(l) + Se(s) → H_2SeO_3(aq) + 6H^+ {+} 6e^-\]

Reduction:

\[Br_2(l) → Br^-(aq)\]

Balance elements:

\[Br_2(l) → 2Br^-(aq)\]

Balance charges/add electrons:

\[2e^- + Br_2(l) → 2Br^-(aq)\]

Balance both reactions electrons:

\[3H_2O(l) + Se(s) → H_2SeO_3(aq) + 6H^+(aq)+ 6e^-\]

\[3(2e^{-} + Br_2(l) → 2Br^{-}(aq)\] = \[6e^{-} + 3Br_2(l) → 6Br^{-}(aq)\]

Add reactions together:

\[3H_2O(l) + Se(s) + 6e^{-} + 3Br_2(l)→ H_2SeO_3(aq) + 6H^{+}(aq)+ 6e^{-}+ 6Br^-(aq)\]

Overall reaction:

\[3H_2O(l) + Se(s) + 3Br_2(l)→ H_2SeO_3(aq) + 6H^+(aq)+ 6Br^-(aq)\]

This reaction is spontaneous because the cell potential is 1.809 V which is positive and the cell potential needs to be positive for it to be spontaneous. This is found by subtracting the cell potential of the cathode by the cell potential of the anode and resulting in a positive number.

b)

\[NO_3^−(aq) + S(s) → HNO_2(aq) + H_2SO_3(aq)\]\

We find that the oxidation number of S(s) is zero. From NO₃⁻(aq), N has an oxidation number of +5 and O₃ has an oxidation number of -2. From HNO₂(aq), H has an oxidation number of +1, N has an oxidation number of +3, and O₂ has an oxidation number of -2. From H₂SO₃(aq), H₂ has an oxidation number of +1, S has an oxidation number of +4, and O₃ has an oxidation number of -2. We can see now that S(s) is oxidized to H₂SO₃(aq) and NO₃⁻(aq) is being reduced to HNO₂(aq). Now, we can follow the steps given above to balance each reaction and figure out the overall reaction.

Oxidation:

\[S(s) → H_2SO_3(aq)\]

Balance oxygen:

\[3H_2O(l) + S(s) → H_2SO_3(aq)\]

Balance hydrogen:

\[3H_2O(l) + S(s) → H_2SO_3(aq) + 6H^{+}(aq)\]

Balance charges/add electrons:

\[3H_2O + S(s) → H_2SO_3(aq) + 6H^{+} (aq)+ 6e^{-}\]

Reduction:

\[NO_3^{-}(aq) → HNO_2(aq)\]

Balance oxygen:

\[NO_3^{-}(aq) → HNO_2(aq) + H_2O(l)\]

Balance hydrogen:

\[3H^{+}(aq) + NO_3^{-}(aq) → HNO_2(aq) + H_2O(l)\]

Balance charge/add electrons:

\[2e^{-} + 3H^{+}(aq) + NO_3^{-}(aq) → HNO_2(aq) + H_2O(l)\]

Balance both reactions electrons:

\[3H_2O + S(s) → H_2SO_3(aq) + 6H^{+}(aq)+6e^{-}\]

\[3(2e^{-} + 3H^{+}(aq) + NO_3^{-}(aq) → HNO_2(aq) + H_2O(l)) = 6e^{-}+ 6H^{+}(aq) + 3NO_3^{-}(aq) → 3HNO_2(aq) + 3H_2O(l)\]

Add reactions together:

\[6e^{-} + 6H^{+}(aq) + 3NO_3^{-}(aq) + 3H_2O + S(s) →H_2SO_3(aq) + 6H^{+}(aq)+ 6e^{-} + 3HNO_2(aq) + 3H_2O(l)\]

Overall reaction:

\[3NO_3^{-}(aq) + S(s)→H_2SO_3(aq) + 3HNO_2(aq)\]

This reaction is spontaneous because the cell potential is positive. This reaction is spontaneous because the difference between the cell potential of the cathode and anode resulted in a positive value.

c)

\[Fe^{3+}(aq) + Cr^{3+}(aq) → Fe^{2+}(aq) + Cr_2O_7^{2-}(aq)\]

We find that the oxidation number of Fe³⁺ and Cr³⁺ is +3, Fe²⁺ has an oxidation number of +2, Cr₂ has an oxidation number of +6, and O₇ has an oxidation number of -2. From this we can tell that the oxidation reaction includes Cr³⁺(aq) → Cr₂O₇²⁻(aq) and the reduction reaction includes Fe³⁺(aq) → Fe²⁺(aq). Now we can follow the above steps to determine the overall reaction of the acidic soultion.

Oxidation:

\[Cr^{3+}(aq) → Cr_2O_7^{2-}(aq)\]

Balance Cr:

\[2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq)\]

Balance oxygen:

\[7H_2O (l) + 2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq)\]

Balance hydrogen:

\[7H_2O (l) + 2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq) + 14H^{+}\]

Balance charge/add electrons:

\[7H_2O (l) + 2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq) + 14H^{+} + 6e^{-}\]

Reduction:

\[Fe^{3+}(aq) → Fe^{2+}(aq)\]

Balance charge/add electrons:

\[e^{-} + Fe^{3+}(aq) → Fe^{2+}(aq)\]

Balance both reactions electrons:

\[7H_2O (l) + 2Cr^{3+}(aq) →Cr_2O_7^{2-}(aq) + 14H^{+} + 6e^{-}\]

\[6(e^{-}+ Fe^{3+}(aq) → Fe^{2+}(aq)) = 6e^{-} + 6Fe^{3+}(aq) → 6Fe^{2+}(aq)\]

Add reactions together:

\[6e^{-} + 6Fe^{3+}(aq) + 7H_2O (l) + 2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq) + 14H^{+} + 6e^{-} + 6Fe^{2+}(aq)\]

Overall reaction:

\[6Fe^{3+}(aq) + 7H_2O (l) + 2Cr^{3+}(aq) → Cr_2O_7^{2-}(aq) + 14H^{+} + 6Fe^{2+}(aq)\]

This reaction is not spontaneous because the cell potential is negative. Spontineity is determined by subtracting the cell potential of the cathode by that of the anode. Both are found in a redox tower.

Q20.9.6

Electrolysis is the most direct way of recovering a metal from its ores. However, the Na⁺(aq)/Na(s), Mg²⁺(aq)/Mg(s), and Al³⁺(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?

Answer 20.9.6

Electrolysis is the use of an electric current to stimulate a non-spontaneous reaction. Since Na⁺(aq)/Na(s), Mg²⁺(aq)/Mg(s) and Al³⁺(aq)/Al(s) have more standard electrode potentials than the standard electrode potential of water, these elements are more easily oxidized than water. In order for electrolysis to take place, the elements must be less reactive than water. These elements are higher up on the activity series than water which shows that they are more reactive. Therefore, electrolysis will not occur. No other reaction will occur because the elements are already in an oxidized state, it cannot be oxidized even further. This can be determined by looking at a table of standard reaction potentials. Sodium, Magnesium, and Aluminum have lower values than water.

Q14.5.2

For any given reaction, what is the relationship between the activation energy and each of the following?

1. electrostatic repulsions
2. bond formation in the activated complex
3. the nature of the activated complex

Answer 14.5.2

• Electrostatic repulsion vs E_a: Activation energy is the minimum amount of energy needed to overcome electrostatic repulsion so that new chemical bonds can be formed.
• Bond formation vs E_a: Any molecule that can overcome the activation energy will form bonds and form arrangements of atoms called the activated complex.
• Nature of the activated complex vs E_a: If the starting energy to the energy of the activated complex is high, then the energy of the activated complex will be high.

We are able to answer the above question keeping these themes in mind.

1. Activation energy signifies the minimum amount of energy reactants must have for a reaction to occur. A reaction can only occur if the reactants have energy to overcome electrostatic repulsion. This will be a result of electrostatic repulsion (interaction between electrical double layers of surrounding particles.)

2. In an activated complex, activation energy is used to initiate bond formation.

3. Initially, the energy of an activated complex will be high so long as the energy of the activated complex is high.