Extra Credit 2

Q.19.1A

From the following observations, estimate the E° from the half reaction M⁺(aq)+e^-→M(s):

1. The metal M reacts with H₂SO₄(aq); but not with HI(aq); M displaces Au⁺(aq), but not Fe³⁺(aq).
2. The metal M reacts with HI(aq), producing H₂(g), but displaces neither Al³⁺(aq) nor Na⁺(aq).

S.19.1A

In order to complete this problem you will be referring to 11.2: Standard Reduction Potential to look at the chart it has with the E°_cell for half-reactions.

NOTE: If a metal dissolves in a certain solution, the standard reduction potential is smaller than the given E°_cell for the half reaction. In contrast, if the metal does not dissolve in a certain aqueous solution, its standard reduction potential will be greater than the E^o_cell for the half reaction.

1. Step 1: Since the metal dissolves in H₂SO₄, it can be determined that it has a reduction potential that is smaller than E°_{SO^2-4(aq)/SO2(g)}=0.17V. This can be determined by the placement of the half reactions on the chart. You want to look at all the possible half reactions that it does and does not react with and look at what there E°_cell is. Based off of the order of the chart it can be determined whether or not it will react with the items below or above the half reaction.

Step 2: It does not dissolve in HI, therefore it has to have a standard cell potential that is greater than E°_{H⁺(aq)/H2(g)}=0.

Step 3: It displaces Au⁺(aq) from solution, so its' standard reduction potential has to be smaller than E°_{Au⁺(aq)/Au(s)}=1.68 V.

Step 4: It does not displace Fe³⁺ thought, so it has to be larder than E°_{Fe³⁺(aq)/Fe²⁺(s)}=0.769V.

Step 5: Analyze all the information and determine the < and >. The minimum value is zero, so the standard reduction potential must be at least 0V. When comparing the other values, one can see that the largest value you can have possible is the standard reduction potential from step one, which is 0.17 V. Therefore, these two values would yield the range of E^o_cell that the metal could have.

Answer: 0V < E°_cell< 0.17V

2. For this problem you want to follow the same steps as above, determine what it can and cannot be greater than and use what you find to determine the < and >.

Step 1: The metal does dissolve in HI(aq) so the reduction potential must be smaller than E°_{H⁺(aq)/H2(g)}=0 V.

Step 2: It does not displace Al³⁺(aq), so the reduction potential is larger than E°_{Al³⁺(aq)/Al(s)}=-1.676V.

Step 3: In addition, it does not displace Na²⁺(aq), so the reduction potential is larger then E°_{Na⁺(aq)/Na(s)}=-2.7144 V

Step 4: Analyze all the information and determine the < and >

Answer: -1.676V<E°_cell<0V

Pay attention to the fact that it will not be the smallest to largest numbers; the idea is to determine what E°_cells form the boundaries.

Q19.27B

Find E°cell, ΔG°, K , and whether reaction goes to completion at 298 K and all substances at standard states for:

3HClO₂(aq) + 2Cr³⁺(aq) + 4H₂O(l) → 3HClO(aq) + Cr₂O₇^2-(aq) + 8H₃O⁺(aq)

S19.27B

First we need to find the oxidation and reduction reactions. If you need a reminder as of how to do this you can look at: Half-Reactions. In order to do this we must determine what is being reduced and what is being oxidized and must assign oxidation numbers (22.6: Assigning Oxidation Numbers).

• Oxidation Half-reaction: 2Cr³⁺(aq) + 7H₂O(l) → Cr₂O₇^2-(aq) + 14H₃O⁺(aq) + 6e^-

You can determine that this is the oxidation half-reaction because the charge of chromium goes from +3 on the reactant side to +6 on the product side. A quick shortcut: You can also tell it's the oxidation half reaction because of the addition of electrons on the product side.

• Reduction Half-reaction: 3HClO₂(aq) + 6H₃O⁺(aq) + 6e^- → 3HClO(aq) + 3H₂O (l)

This is the reduction half-reaction because chlorine goes from +3 on the reactant side to +1 on the reactant side. A quick shortcut: You can also tell it's the reduction half reaction because of the addition of electrons on the reactant side.

In order to find out E°_cell , you must look at the a standard reduction potential table, one can be found here 11.2: Standard Reduction Potential, and locate the half reactions on it. If what you are looking for isn't on one list it can be found on another.

• Oxidation Reaction: E°_cell = +1.33 V
• Reduction Reaction: E°_cell = +1.674 V

To find the total energy, E°_cell , subtract the cathode from the anode.

E°_cell = E_cathode - E_anode

*cathode= reduction and anode= oxidation

E°_cell = +1.674V - (+1.33V) = +0.344 V

Now that E°_cell has been determined, now you have to find ΔG°.

ΔG°= -nFE°_cell

• F is a constant = 96,485 J
• E°_cell was already determined= +0.344 V
• n is equal to the moles of electrons transferred. In order to determine this, refer back to the reduction and oxidation half reactions already found. It is clear that 6e^- were transferred. Therefore, n=6

So, ΔG°= (-6e^-)((96485 C / (mol e^-)) (+0.344V) = - 1.99 x 10⁵ J

ΔG° = -1.99 x 10⁵ kJ

Next, determine K.

lnK_eq= - ΔG°/ RT

• R is a constatant = 8.3145 J K^-1mol^-1
• ΔG°= -1.99 x 10⁵ J (-199.15 kJ)
• T is given in the problem: T= 298 K

lnK_{eq =} - (-199.15 KJ) (1000) / (8.3145 J K^-1mol^-1) (298K) = 80.4

In order to cancel out ln we must do e^ln . So we take e^x on both sides.

e^lnK= e^80.4

K= 8.02 x 10³⁴

K is the equilibrium constant. When K> 1 then the products dominate and if K< 1 than the reactants dominate. Since K is a very large number, the products certainly dominate, meaning the reaction will go to completion.

Q20.11C

Complete and balance the equation: Cr₂O₃(s) + Al(s) →

S20.11C

First, determine what the products will be.

• Al will bond to O₃ to form a metal-oxide; it will not bond to Cr because than it would be two metals. Since it is a metal oxide, it will form a solid.
• Since Al bonds to O₃, Cr will not bond with anything. As a result, it will also be a solid.

So, Cr₂O₃(s) + Al(s) → Al₂O₃(s) + Cr(l)

*there are 2 Al because O has an oxidation number of -2, therefore it is -6 and Al has an oxidation number of +3, but it needs to equal 0 so we need to Al to give us +6.

Now, balance the equation.

• Since there are 2 Cr we must put a 2 in front of the Cr on the product side.
• There are 2 Al on the product side so a 2 is needed on the reactants side.

The final balanced equation is:

Cr₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s)

Q21.21A

Predict:

1. Which of the following complex ions is paramagnetic and which is diamagnetic? [CoCl₂]^2- and [Ni(CN)₄]^2-
2. What is the number of unpaired electrons in the tetrahedral complex, [CoCl₂]^2-?

S21.21A

Reminder: Diamagnetic is when the valence shell is filled completely with electrons, all are paired. While with paramagnetic there are unpaired electrons.

1. Start with [CoCl₂]^2-

- Find the charge of the cobalt ion.

• There are 2 Cl^- ions
• The overall charge is 2^-
• Co must have no charge

- Now determine the d electrons of Co. Since Co has no charge, we will count the d electrons on the periodic table without having to subtract or add. Co= d⁷.

- Next, determine if it is weak or strong field by referring to spectrochemical series. Based off of the list, Cl^- has a weak attractive force, so this has a weak ligand field. Therefore, electrons will fill all the rows before pairing electrons and then pair with leftover electrons. Draw a crystal field diagram to determine if it is diamagnetic or paramagnetic.

• After drawing it, it becomes clear that it is paramagnetic due to the three unpaired electrons.

Now, repeat this process again but for [Ni(CN)₄]^2-

- Find the charge of the Nickel ion

• There are 4 CN^- ions
• The overall charge is 2-
• Ni must have a charge of +2

- Determine the d electrons of Co. Since Ni has a charge of 2, you must subtract two electrons. You always subtract from the s orbitals before the d orbitals. So Ni has 8 electrons in the d orbital.

- Based off of the spectrochemical series, determine if it is weak or strong field. CN- is strong field, therefore it will fill all of the bottom before the top.

• After drawing it, it becomes clear that it is also paramagnetic due to the two unpaired electrons.

2. Part two is asking how many unpaired electrons there are for [CoCl₂]^2-. By looking at the crystal field splitting diagram it is evident that there are three unpaired electrons.

Q24.33A

The reaction A+B→C+D is second order in A and zero order in B. The value of k is 0.0107 M^-1 min^-1. What is the rate of this reaction when [A]=0.106M and [B]=3.73M?

S24.33A

In order to find the rate you multiply k by A and B.

R= k[A]^m[B]ⁿ

• But, since B is zero order, it is going to equal one no matter what value that is given.
• Also, since A is second order, you must square the value given.

*If this isn't ringing any bells and you need a little bit of a reminder of how to determine this, look at 14.3: Determining Rate Laws from Initial Rates (Differential Rate Laws)

Therefore,

R= (0.0107M^-1min^-1)[0.106M]²[3.73M]⁰ = 1.20x10^-4 M/min

Q25.11A

Write equations for the following nuclear reactions:

1. Bombardment of ²³⁴Th with α particles to produce ²³⁸U
2. Bombardment of ²³⁴Pa with hydrogen atoms to produce ²³⁵U
3. Bombardment of ²³⁸U with neutrons to produce ²³⁹U and γ rays and then β-decay of ²³⁹U

S25.11A

1. Step 1: To begin, you want to determine the number of protons that Th has. By locating Th on the periodic table it can be determined that it has 90 protons.

Step 2: Since it is the bombardment of an alpha particle, an alpha particle is another reactant. An alpha particle has no charge but has a molar mass of about 4 and 2 protons. He is used as an alpha.

Step 3: The molar mass and protons need to be equal on both sides of the equation, so add the molar masses from Th and He, 234+4=238, and add the protons from Th and He, 90+2=92.

Step 4: Now look at the protons you have, 92, and locate the element on the periodic table that has 92 protons. U has 92 protons.

The final equation is: ²³⁴₉₀Th + ⁴₂He → ²³⁸₉₂U

2. Follow similar steps at before.

Step 1: Determine the number of protons Pa, 91.

Step 2: Hydrogen has 1 proton and a molar mass of 1.

Step 3: Balance the molar masses and protons. 234+1=235, so 235 neutrons. 91+1= 92, so 92 protons.

Step 4: Look at periodic table to find the element with 92 protons; U.

The final equation is: ²³⁴₉₁Pa + ¹₁H → ²³⁵₉₂U

3a. For part a, focus on the bombardment of neutrons to produce ²³⁹U and γ rays.

Step 1: A neutron has a molar mass of 1 and no protons.

Step 2: γ has no molar mass nor protons.

Step 3: Determine the protons for ²³⁸U. Which is 92.

Step 4: Add the molar masses together, 238+1=239. This finds ²³⁹U, one of the products.

Step 5: The number of protons for U clearly does not change, so it has 92.

The final equation is: ²³⁸₉₂U + ¹₀n → ²³⁹₉₂U + γ

3b. Part b focuses on β-decay of ²³⁹U.

Step 1: Determine protons of U, 92.

Step 2: ß particles have no molar mass but have -1 protons. β-decay takes place on the products side. Therefore, the other product must have one more proton than U to ensure that the number of protons will be the same on both sides. So the other product will have 93 protons.

Step 3: Look at the periodic table to see what element has 93 protons; Np.

Step 4: Balance out molar masses. ß particles have no molar mass so 239+0=239.

The final equation is: ²³⁹₉₂U → ²³⁹₉₃Np + ⁰_-1β

*If any of this was confusing refer to Nuclear Chemistry.

Q25.42B

Which isotope is more likely to be the most abundant in nature? Explain.

a. ¹⁴₆C vs. ¹²₆C

b. ³⁵₁₇Cl vs. ³⁶₁₇Cl

c. ⁹⁶₄₄Ru vs. ⁹⁵₄₄Ru

S25.42B

a. In order to determine which will be most abundant in nature, you must figure out which one is more stable. The more stable the isotope, the more abundant it is going to be. You can determine which one is more stable by figuring out the ratio of neutrons to protons. For elements with Z ≤ 20, a 1:1 ratio of protons to neutrons is favorable. Since C is less than 20, this ratio would be favored. In order to determine the ratio, subtract the molar mass from the protons for both isotopes. 14-6= 8 while 12-6=6. Therefore, ¹²₆C will be more abundant in nature due the protons and neutrons both equaling 6, yielding a 1:1 ratio.

b. For isotopes over 20, the best way to determine which is most abundant is based off of the fact that an even amount of neutrons and an even amount of protons will be the isotopes that are more stable, therefore are more abundant. The more stable in nature, the more abundant they will be. Since both of these isotopes have odd numbers of protons, that does not help narrow down the answer. So look at the neutrons. In order to determine the neutrons subtract the molar masses from the protons: 35-17=18 and 36-17=19. Since ³⁵₁₇Cl has an even number of neutrons, this isotope is more abundant in nature.

c. This problem is similar to part b considering Z is greater than 20. It is evident that they both have an even number of protons, making them both semi stable. To determine which one is most stable you must find the number of neutrons. 96-44=52 and 95-44=51. Since both the protons and neutrons are even numbers, ⁹⁶₄₄Ru is most abundant in nature since it is most stable.

Q21.2.31

Calculate the amount of energy that is released by the neutron-induced fission of ²³⁵U to give ¹⁴¹Ba, ⁹²Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.

S21.2.31

In order to calculate the amount of energy that is released you are going to use: E= mc²

• m is the difference in mass
• c is a constant

Step 1: Write out the equation:

²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n

Step 2: Determine the masses of each reactant and product

• ²³⁵U= 235.0439
• ¹⁴¹Ba= 140. 9144
• ⁹²Kr= 91.9262
• n= 1.008664

Step 3: Plug them in to find the difference in mass

Difference in mass= M_products - Mreactants

(140.9144 amu + 91.9262 amu + (3 x 1.008664 amu))-(235.0439 amu +1.008644 amu)= -.185972 amu

Step 4: Plug everything into the equation

E=(-.185972 amu)(3.00 x 10⁸ m/s)²(1.6606x10^-27 kg/amu)(1 J*s² / kg*m²) = -2.78 x 10^-14 J/mol

E= (-.185972 amu)(931.5 MeV/amu) = -173.2 MeV/atom