Extra Credit 19

Introduction:

On this page, the original questions are indicated with black Q's, the student-created tutorial is provided with a green A. Corrections by Lauren are made in purple font.

This page was created by Gopika Prabhu, and edited by Lauren Reynolds for Dr. Larsen's Chem 2C class (2017 Spring).

Q17.2.8

An aqueous solution containing the following ions has 2.15 A of current passed through for 60 min, how much metal (in g) is formed at the cathode in each solution?

1. \(Zn^{2+}\)
2. \(Al^{3+}\)
3. \(Ag^+\)
4. \(Ni^{2+}\)

A17.2.8

To solve this problem, use the equation \[Q=I\times t=n\times F\] where Q is charge (in Coulombs), t is time (seconds), I is current (in Amperes), n is moles of electrons, and F is Faraday's constant. In this case, the current passes through the solution for a total of 60 minutes, which is equivalent to 3600 seconds. We are given the current: 2.15 A. Also, note that the Faraday's constant is 96485 C/mol. In order to find how much metal is formed at the cathode in each solution, we have to solve for the moles of electrons transferred during the reaction, and then convert to grams of the metal using dimensional analysis. mol. In order to find how much metal is formed at the cathode in each solution, we have to solve for the moles of electrons transferred during the reaction, and then convert to grams of the metal using dimensional analysis.

First, solve for the number of electrons given in the equation.

\[n=I\tfrac{t}{F}=2.15\;C/s\tfrac{60 s}{96485\;C/mol}=1.34\times 10^{-3}\;mol\]

CORRECTED VERSION: \[n=I\tfrac{t}{F}=2.15\;C/s\tfrac{3600 s}{96485\;C/mol}=8.02\times 10^{-2}\;mol\]

A) For this question, we want to first write the half-reaction for the formation of Zn from its ion. \[Zn^{2+}+2e^-\longrightarrow\ Zn(s)\] Now, we know that the formation of Zn(s) requires the supply of 2 moles of electrons for every 1 mole of Zinc plated. Now that you know the number of moles of electrons, you can solve for the number of grams of Zinc using dimensional analysis. Now, we know that the formation of Zn(s) requires the supply of 2 moles of electrons for every 1 mole of Zinc plated. Now that you know the number of moles of electrons, you can solve for the number of grams of Zinc using dimensional analysis

\[1.34\times 10^{-3}\;mol\;e^-\times \dfrac{1\;mol\;Zn (s)}{2\;mol\;e^-}\times \dfrac{65.38\;g\;Zn}{1\;mol\;Zn}=0.0437\;g\;Zn\]

CORRECTED VERSION: \[8.02\times 10^{-2}\;mol\;e^-\times \dfrac{1\;mol\;Zn (s)}{2\;mol\;e^-}\times \dfrac{65.38\;g\;Zn}{1\;mol\;Zn}=2.62174\;g\;Zn\]

B) For this question, we want to first write the half-reaction for the formation of Al from its ion. \[Al^{3+}+3e^-\longrightarrow\ Al(s)\] Now, we know that the formation of Al(s) requires the supply of 3 moles of electrons for every 1 mole of Al plated.longrightarrow\ Al(s)\] Now, we know that the formation of Al(s) requires the supply of 3 moles of electrons for every 1 mole of Al plated.

Now that you know the number of moles of electrons, you can solve for the number of grams of Al using dimensional analysis

\[1.34\times 10^{-3}\;mol\;e^-\times \dfrac{1\;mol\;Al (s)}{3\;mol\;e^-}\times \dfrac{26.98\;g\;Al}{1\;mol\;Al}=0.0120\;g\;Al\]

CORRECTED VERSION: \[8.02\times 10^{-2}\;mol\;e^-\times \dfrac{1\;mol\;Al (s)}{3\;mol\;e^-}\times \dfrac{26.98\;g\;Al}{1\;mol\;Al}=0.721265\;g\;Al\]

C) For this question, we want to first write the half-reaction for the formation of Ag from its ion. \[Ag^{+}+e^-\longrightarrow\ Ag(s)\] Now, we know that the formation of Ag(s) requires the supply of 1 mole of electrons for every 1 mole of Ag plated.longrightarrow\ Ag(s)\] Now, we know that the formation of Ag(s) requires the supply of 1 mole of electrons for every 1 mole of Ag plated.

Now that you know the number of moles of electrons, you can solve for the number of grams of Ag using dimensional analysis

\[1.34\times 10^{-3}\;mol\;e^-\times \dfrac{1\;mol\;Ag (s)}{1\;mol\;e^-}\times \dfrac{107.87\;g\;Ag}{1\;mol\;Ag}=0.144\;g\;Ag\]

CORRECTED VERSION: \[8.02\times 10^{-2}\;mol\;e^-\times \dfrac{1\;mol\;Ag (s)}{1\;mol\;e^-}\times \dfrac{107.87\;g\;Ag}{1\;mol\;Ag}=8.6512\;g\;Ag\]

D) For this question, we want to first write the half-reaction for the formation of Ni (s) from its ion. \[Ni^{2+}+2e^-\longrightarrow\ Ni(s)\] Now, we know that the formation of Ni(s) requires the supply of 2 moles of electrons for every 1 mole of Ni plated.longrightarrow\ Ni(s)\] Now, we know that the formation of Ni(s) requires the supply of 2 moles of electrons for every 1 mole of Ni plated.

Now that you know the number of moles of electrons, you can solve for the number of grams of Ni using dimensional analysis

\[1.34\times 10^{-3}\;mol\;e^-\times \dfrac{1\;mol\;Ni (s)}{2\;mol\;e^-}\times \dfrac{58.69\;g\;Ni}{1\;mol\;Ni}=0.0392\;g\;Ni\]

CORRECTED VERSION: \[8.02\times 10^{-2}\;mol\;e^-\times \dfrac{1\;mol\;Ni (s)}{2\;mol\;e^-}\times \dfrac{58.69\;g\;Ni}{1\;mol\;Ni}=2.3535\;g\;Ni\]

STUDENT'S ERROR: Failure to convert time in min to time in sec.

Q19.1.17

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

1. \(Fe(s)+H_2SO_4(aq)\longrightarrow\ ?\)
2. \(FeCl_3(aq)+NaOH(aq)\longrightarrow\ ?\)
3. \(Mn(OH)_2(s)+HBr(aq)\longrightarrow\ ?\)
4. \(Cr(s)+O_2(g)\longrightarrow\ ?\)
5. \(Mn_2O_3(s)+HCl (aq)\longrightarrow\ ?\)
6. \(Ti(s)+F_2(g)\longrightarrow\ ?\)

A19.1.17

1. For this problem, you want to look at the activity series. Because Fe is higher up on the activity series than H, you know that the oxidation of the Fe (s) will displace the H ions out of solution. Now, because you know that this is a redox reaction, the first step you should take is constructing two half reactions to balance final equation. \[\underbrace{Fe(s)\longrightarrow\ Fe^{2+} (aq)+ 2e^-}_{\text{Oxidation}}\] \[\underbrace{H_2SO_4(aq)+2e^-\longrightarrow\ SO_4^{2-}(aq)+H_2(g)}_{\text{Reduction}}\] From this, you should see that the electrons cancel out when you combine the two half reactions. From this, you should see that the electrons cancel out when you combine the two half reactions. Now the SO₄^2- , a spectator ion in second half reaction, can be written as an aqueous form of FeSO₄ when combined with the aqueous iron ions. Finally, you get: \[Fe(s) +H_2SO_4 (aq)\longrightarrow\ FeSO_4(aq)+H_2(g)\] A side note, the H⁺ ions are sometimes written as H₃O^-, but if you use the H₃O^- ions in your balancing, you must remember to balance the O accordingly. Your products should contain 2 more moles of water, but the same stoichiometric amounts of H₂ and FeSO₄ .
2. At a glance, the first thing you should notice is that there are two metals reacting. Your next step should be to either consult an Activity series table, or to examine the metals' standard reduction potentials. Fe is lower on the activity series than Na, so we know that the higher metal pushes the lower metal out of solution. If we consider this a double-displacement reaction, we can try switching the anions and see if Fe precipitates. This proposed reaction (once balanced) should look like \[FeCl_3(aq)+3NaOH(aq)\longrightarrow\ Fe(OH)_3(?)+3NaCl(aq)\] After consulting solubility rules, we discover that Fe(OH)₃ is insoluble, therefore we know that Fe is solid and is thus displaced out of solution. \[FeCl_3(aq)+3NaOH(aq)\longrightarrow\ Fe(OH)_3(s)+3NaCl(aq)\]
3. Again, the first thing you should notice is that there are two metals reacting here. After consulting the Activity Series table, we see that Mn is above H, so Mn ions will displace H ions out of solution. Try writing a double-displacement reaction where Mn and H switch anions and see if anything precipitates out. The proposed balanced reaction should look like \[Mn(OH)_2(aq)+2HBr(aq)\longrightarrow\ MnBr_2 (aq)+2H_2O (l)\] Notice that water is not aqueous, rather it is considered liquid. This means that H (in the form of water) precipitated out of aqueous solution, and only Mn cations remain. As mentioned above, add 2 waters to the product side if you used Hlongrightarrow\ MnBr_2 (aq)+2H_2O (l)\] Notice that water is not aqueous, rather it is considered liquid. This means that H (in the form of water) precipitated out of aqueous solution, and only Mn cations remain. As mentioned above, add 2 waters to the product side if you used H₃O^- instead of H⁺ to balance for the O.
4. This is a redox reaction, as both Cr and O are currently having oxidation numbers of 0, but we know that O is very electronegative. The most stable ion of Cr is Cr⁶⁺ , where Cr has the electron configuration of Ar. Knowing this, we can design two half reaction equations and balance them. \[Cr(s)\longrightarrow\ Cr^{6+}+6e^-\; Oxidation\] \[O_2(g)+4e^-\longrightarrow\ 2O^{2-}\; Reduction\] Now we must balance them both so that their electrons cancel out and combine anions with cations. Multiply the Cr half reaction by 2 and the O half reaction by 3 so that you can cancel out 12 electrons. Then, when you combine the equation, also combine anions with cations and you get: \[2Cr(s)+3O_2(g)\longrightarrow\ Cr_2O_3(s)\] CORRECTION: This is a synthesis reaction, meaning the Cr and the O₂ combine to form one compound. In this case, a metal is reacting with oxygen, which forms a metal oxide. Since Cr has several oxidation states, this reaction has several outcomes. One example is \[4Cr(s) + 3O_2 (g) \longrightarrow\ 2Cr_2O_3 (s)\] In this case, chromium has a +3 oxidation state, which is one of the most common oxidation states observed for Chromium. Other common oxidation states of Chromium are +2 and +6.
5. The first thing you should notice is that this reaction involves a metal oxide and hydrochloric acid. Again, we can set up an equation with a double displacement reaction as our proposed reaction \[Mn_2O_3(s)+?HCl(aq)\longrightarrow\ ?MnCl_? (aq)+?H_2O (l)\] Balancing this reaction is a bit tricky, so we will do it step-by-step. Examining the reactants, we find that there are 3 O^2-, meaning that it has to be balanced by a +6. Because there are 2 Mn, each Mn must have an oxidation state of +3. This means on the product side there are 2 Mn. Since this is not a redox equation, we can write that the product Mn also has an oxidation state of +3, so there must be 3 Cl^- to counter that. If there are 3 Cl^-, there must be 6 total Cl^-, so there must be 6HCl reactant. Now we know that there are 6H⁺ ions reacting with 3O^2-ions, which makes 3 H₂O. This means our final equation would be:\[Mn_2O_3(s)+6 HCl(aq)\longrightarrow\ 2MnCl_3 (aq)+3H_2O (l)\] As mentioned previously, if you used H₃O⁺, you must balance your produced water accordingly.
6. In this problem we see a halogen reacting with a metal. This is obviously a redox equation, because both reactants have an oxidation number of 0, but we know that both are reactive with each other as Ti has a relatively high metallic character and F has a very electronegative character based on their positions on the periodic table. Since Ti has a relatively high metallic character, it wants to lose electrons and be oxidized. On the other hand, F is a nonmetal that wants to gain electrons (be reduced). In order for Ti to be stable, it must ionize to become Ti⁴⁺. The first thing we are going to do, based on all this information, is design two half reactions and balance them. \[Ti\longrightarrow\ Ti^{4+}+4e^-\; Oxidation\] \[2F_2(g)+2e^-\longrightarrow\ 4F^-\; Reduction\] Combining these equations, we get the total balanced equation \[Ti(s)+2F_2(g)\longrightarrow\ TiF_4(g)\] \[2F_2(g)+2e^-\longrightarrow\ 4F^-\; Reduction\] Combining these equations, we get the total balanced equation \[Ti(s)+2F_2(g)\longrightarrow\ TiF_4(g)\]

Q19.3.9

Trimethylphosphine, P(CH₃)₃, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH₃)₃ can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?

A19.3.9

We know that the molecular mass of the compound is 270 grams/mol. This means we can multiply the percentages of mass of each element/ligand to find the mass of the Ni, Cl, and P(CHmol. This means we can multiply the percentages of mass of each element/ligand to find the mass of the Ni, Cl, and P(CH₃)₃ .

\(Ni\) \[270g\times 0.215\;Ni=58.05 g\;Ni\]

\(Cl\) \[270g\times 0.26\;Cl=70.2 g\;Cl\]

\(P(CH_3)_3\) \[270g\times 0.525\;P(CH_3)_3=141.74g\;P(CH_3)_3\]

Now we know the masses of each in the molecule, so divide each by its atomic mass as given by the periodic table. This will give you the number of moles of each you need to make this compound.

\[58.05 g\;Ni\times \dfrac{1\;mol\;Ni}{58.693g\;Ni}\approx 1mol\;Ni\]

\[70.2 g\;Cl\times \dfrac{1\;mol\;Cl}{35.453g\;Cl}\approx 2mol\;Cl\]

\[141.74g\;P(CH_3)_3\times \dfrac{1\;mol\;P(CH_3)_3}{76.0773g\;P(CH_3)_3}\approx 2mol\;P(CH_3)_3\]

So we find that we have in one molecule, 1 mol Ni, 2 mol Cl, and 2 mol P(CHmol Ni, 2 mol Cl, and 2 mol P(CH₃)₃ . From the description above, trimethylphosphine appears to be monodentate, and a consultation with a table containing common neutral ligands confirms this. This means that this complex has a coordination number of 4, so the geometry is either tetrahedral or square planar. Because it has no isomers, this molecule cannot be square planar, so it is tetrahedral. The problem states that the blue compound has no isomers. If this were a square planar complex, there would be both cis and trans iomers present, so we can rule out this geometry. This leads us to conclude that the geometry of this complex is tetrahedral since tetrahedral complexes cannot have isomers for MA₂B₂ complexes since each ligand is 109.5 degrees from the metal center. The molecular formula is [NiCl₂(P(CH₃)₃)₂].

Q12.4.9

What is the half-life for the decomposition of O₃ when the concentration of O₃ is 2.35 × 10⁻⁶ M? The rate constant for this second-order reaction is 50.4 L/mol/h.mol/h.

A12.4.9

We know that the half-life equation is \[t_{1/2} = \dfrac{1}{k[A]_0­}\] for second-order reactions. A₀ is the initial concentration (2.35 × 10⁻⁶ M) and k is (50.4 L/mol/h). Plug these two values into the equation and you get: \[t_{1/2} = \dfrac{1}{(50.4 Lm^{-1}h^{-1})\times (2.35\times 10^{-6}M))}=8440 hrs\] 8440 hours is the amount of time it takes for 2.35 × 10⁻⁶ M to decompose by half. This answer has been adjusted to 3 significant figures based on the lowest number of significant figures in the provided values.

Q21.2.4

For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.

1. \(^{34}_{14}Si\)
2. \(^{36}_{15}P\)
3. \(^{57}_{25}Mn\)
4. \(^{121}_{56}Ba\)

A21.2.4

For all of these, note that the isotope is a neutral atom, not a cation nor an anion. This means that the oxidation charge will be zero. No electrons were lost nor gained, so the number of electrons should be identical to the number of protons.

1. Using the periodic table, you know that Si has 14 protons. In order to find the number of neutrons, subtract the number of protons from the given mass number, 34. Because the difference between 34 and 14 is 20, this isotope has 20 neutrons. Since Si is not ionized we can assume that there are 14 electrons (same as protons).
2. From the given number (and the periodic table), you know that P has 15 protons. To find the number of neutrons, subtract the atomic number from the mass number given, 36, and you get 21 neutrons because 36-15=21. Since P is not ionized, we can assume that there are 15 electrons (same as protons).
3. Mn has 25 protons by definition since its atomic number is 25 on the periodic table. Subtract the atomic number from the mass number given, 57, and you get 32 neutrons since 57-25=32. Since Mn is not ionized, we can assume that there are 25 electrons (same as protons).
4. First, determine the number of protons Ba has. Since Ba has an atomic number of 56, Ba has 56 protons by definition. Subtract the atomic number from the mass number given, 121, and you get 65 neutrons since 121-56=65. Since Ba is not ionized, we can assume that there are 56 electrons (same as protons).

Q21.5.7

Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.

A21.5.7

When Uranium undergoes fission after being bombarded by neutrons, a lot of energy is released. Fission is the process by which a heavier atom is split into two lighter ones. This energy heats up water, which releases steam to turn turbines. The mechanical energy derived using an electrical generator hooked up to the turbine is then used to supply energy we can use.

Q20.4.5

Write a cell diagram representing a cell that contains the Ni/Ni²⁺ couple in one compartment and the SHE in the other compartment. What are the values of E°_cathode, E°_anode, and E°_cell?

A20.4.5

To write the cell diagram, you first want to write the half-reactions corresponding to the cell to determine the natural anode and cathode, and determine the reactants and products. Our half reactions are the following: \[Ni(s)\longrightarrow\ Ni^{2+}(aq)+2e^-\] \[2H^+(aq)+2e^-\longrightarrow\ H_2(g)\] To determine spontaneity, we consult the Standard Reduction Potentials table and find that Ni(s) spontaneously oxidizes itself (as it has an E^º value of -0.257 V) so it must occur at the anode. This puts the SHE at the cathode because it has a more positive reduction potential of 0.00 V. Now we know that E^º_anode=-0.257 V, and E^º_cathode= 0.00 V. Using the equation \[E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.00V-(-0.257)V=0.257V\]

Now we can use the two half reactions to write our cell diagram, and we know where everything goes now. Placing the half-reaction products on the outside, and the reactants on the inside, we can write

\[Ni(s)\mid Ni^{2+}(aq)\parallel H^+(aq, 1.00 M)\mid H_2(g, 1.00atm)\mid Pt(s)\]

Make sure you indicate the molarities and pressures of the hydrogen ions and hydrogen gas, as they must be 1.0M and 1 atm in order for this to be a SHE. Also, since there is no electrode on the SHE side due to the hydrogen ions being in aqueous solution and hydrogen gas being produced, make sure to include a Pt(s) electrode. Platinum is a good choice for an electrode in this case because it conducts electricity, but does not participate in the reaction. As usual, the single bar indicates change in phase, and the double bars indicate the salt bridge.

Q20.7.3

What type of battery would you use for each application and why?

1. powering an electric motor scooter
2. a backup battery for a smartphone
3. powering an iPod

A20.7.3

1. Here you want a secondary battery that can be discharged and recharged numerous amount of times. It should be easily replaceable, as most consumers don't want to buy a new motor scooter every time the battery dies. Most automobiles opt for a lead storage battery because, with proper care, it lasts many uses, and it is easy to monitor and replace.
2. Lithium-iodine batteries would work best in a portable, backup battery for a phone because they are long-lasting and reliable. They provide enough power to charge a phone. While lithium-iodine batteries are primary batteries, they do not contain water, and do not corrode as quickly as its peers. Because of this, it will last longer than other batteries. It discharges slowly, ensuring that it does not fry your phone's circuits.
3. For an iPod, you want a battery that is reversible so you can charge it back up easily. You want a secondary battery to do this. Here are a list of common secondary batteries: NiCad (Nickel-Cadmium), NiMH (Nickel Metal Hydride), and a rechargeable lithium ion battery. These batteries are all lightweight and are practical in a small device such as an iPod.