Extra Credit 19

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Q17.2.8

For each of the following reactions, list the substance reduced, the substance oxidized, the reducing agent, and the oxidizing agent.

6H⁺ + MnO₄^- + 5SO₃²^- → 5SO₄²^- + 2Mn²⁺ + 3H₂O
8H⁺ + Cr₂O₇²^- + 6HI → 2Cr³⁺ + 3I₂ + 7H₂O
3Cl₂ + 6OH^- → ClO₃^- + 5Cl^- + 3H₂O

S17.2.8

oxidizing agent = substance (reactant) reduced = species that gained electrons

reducing agent = substance (reactant) oxidized =species that lost electrons

Read Oxidation States (Oxidation Numbers) for rules on assigning oxidation states.

OIL RIG

oxidation is loss, reduction is gain

1. Mn in MnO₄^- has an oxidation state of (+7) on the reactant side and oxidation state of 0 on the product side. Therefore, it is reduced because it gained electrons. MnO₄^- is the reduced and is the oxidizing agent. S in SO₃^- is +5 and is +7 in SO₄^-. It lost 2 electrons, SO₃^- is oxidized and is the reducing agent.

Water is not reduced or oxidized because its charge did not change.

2. Cr in Cr₂O₇²^- has an oxidation state of +6 is reduced to Cr³⁺by gaining 3 electrons. Cr₂O₇²^-is reduced and is the oxidizing agent. I in HI has an oxidation state of -1 and is oxidized to I₂which has an oxidation state of 0 because it lost one elctron. HI is oxidized and is the reducing agent.

Water is not reduced or oxidized because its charge did not change.

3. Cl₂ is both reduced and oxidized which means it is both the reducing agent and the oxidizing agent.

Since Cl in Cl₂has an oxidation state of 0, it is reduced and gained one electron to become Cl^-.

Again, Cl in Cl₂ has an oxidation state of 0, but this time it lost 7 electrons and became oxidized to ClO₃^- where the Cl has an oxidation state +7

Water is not reduced or oxidized because its charge did not change

Q19.1.7

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.)

Fe(s)+H₂SO₄(aq)⟶
FeCl₃(aq)+NaOH(aq)⟶
Mn(OH)₂(s)+HBr(aq)⟶
Cr(s)+O₂(g)⟶
Mn₂O₃(s)+HCl(aq)⟶
Ti(s)+xsF₂(g)⟶

S19.1.7

Make sure to check the 3.2 Solubility Tables & the Solubility of Ionic Compounds to determine the phase of the product formed.

1. Fe(s)+H₂SO₄(aq)⟶FeSO₄(aq)+H₂(g)

This is a single replacement reaction, so replace the Fe with H. Since SO₄ has a 2- charge, it will cancel out with the 2+ change in the Fe ion and form FeSO₄.

2. FeCl₃(aq)+NaOH(aq)⟶Fe(OH)₃(aq)+NaCl(aq)

This is a double replacement reaction. So the Na and Fe (cations) will exchange with each other and form new compound with Cl^- and OH^-(anions.

3. Mn(OH)₂(s)+2HBr(aq)⟶MnBr₂(aq)+ H₂O(l)

This is also a double replacement reaction. So the Mn and H will exchange place and form ne compound with OH^- and Br⁺. Then, we have to balance the reaction by adding a coefficient 2 to HBr.

4. 4Cr(s)+3O₂(g)⟶ 2Cr₂O₃(s)

This is a synthesis reaction. So the Cr and O₂combine together to form new compound. The new compound formed is not soluble, so the phase is solid. Then, balance the reaction.

5. Mn₂O₃(s)+HCl(aq)⟶MnCl₃(aq)+H₂O(l)

This is also a double replacement reaction. Mn and H exchange space and form new compound with Cl and O.

6. Ti(s)+XsF₂(g)⟶TiF(aq)+xs²⁺(aq)

This is a single replacement reaction. So Ti get to form with F and procude a new product. And xs become an ion.

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?

S19.3.9

The formula is [NiCl₂(P(CH₃)₃)₂] and its geometry is tetrahedral.

Since the compound has 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH₃)₃, multiplying each percentage with the total mass will give how many gram each elements are in this coumpound.

270 x 21.5% = 58.05g Ni,

270 x 26.0% = 70.2g Cl,

270 x 52.5% = 141.75g P(CH₃)₃

Then, we can find their mole by using their molar mass

58.05g Ni x 1mol/58.6g = 1mol Ni

70.2g Cl x 1mol/35.4g = 2mol Cl

141.75g P(CH₃)₃x 1mol/72g= 2mol P(CH₃)₃

Then, use their mole to write the formula which is [NiCl₂(P(CH₃)₃)₂].

Cl^- is a weak field ligand meaning it absorbs low energy wavelengths; the compound is blue which means it absorbs orange light which is low energy. Weak field ligands have high spin complexes. From its formula, it shows that Ni is the center atom with 4 ligand binding to it meaning it can be square planar or tetrahedral. The complex is tetrahedral because tetrahedral complexes have a small crystal field splitting energy which means it favors high spin complexes.

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.

S12.4.9

Since the reaction is second order, its half-life is

\[t_{1/2}=\dfrac{1}{(50.4M^{-1}/h)[2.35×10^{-6}M]}\]

So, half-life is 8443 hours.

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.

S21.2.4

The isotopes in Question 13 are the following:

a. \( ^{34}_{14}\text{X}\)

b. \( ^{36}\text{P}\)

c. \( ^{57}_{X}\text{Mn}\)

d. \( ^{121}_{56}\text{X}\)

a) The total number of proton is 14. The total number of neutron is 20 because the neutrons number= mass number-proton number. And the electron number is 14 because electron number equals to proton number.

b)Since the element is P we can find its proton number at the periodic table, which is 15. Since neutrons number= mass number-proton number, so 36-15=21, so neutron number is 21. And electron number is 15 because electron number equals to proton number.

c)Since the element is Mn we can find its proton number at the periodic table, which is 25. Since neutrons number= mass number-proton number, so 57-25=32, so neutron number is 32. And electron number is 25 because electron number equals to proton number.

d)The proton number is 56 which is given in the question. Since neutrons number= mass number-proton number, so 121-56=65, so neutron number is 65. And electron number is 56 because electron number equals to proton number.

Q21.5.7

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

S21.5.7

Uranium is radioactive, meaning it is in an energized state and wants to move lower in energy to a more stable form. This occurs in a process called fission where uranium atoms get split apart into two separate atoms. This process releases heat. Then, water collects the energy and turns it into steam by giving this energy to cooler water. Then, the steam will blow pass the turbine, turning it and powering the electrical generator.

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?

S20.4.5

The cell diagram is Ni(s)∣Ni²⁺(aq)∥H⁺(aq)∣H₂(g)∣Pt(s)

First, I write out the half reactions and find their cell potential.

E°_anode: Ni²⁺(aq)+2e^-→Ni(s) cell potential:−0.25 V

E°_cathode: 2H⁺(aq)+2e^-→H₂(g) cell potential:0.000 V

Since Ni reaction has a lower (more negative) reduction potential than H reaction, then, it should be oxidized, which makes it the anode and placed on the left side of the cell diagram. Then the H reaction will be reduced, which is the cathode and place on the right diagram. It is standard that oxidation occurs at the anode which is the right cell in the cell diagram, and reduction occurs t the cathode which is the left cell in the cell diagram.

E°_cell of this reaction can be found with this formula

E^o_cell = E^o_cathode - E^o_anode so, E°_cell=0-(-0.25) E°_cell=0.25V

Q20.7.3

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

powering an electric motor scooter
a backup battery for a smartphone
powering an iPod

S20.7.3

powering an electric motor scooter
a backup battery for a smartphone
powering an iPod

1. The battery for powering an electric motor scooter should be rechargeable, so that the scooter wouldn't be for a one-time use. Secondary batteries would be best suited for this application because they can be recharged. We could use a Lead Acid storage battery commonly used in cars for this situation, because you would need a rather large battery not like Nickel-Cadmium or Nickel-metal hydride battery.

2. a backup battery for a smartphone would probably be a primary battery, one that is not rechargeable. The reason being that primary batteries can be stored for long periods of time and still maintain the same capacity as before. Alkaline batteries would be a good choice because they produce more current than a zinc-carbon battery.

3. A secondary battery would be the best for powering an iPod because they are rechargeable. The best type of secondary battery would be the lithium ion battery because it is light and portable.

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