Extra Credit 4

Q17.1.4

For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.

a.Cl^-⟶Cl₂

Solution:

First: determine the oxidation number of Cl atom in each species

Cl^-= -1

Cl₂=0

Second: Determine whether electrons are gained or lost

electrons are lost since the oxidation state increases from -1 to 0.

Third: Use OIL RIG (Oxidation is loss; Reductions is gain) to determine whether it its being oxidized or reduced

Answer: Oxidation since electrons are lost

b. Mn²⁺⟶MnO₂

Solution:

First: determine the oxidation number of Mn in each species

Mn²⁺= +2

MnO₂=+4 (Since Oxygen has a -2 charge and there are 2, the Mn needs to be +4 in order to keep the overall charge of molecule at 0)

Second: Determine whether electrons are gained or lost

electrons are lost since the oxidation state increases from +2 to +4.

Third: Use OIL RIG (Oxidation is loss; Reductions is gain) to determine whether it its being oxidized or reduced

Answer: Oxidation since electrons are lost

c.H₂⟶H⁺

Solution:

First: determine the oxidation number of H atom in each species

H₂= 0

H⁺=+1

Second: Determine whether electrons are gained or lost

electrons are lost since the oxidation state increases from 0 to +1.

Third: Use OIL RIG (Oxidation is loss; Reductions is gain) to determine whether it its being oxidized or reduced

Answer: Oxidation since electrons are lost

d. NO₃^-⟶NO

First: determine the oxidation number of N in each species

NO₃^-= +5 (Since Oxygen has a -2 charge and there are 3, the N needs to be +5 in order to keep the overall charge of molecule at -1)

NO=+2 (Since Oxygen has a -2 charge, the N must have a +2 oxidation state in order to keep the overall charge of molecule at neutral)

Second: Determine whether electrons are gained or lost

electrons are gained since the oxidation state of N decreased from +5 to +2.

Third: Use OIL RIG (Oxidation is loss; Reductions is gain) to determine whether it its being oxidized or reduced

Answer: Reduction since electrons are gained

Q19.1.2 All correct

Write the electron configurations for each of the following elements and its ions:

a. Ti

First: Locate the element on the periodic number and record its atomic number. The atomic number tells you about the number of protons and thus indicates the number of electrons before ionization since they are equal.

Ti=22 electrons

Second: Identify the Noble Gas in the row above the element of interest and write out its configuration. These are consider the inner electrons in the element and are not typically involved in ionization which is why it's okay to use a shortcut to denote those electrons.

Ar= 1s²2s²2p⁶3s²3p⁶ (18 electrons are present in this configuration)

Third: Look at the 4th row (Where Ti is located) and identify which orbitals are filled.

Two electrons will go to the 4s orbital and two electrons will go to the 3d orbital.

Fourth: Write the configuration for the element.

[Ar]4s²3d²

b.Ti²⁺

First: Locate the element on the periodic number and record its atomic number. The atomic number tells you about the number of protons and thus indicates the number of electrons before ionization since they are equal.

Ti=22 electrons

Second: Look at the charge and adjust the total electrons in the ion.

22 electrons - 2 electrons = +2 overall charge

20 electrons

Third: Identify the Noble Gas in the row above the element of interest and write out its configuration. These are consider the inner electrons in the element and are not typically involved in ionization which is why it's okay to use a shortcut to denote those electrons..

Ar= 1s²2s²2p⁶3s²3p⁶ (18 electrons are present in this configuration)

Fourth: Look at the 4th row (Where Ti is located) and identify which orbitals are filled.

Two electrons will go to the 3d orbital. They do not go to the s orbital since transition metals tend to lose their s orbital electrons before their d orbital for stability reasons.

Fifth: Write the configuration for the element.

[Ar]3d²

c.Ti³⁺

First: Locate the element on the periodic number and record its atomic number. The atomic number tells you about the number of protons and thus indicates the number of electrons before ionization since they are equal.

Ti=22 electrons

Second: Look at the charge and adjust the total electrons in the ion.

22 electrons - 3 electrons = +3 overall charge

19 electrons

Third: Identify the Noble Gas in the row above the element of interest and write out its configuration.These are consider the inner electrons in the element and are not typically involved in ionization which is why it's okay to use a shortcut to denote those electrons..

Ar= 1s²2s²2p⁶3s²3p⁶ (18 electrons are present in this configuration)

Fourth: Look at the 4th row (Where Ti is located) and identify which orbitals are filled.

One electron will goes to the 3d orbital. They do not go to the s orbital since transition metals tend to lose their s orbital electrons before their d orbital for stability reasons. After the s orbital is empty, it starts taking electrons from the d orbital.

Fifth: Write the configuration for the element.

[Ar]3d¹

d.Ti⁴⁺

First: Locate the element on the periodic number and record its atomic number. The atomic number tells you about the number of protons and thus indicates the number of electrons before ionization since they are equal.

Ti=22 electrons

Second:Look at the charge and adjust the total electrons in the ion.

22 electrons - 4 electrons = +4 overall charge

18 electrons

Third: Identify the Noble Gas in the row above the element of interest and write out its configuration. These are consider the inner electrons in the element and are not typically involved in ionization which is why it's okay to use a shortcut to denote those electrons..

Ar= 1s²2s²2p⁶3s²3p⁶ (18 electrons are used in this configuration)

Fourth: Look at the 4th row (Where Ti is located) and identify which orbitals are filled.

No electrons go to the outer orbitals since they are all removed.

Fifth: Write the configuration for the element.

[Ar]

Q19.2.4

Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.

a.[Pt(H₂O)₂Br₂] (square planar)

First: identify the number of ligands.

4= 2x H₂O, 2x Br^-

Second: Identify potential isomers given the geometry

It can have a geometric isomers since it has two sets of two ligands. If the two same ligands (like H₂O) are next to each other, it is considered cis. If the H₂O ligands have a 180 degree angle between them, then it is trans. There are no optical isomers since square planar has a plane that dissect all of the ligands hence gives it a line of symmetry. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the respective isomers

b.[Pt(NH₃)(py)(Cl)(Br)] (square planar, py = pyridine, C₅H₅N)

First: identify the number of ligands.

4= 1x NH₃, 1x py, 1x Cl^-, 1x Br^-

Second: Identify potential isomer

Since it has four different ligands, it does not have any geometric isomers since no ligands have pairs to form 90 degree and 180 degree bonds. There are no optical isomers since square planar has a plane that dissect all of the ligands hence gives it a line of symmetry. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the complex

c.[Zn(NH₃)₃Cl]⁺ (tetrahedral)

First: Identify all ligands

4 ligands= 3x NH₃, 1x Cl^-

Second: Identify all potential isomers

There are no geometric isomers when there are three of the same ligands and a fourth different ligand. There are no optical isomers since there is a plane of symmetry. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the complex

d.[Pt(NH₃)₃Cl]⁺ (square planar)

First: Identify all ligands

4 ligands= 3x NH₃, 1x Cl^-

Second: Identify all potential isomers

There is no geometric isomers since there is no pair of identical ligands that can be rearranged at 90 degrees away from each other and also 180 degrees away from each other. There are no optical isomers since there is a plane of symmetry that dissects all of the ligands. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the complex

e.[Ni(H₂O)₄Cl₂]

First: Identify all ligands

6 ligands= 4x H₂O, 2x Cl^-

Second: Identify all potential isomers

There is a geometric isomers since there is the Chlorine ligands t can be rearranged at 90 degrees away from each other and also 180 degrees away from each other. There are no optical isomers since there are planes of symmetry. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the complex

f.[Co(C₂O₄)₂Cl₂]³⁻ (note that C2O2−4C2O42− is the bidentate oxalate ion, −O2CCO^-2)

First: Identify all ligands

4 ligands= 2x C₂O₄, 2x Cl^-

Second: Identify all potential isomers

There is a geometric isomers since the Chlorine and C₂O₄ ligands can be rearranged at 90 degrees away from each other and also 180 degrees away from each other. There are optical isomers since there is no plane of symmetry or a way to move the mirror image of the complex to be stacked on the original. Additionally, there are no structural isomers because the coordinate complex is not bonded with another species and does not have ambidentate ligands.

Third: Draw the complex

Q12.3.17

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:

\[\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g)]

Use latex:

Determine the rate equation, the rate constant, and the orders with respect to each reactant from the following data:

First: In order to determine the rate equation of this reaction, we require the experimental data to determine the order in respects with each reactant. For each reactant, we need to see the relationship between concentration and rate when we change the concentration of one and keep the other constant and ten observe the change in the rate. If the rate is unchanged when the initial concentration is changes, then it is zero order. If the rate changes with the same factor as the concentration then it is first order. If the rate changed with a factor double the factor of the concentration change then it is second order.

When NO is doubled from .3 to .6 (and the concentration of H₂ is kept constant), the rate changes with a factor of 4. This indicates that NO is second order. When H₂ is double from .35 to .7 (and the concentration of NO is kept constant), the rate changes with a factor of 2. This indicates a first order relationship.

Second: Use the orders of each reactant to write the rate law

Rate=k[NO]²[H₂]

Third: To calculate the rate constant, use one of the trials to plug in the values for the rate and the initial concentration.

2.835E-3=k[.3]²[.35]

k=.09 M^-2.S^-1

Q12.6.8

Nitrogen(II) oxide, NO, reacts with hydrogen, H₂, according to the following equation:

2NO+2H₂⟶N₂+2H₂O

What would the rate law be if the mechanism for this reaction were:

2NO+H₂⟶N₂+H₂O₂(slow)

H2O2+H2⟶2H2O(fast)

First: Since the slow step is first, we use the first step rather than the overall equation to determine the rate law. This is so because the second reaction uses then products from the first reaction, so it can only go as fast as the reactant are being produced. We use the reactant and coefficients of the first reaction to formulate the rate law.

In this case, rate=k[NO]²[H₂]

Q21.4.20

What is the age of mummified primate skin that contains 8.25% of the original quantity of ¹⁴C?

First: Use the half life of carbon-14 to determine k using the equation: t_1/2=ln(2)/k. We use this equation because radioactive decay is first order.

5730 years=ln(2)/k

k=1.21E-4

Second, use this constant to calculate the amount of time that has passed using this equation: ln(A₀/A)=-kt

ln(.0825)=-1.21E-4t

t= 20625 years

Q20.3.8

Consider the following spontaneous redox reaction: NO₃⁻(aq) + H⁺(aq) + SO₃²⁻(aq) → SO₄²⁻(aq) + HNO₂(aq).

1.Write the two half-reactions for this overall reaction.

When writing the half reactions you must pay attention to the species involved in the electron transfer. In this equation, electrons are being transferred from Sulfur to Nitrogen. Separate the species involving Sulfur in one equation and separate the species containing Nitrogen in another equation. To balance the two equations, start be balance species that are not Hydrogen or Oxygen. Second, add water to balance the Oxygens. Third, balance the Hydrogens by adding H⁺. Lastly, balance the charge by adding electrons as needed. Oxidation: H₂O+SO₃²⁻(aq) → SO₄²⁻(aq)+2e^-+2H⁺

Reduction:2e^-+3H⁺+NO₃⁻(aq)→ HNO₂(aq)+H₂O

2.If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?

Anodes carry out oxidation while cathodes carry out the reduction. In this case, the anode and cathode would be inert (Platinum) since the oxidation and reduction happens within the same state (aqueous). The anode would be submerged with the solution containing SO₃²⁻ and the cathode would be submerged in the solution containing NO₃⁻.

3.Which electrode is negatively charged, and which is positively charged?

The anode would be negative since it accepts the electrons from SO₃²⁻ while the cathode would be negative because the NO₃⁻ in the solution would be taking the electrons from the electrode. This is not, however, always the case, anode and negative charged electrode are not synonymous (same with cathode and positive).

Q20.5.19

Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.

1.Write the balanced chemical equation for the oxidation of Fe²⁺ by Ce⁴⁺.

Since Fe²⁺ is being oxidized that means it will lose an electon. When it loses an electron the charge will increase to Fe³⁺. The Ce⁴⁺ accepts this electron in the reduction part of this reaction to become Ce³⁺.

Fe²⁺+Ce⁴⁺→Fe³⁺+Ce³⁺

2.Plot the data and then locate the endpoint.

The endpoint appears to be 10 mL. This is so because the that is the steepest point in the curve above.

3.How many millimoles of Fe²⁺ did the solution being titrated originally contain?

We can find this value by using the endpoint identified in part 2. The endpoint indicated the the moles of Ce⁴⁺ equals the moles of Fe²⁺. We are give the molarity and the volume of titrant used, this means that we can calculate the number of moles of Ce⁴⁺ that were added (multiply the volume in Liters by the molarity).The answer will be in moles so to convert it to millimoles, multiply by 1000.

10mL x 1L/1000mL= .01 L

.01 L x .1 mol/L=.001 mol of Ce⁴⁺ x 1000 millimoles/1 mole

1 millimole of Ce⁴⁺= 1 millimole of Fe²⁺