Extra Credit 16

Q17.2.5

Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

From Q17.2.4

a. \(Al_{(s)}+Zr^{4+}_{(aq)}\rightarrow Al^{3+}+Zr_{(s)}\)

b. \(Al^+_{(aq)}+NO_{(g)}\rightarrow Ag_{(s)}+NO^-_3{(aq)}\) (acidic solution)

c. \(SiO^{2-}_3{(aq)}+Mg_{(s)}\rightarrow Si_{(s)}+Mg(OH)_2{(s)}\) (basic solution)

d. \(ClO^-_3{(aq)}+MnO_2{(s)}\rightarrow Cl^-_{(aq)}+MnO^-_4{(aq)}\) (basic solution)

Answer Q17.2.5

Definitions-

Oxidized species: the compound or atom that loses an electron.

Reduction species: the compound or atom that gains an electron.

Oxidizing Agent: the species that inhibits oxidation to occur, basically, the reduction species.

Reducing Agent: the species that inhibits reduction to occur, basically the oxidation species.

Oil Rig

Oxidation is losing Reduction is gaining

Reducing agent Oxidizing agent

a. Separate the equation by their atoms balance the equation by adding electrons and determine which is the oxidation equation and which is the reduction equation

\(Al_{(s)}\rightarrow Al^{3+}_{(aq)}+3e^-\) \(4e^-+Zr^{4+}_{(aq)}\rightarrow Zr_{(s)}\)

Since Al(s) lost three electrons it is the oxidation species and Zr⁴⁺(aq) gained four electrons so it is the reduction species.

The oxidizing agent is Zr⁴⁺(aq) and the reducing agent is Al(s).

b. Separate the equation by their atoms

\(Ag^+_{(aq)}\rightarrow Ag_{(s)}\) \(NO_{(g)}\rightarrow NO^-_{(aq)}\) (Acidic Solution)

Balance the equation by adding electrons

\(e^-+Ag^+_{(aq)}\rightarrow Ag_{(s)}\)

Since it is an acidic solution balance by using H₂O

\(2H_2O+NO_{(g)}\rightarrow NO^-_3(aq)\)

Balance the other side with H⁺ ions

\(2H_2O+NO_{(g)}\rightarrow NO^-_3(aq)+4H^+\)

Balance with electrons

\(2H_2O+NO_{(g)}\rightarrow NO^-_3(aq)+4H^++3e^-\)

Since NO(g) lost three electrons, it is the oxidation species and Ag⁺(aq) gained an electron so it is the reduction species.

The oxidizing agent is Ag⁺(aq) and the reducing agent is NO(g).

c. Separate the equation by their atoms

\(SiO^{2-}_3\rightarrow Si_{(s)}\) \(Mg_{(s)}\rightarrow Mg(OH)_2(s)\) (Basic Solution)

Balance by adding H₂O and H⁺

\(3H^++SiO^{2-}_3\rightarrow Si_{(s)}+3H_2O\) \(2H_2O+Mg_{(s)}\rightarrow Mg(OH)_2(s)+2H^+\)

Balance with electrons

\(4e^-+6H^++SiO^{2-}_3\rightarrow Si_{(s)}+3H_2O\) \(2H_2O+Mg_{(s)}\rightarrow Mg(OH)_2(s)+4H^++4e^-\)

Since Mg(s) lost four electrons, it is the oxidation species and SiO₃^2- gained four electrons so it is the reduction species.

The oxidizing agent is SiO₃^2- and the reducing agent is Mg(s).

Note= to find the reduction/oxidation you can stop here, however, if you are balancing for a redox reaction make sure to add OH^- ions on both sides to cancel out the H⁺ ions since it is in a basic solution

d. Separate the equation by their atoms

\(ClO^-_3(aq)\rightarrow Cl^-_{(aq)}\) \(MnO_2(s)\rightarrow MnO^-_4\) (basic solution)

Balance by adding H₂O and H⁺ ions

\(6H^++ClO^-_3(aq)\rightarrow Cl^-_{(aq)}+3H_2O\) \(2H_2O+MnO_2(s)\rightarrow MnO^-_4+4H^+\)

Balance with electrons

\(6e^-+6H^++ClO^-_3(aq)\rightarrow Cl^-_{(aq)}+3H_2O\) \(2H_2O+MnO_2(s)\rightarrow MnO^-_4+4H^++3e^-\)

Since MnO₂(s) lost three electrons, it is the oxidation species and ClO₃^-(aq) gained six electrons so it is the reduction species.

The oxidizing agent is ClO₃^-(aq) and the reducing agent is MnO₂(s).

Q19.1.14

A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

Answer Q19.1.14

To find the percent mass of chloride we need to use the percent mass equation

(mass of element)/(mass of the compound)*100=percent mass

For the mass of the element, we focus on the number of what element is being asked for. In this case, the percent mass of Chloride is being asked. The number for Chloride is 2.5624 grams (which is given in the problem). For the mass of the compound, we look for the total mass of the compound, in this case 3.03707 (also given in the problem).

So the problem would look like:

\(\dfrac{2.5624 \text{grams}Cl^-}{3.03707 \text{grams} AgCl}\times 100=\text{percent mass of chlorine}\)

84.4% was the mass of chlorine

The identity of the salt can be determined by writing the chemical equation of what was happening in the problem.

Cl^- (s) + AgNO₃(aq) → AgCl(s) + NO₃^-

^{So the identity of the salt in this case was \(AgCl_{(s)}\).}

Q19.3.6

How many unpaired electrons are present in each of the following?

1. [CoF₆]^3- (high spin)
2. [Mn(CN)₆]^3- (low spin)
3. [Mn(CN)₆]^4- (low spin)
4. [MnCl₆]^4- (high spin)
5. [RhCl₆]^3- (low spin)

Q19.3.6 Answer

1. First we need to find the charge of the transition metal Co

[CoF₆]^3-

F=-1*6= -6

x+-6=-3 x=+3

Co has a +3 charge

Now we need to build an octahedral diagram (since 6 Flourines are attached to the cobalt)

We look at a periodic table and determine how many electrons Co will have. Co has 7 d electrons, but we take 2 away from the s orbital and one from the d orbital

Co³⁺ will have 6 electrons, and since the ligand is a weak field ligand, it will be a high spin diagram and will look like this

There are 4 unpaired electrons in this system.

2. First we need to find the charge of the transition metal Mn

[Mn(CN)₆]^3-

CN has a charge of -1 -1*6=-6

x+-6=-3 x=+3

Mn has a +3 charge

This one is also octahedral so and octahedral diagram is expected.

We look at a periodic table and determine how many electrons Mn will have. Mn has 5 d electrons, but we will take 2 away from the s orbital and one away from the d orbital

Mn³⁺ will have 4 electrons, therefore the low spin diagram will look like this

It is a low spin diagram, because CN^- is a strong field ligand

There will be 2 unpaired electrons for this system

3. First we need to find the charge of the transition metal Mn

[Mn(CN)₆]^4-

CN has a charge of -1 -1*6=-6

x+-6=-4 x=+2

Mn has a charge of +2

This one is also octahedral so and octahedral diagram is expected.

We look at a periodic table and determine how many electrons Mn will have. Mn has 5 d electrons, but we take 2 away from the s orbital.

Mn²⁺ will have 5 electrons, therefore the low spin diagram will look like this

The diagram that is circled.

This is a low spin diagram, because CN^- is a strong field ligand

Therefore there is one unpaired electron in the system.

4. First we need to find the charge of the transition metal Mn

[MnCl₆]^4-

Cl has a charge of -1 -1*6=-6

x+-6=-4 x=+2

Mn will have a charge of +2

This one is also octahedral so and octahedral diagram is expected.

We look at a periodic table and determine how many electrons Mn²⁺ will have. Mn has 5 d electrons, but we will again take 2 away from the s orbital

Mn²⁺ will have 5 electrons, therefore the high spin diagram will look like this.

This is a high spin diagram, because Cl^- is a weak field ligand

Therefore there is five unpaired electrons in the system.

5. First we need to find the charge of the transition metal Rh

[RhCl₆]^3-

Cl has a charge of -1 -1*6=-6

x+-6=-3 x=+3

Rh has a charge of +3

This one is also octahedral so and octahedral diagram is expected.

We look at a periodic table and determine how many electrons Rh³⁺ will have. Rh has 7 d electrons, but we will take 2 away from the s orbital and 1 from the d orbital

Rh³⁺ will have 6 electrons, therefore the low spin diagram will look like this

Therefore there are 0 unpaired electrons for the system.

Q12.4.6

What is the half-life for the first-order decay of phosphorus-32? (³²₁₅P⟶³²₁₆S+e−) The rate constant for the decay is 4.85 × 10^-2 day^-1

^{Q12.4.6 Answer}

The first order for the half-life for the reaction is t_1/2=ln(2)/K where K is the rate constant.

Plug everything into the equation and solve for half-life t_1/2.

Plug 4.85 × 10^-2 day^-1 in for K and solve

\(t_{1/2}=\dfrac{ln(2)}{4.85\times 10^{-2}\text{day}^{-1}}\)

The half-life of phosphorous-32 is 14.29.

t_1/2=14.29 days

Note: Make sure to add units to the end of your answer!!

Q21.2.1

Write the following isotopes are in hyphenated form (e.g., “carbon-14”)

1. \(^{24}_{11}Na\)
2. \(^{29}_{13}Al\)
3. \(^{73}_{36}Kr\)
4. \(^{194}_{77}Ir\)

Q21.2.1 Answer

\(^A_ZE\)

A is mass number

Z is atomic number

E is the element

1. Identify the name of the element in the isotope by using a periodic table.

\(^{24}_{11}Na\)

Na is identified as Sodium

For the number that comes after the hyphen, it is the mass number of the isotope

Therefore the isotope is called Sodium-24

2. Identify the name of the element in the isotope by using a periodic table.

\(^{29}_{13}Al\)

Al is identified as Aluminum

For the number that comes after the hyphen, it is the mass number of the isotope

Therefore the isotope is called Aluminum-29

3. Identify the name of the element in the isotope by using a periodic table.

\(^{73}_{36}Kr\)

Kr is identified as Krypton

For the number that comes after the hyphen, it is the mass number of the isotope

Therefore, the isotope is called Krypton-73

4. Identify the name of the element in the isotope by using a periodic table.

\(^{194}_{77}Ir\)

Ir is identified as Iridium

For the number that comes after the hyphen, it is the mass number of the isotope.

Therefore, the isotope is called Iridium-194

Q21.5.4

Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.

Q21.5.4 Answer

Nuclear reactions have two types of reactions, fission and fusion.

Definitions:

Fission: A nuclear reaction in which heavy nuclei disintegrates to form smaller nuclei

Fusion: A nuclear reaction in which light nuclei combine to form a heavier nucleus.

in order for a nuclear chain reaction to occur it has to have a reaction, either fission or fusion. This may be caused by a single reaction of two nuclei that collide for a chain reaction to occur. High temperatures are required to give the nuclei enough kinetic energy to overcome the very strong repulsion from the nucleus because of their positive charges. To avoid an explosion from occurring, scientist use control rods that absorb excess neutrons that are produced during the reactions, and this than results in stability.

Q20.4.2

List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.

Q20.4.2 Answers

First we have to look at the equations that are involved in electrochemical cells.

These are:

\(\Delta G=-nFE\)

\(\Delta G=\Delta G^o-RT(ln(Q))\)

\(E=E^o-\dfrac{RT}{nF}ln(Q)\)

ΔG is Gibbs free energy

n is the number of electrons changed

F is Faraday's Constant

E is cell potential

ΔGº is standard Gibbs Free Energy

R is the rate constant

T is temperature

Q is equilibrium constant

Eº is cell potential standard

One factor that can change the measured potential is the number of electrons (n). If the number of electrons changed to being a smaller number than the original n, the cell potential will be increased. Conversely, if n is increased, the opposite will happen. The cell potential will then decrease.

Another factor that could change the measured potential is Temperature. If temperature is increased, then the cell potential is more likely to be increased as well. When temperature is decreased, the cell potential will most likely decrease as well.

Q20.5.30

Hydrogen gas reduces Ni²⁺ according to the following reaction: Ni²⁺(aq) + H₂(g) → Ni(s) + 2H⁺(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.

1. What is K for this redox reaction?
2. Is this reaction likely to occur?
3. What conditions can be changed to increase the likelihood that the reaction will occur as written?
4. Is the reaction more likely to occur at higher or lower pH?

Q20.5.30 Answers

1. What is K for this redox reaction?

The procedure: We want to find K and to do this we have to use the equation ΔGº=-RT(lnK). First we need to find ΔGº by using the equation ΔGº=-nFEº_cell.

To find ΔGº: We first need to find n (the number of electrons). We can do this by separating the redox reactions and balancing electrons.

2e^-+ Ni²⁺(aq) → Ni(s) H₂(g) → 2H⁺(aq) + 2e^-

Due to the redox reaction, the number of electrons (n) is 2, for F, we use the Faraday Constant 96485 C/V, and Eº_cell is given in the question, -0.25 V.

Plugging everything into the equation

\(-(2mole^-)(96485C/V)(-0.25V)= ΔGº\)

ΔGº= 48242.5 J/mol

Using ΔGº in the next equation, the Faraday Constant 96485 C/V and room temperature 298ºK (assuming system is at room temperature):

\(48242.5 J/mol=-(8.314 J/mol\times K)(298 K)(ln{K})\)

\(-19.47)=ln{K}\)

e^{-19.47}=k

K= 3.5*10^-9

2. Is this reaction likely to occur?

If ΔGº is positive, the system is nonspontaneous or is not favorable to conditions while if ΔGº is a negative number, then the reaction is spontaneous and the system is favorable.

It is not likely to occur because the ΔGº for the system is a positive number. If Gibbs Free Energy is positive, it is nonspontaneous.

3. What conditions can be changed to increase the likelihood that the reaction will occur as written?

We know that the reaction is an endothermic reaction because ΔH (enthalpy) is positive (54 kJ/mol) and according to this table

if temperature increases, the products would be favored and there is a possibility that ΔGº will become negative becoming spontaneous in the process.

4. Is the reaction more likely to occur at higher or lower pH?

The higher the pH, the more OH^- ions there are on the products side of the reaction and that makes the system basic. The lower the pH the more H⁺ ions there are on the product side of the reaction and that makes the system acidic.

The reaction would be more likely to happen at a lower pH due because there are H⁺ ions on the products side of the reaction. The more H⁺ ions there are, the more acidic the solution is, therefore resulting in a low pH to occur.