Extra Credit 30

Q17.4.3

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

1. Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)
2. The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br₂(l) is the same as that of Br₂(aq).

1) 1. Separate the equation into two half reactions. One will be the oxidizing reaction and the other will be the reducing reaction.

2Ag⁺_(aq) + 2e^- ---> 2Ag_(s) Oxidized

Hg_(l) + S^2-_(aq) ---> HgS_(s) + 2e^- Reduced

2. Find the number of elections transferred: 2 elections

3. Find the standard cell potential using the reduction potential table. Use this equation: E^ocell=E^ocathode - E^oanode

E^ocell=0.7V - (-0.7994V) = 1.5V

4. Plug everything needed into the Nernst equation

E_cell = E^o_cell - [(RT)/(nF)]*ln Q

= 1.5V - [(8.314*298.15)/(2*96485)]*ln(1/[0.1][0.25]²)

=1.43V

5. Because E_cell is a positive number, I now that this reaction is spontaneous

2) 1. Make the electrical cell. Then split the reaction into its two oxidation/Reduction reactions.

2Al_(s)+3Ni²⁺_(aq) ---> 3Ni_(s)+2Al³⁺_(aq)

Al_(s) l Al³⁺(0.015M) ll Ni²⁺(0.25M) l Ni_(s)

Ni²⁺_(aq) + 2e^- ---> Ni_(s) Reduced

Al_(s) ---> Al³⁺_(aq) + 3e^- Oxidized

2. Find the number of elections transferred: 6 elections, because the lowest common denominator between 2 and 3 is 6.

3. Find the standard cell potential using the reduction potential table. Use this equation: E^ocell=E^ocathode - E^oanode

E^ocell=-0.25V - (-1.66V) = 1.41V

4. Plug everything needed into the Nernst equation

E_cell = E^o_cell - [(RT)/(nF)]*ln Q

= 1.41V - [(8.314*298.15)/(6*96485)]*ln([0.015]²/[0.25]³)

=1.428V

5. Because E_cell is positive, I know this reaction is spontaneous

3) 1. Make the electrical cell. Then split the reaction into its two oxidation/Reduction reactions.

Pt_(s) l Br^-(0.11M), Br₂(1.0M) ll Al³⁺(0.023M) l Al(s)

Al³⁺_(aq) + 3e^- ---> Al_(s) Reduced

2Br^-_(aq) ---> Br²_(aq) + 2e^- Oxidized

2. Find the number of elections transferred: 6 elections

3. Find the standard cell potential using the reduction potential table. Use this equation: E^ocell=E^ocathode - E^oanode

E^ocell= -1.66V - 1.066V = -2.726V

4. Plug everything needed into the Nernst equation

E_cell = E^o_cell - [(RT)/(nF)]*ln Q

= -2.726V - [(8.314*298.15)/(6*96485)]*ln([1.0]/[0.023]²[0.11]⁶)

=-2.815V

5. Because E_cell is negative, I know this reaction is non-spontaneous

12.1.3

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Cl2(g)+3F2(g)⟶2ClF3(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl₂ and F₂ and the formation of ClF₃.

1. In the this simple reaction R--->P we write the rate expressions as so.

rate= d[P]/dt = -d[R]/dt

2. Also, if the reaction is like this 2R--->P we write the rate expressions as so.

rate= d[P]/dt = -0.5d[R]/dt

3. So using this information we can write the expression for this equation as so.

rate= 0.5d[ClF₃]/dt = -d[Cl₂]/dt = -d[F₂]/3dt

12.5.1

Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

The collision theory states that for a chemical reaction to occur the reacting particles must collide with one another. Requirements must be met in order for the reaction to proceed.

1. Reactants must have enough energy when they collide to react because there is activation energy to overcome. That energy can be small or large, but it must be enough so that molecules can be bumped out of the way properly.

2. Reactants must also be in the correct orientation for the reaction to take place. Atoms must hit each other at the right angle so certain atoms can replace others. This image represents these two concept well.

Collision 1 could work because the hydrogen could brake up the double bond, but all the other orientations will not work. This is because those other orientations are incorrect for the reaction so they will not work.

21.3.5

Write a balanced equation for each of the following nuclear reactions:

1. the production of ¹⁷O from ¹⁴N by α particle bombardment
2. the production of ¹⁴C from ¹⁴N by neutron bombardment
3. the production of ²³³Th from ²³²Th by neutron bombardment
4. the production of ²³⁹U from ²³⁸U by ²H bombardment

1) When writing the balanced equation of a nuclear reactions it is important to identify you reactants, and products along with each of their mass numbers and proton numbers.

Reactant: Nitrogen

Mass number: 14

Proton number: 7

Reactant: Alpha Particle (Helium nucleus)

Mass number: 4

Proton number: 2

Product: Oxygen

Mass number: 17

Proton number: 8

Now, counting the mass numbers on each side of the equation it seems we are short 1 mass number on the products side. Then counting the proton number, we can see we are also down 1 proton on the product side. We then are looking for an atom that has one proton in its nucleus. That would be Hydrogen! So then we can create the full equation now, by adding in the Hydrogen atom to the products.

¹⁴N+⁴He--->¹⁷O+¹H

2) When writing the balanced equation of a nuclear reactions it is important to identify you reactants, and products along with each of their mass numbers and proton numbers.

Reactant: Nitrogen

Mass number: 14

Proton number: 7

Reactant: Neutron

Mass number: 1

Proton number: 0

Product: Carbon

Mass number: 14

Proton number: 6

Now, the total mass number seems to be one short again on the products side. Also, the proton number is one short again on the products side. So I would say we need the Hydrogen nucleus again on the products side. Here is the completed reaction.

¹⁴N+¹n--->¹⁴C+¹H

3) When writing the balanced equation of a nuclear reactions it is important to identify you reactants, and products along with each of their mass numbers and proton numbers.

Reactant: Thorium

Mass number: 232

Proton number: 90

Reactant: Neutron

Mass number: 1

Proton number: 0

Product: Thorium

Mass number: 233

Proton number: 90

The totals for both mass number and proton number are even on both sides of the reaction. This means we dont need to add any extra stuff to the equation. Here is the equation.

²³²Th+¹n--->²³³Th

4) When writing the balanced equation of a nuclear reactions it is important to identify you reactants, and products along with each of their mass numbers and proton numbers.

Reactant: Uranium

Mass number: 238

Proton number: 92

Reactant: Hydrogen

Mass number: 2

Proton number: 1

Product: Uranium

Mass number: 239

Proton number: 92

Totaling the mass number we are one short on the products side. The proton number is also one short on the products side. So we should add the hydrogen nucleus again here. This is the equation.

²³⁸U+²H--->²³⁹U+¹H

20.2.1

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

On the periodic table, starting with the first group on the far left we get an idea of the valence electrons of each atom. Valence electrons are the outer most electrons to that atom. In group 1 we have 1 valence electron, then in group 2, 2 electrons. This continues until you get to group 8 which contain a full outer shell of 8 electrons. A full outer shell creates a very stable atom. So atoms in groups 1-7 will try to either give away or take electrons so to eventually have a full 8 outer shell electrons. The transition metals in between group 2 and 3 have varying valence electron numbers so they are more unpredictable.

1. So, the first group on the periodic table tend to be good oxidants because they only have that one electron on the outside. It is easy for them to give it away during a reaction.

2. The seventh group on the periodic table tend to be good reductants because they only need 1 electron to have a full outer shell. So, they will normally take an electron during a reaction.

20.4.20

Will each reaction occur spontaneously under standard conditions?

1. Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
2. Zn²⁺(aq) + Pb(s) → Zn(s) + Pb²⁺(aq)

A) The first thing to do for both of these equations is to find the standard cell potential.

B) You can do that by splitting them up into the reduction equation and oxidizing equation.

C) After that use the reduction potential table to determine the individual potential values for each smaller equation.

D) Then use the equation E^ocell= E^ocathode - E^oanode (Cathode=Reductor, Anode= Oxidator)

E) Find the number of electrons transferred in the system: n=?

F) Using this value, plug it into this equation to find dG^o

dG^o = -nFE^ocell

G) If dG^o is a negative value then they equation is spontaneous because energy will be released when the reaction comes to completion. However, if it is positive then it is non-spontaneous because energy is needed for it to come to completion.

1) B) Cu_(s) ---> Cu²⁺_(aq) + 2e^- Oxidized

2H⁺_(aq) +2e^- ---> H_2(g) Reduced

C) Oxidized reaction potential: -0.337V

Reduced reaction potential: 0V

D) E^ocell = 0V - (-0.337V) = 0.337V

E) 2 electrons

F) dG^o = -(2)(96485 J/V*mol)(0.337V) = -65030 J/mol

G) This reaction is spontaneous!

2) B) Pb_(s) ---> Pb²⁺_(aq) + 2e^- Oxidized

Zn²⁺_(aq) + 2e^- ---> Zn_(s) Reduced

C) Oxidized reaction potential: 0.126V

Reduced reaction potential: -0.763V

D) E^ocell = -0.763V - 0.126V = -0.889V

E) 2 Electrons

F) dG^o = -(2)(96485 J/V*mol)(-0.889V) = 171550 J/mol

G) This reaction is non-spontaneous!

20.9.5

The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?

1. AlCl₃
2. MgCl₂
3. FeCl₃

For these three salts we can determine how many electrons are transferred based upon the ions involved when they are broken up. In 1. AlCl₃ we can see that it is made up of Al³⁺ and 3Cl^-. So 3 electrons are involved between the two ions. For every electron we get 1/3 of an aluminum ion. so then for this one we can say this...

0.33Al³⁺mol * 26.98g/mol = 8.99grams of Al

For #2 we can say that it brakes up into Mg²⁺ and 2Cl^- ions. So there is 2 electrons involved between the two. So for every electron we get 1/2 of a magnesium ion. so then for this one we can say this...

0.5Mg²⁺mol * 24.31g/mol = 12.16grams of Mg

For #3 we can say that it brakes up into Fe³⁺ and 3Cl^- ions. So there is 3 electrons involved between the two. So for every electron we get 1/3 of an Iron ion. So then for this one we can say this...

0.33Fe³⁺mol * 55.85g/mol = 18.62grams of Fe

14.5.1

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

Activation energy is the amount of energy needed to be overcome in a reaction in order for the product to be reached. Now chemical kinetics is the collection of variables that affect how easily a reaction can proceed. So basically they dictate the intensity of the activation energy for a specific reaction. Now an increase in temperature will speed up the atoms in a solution by making them move faster. This increase in speed of atoms will cause them to collide more often. This then also increases the rate of the reaction. However, the average kinetic energy could still be lower than then activation energy. This is because the increased temperature does not affect kinetic energy, but affects collision rate. So if the temperature is cranked up in a system the rate of a reaction will increase do to faster atom movement. The activation energy will remain the same along with the kinetic energy, but the reaction rate will increase.