Extra Credit 12

Q19.3C

For the electrochemical cells shown below, write the cell reactions and find \(E^o_{Cell}\) using the Standard Reduction Potential table

a. \({Cu(s)|Cu^{2+}_{(aq)}||Ag^+{(aq)}|Ag(s)}\)

b. \({Cu(s)|Cu^{2+}_{(aq)}||Fe^{2+}_{(aq)},\,Fe^{3+}_{(aq)}|Pt(s)}\)

c. \({PbO2(s)|Pb^{2+}_{(aq)}||Mn^{2+}_{(aq)}|MnO2(s)}\)

d. \({Cu(s)|Cu2^+_{(aq)}||{O_2}_{(g)}|{H_2O_2}_{(aq)}|Pt(s)}\)

S19.3C

Electrochemical cells can be displayed in the format above, which is known as a Cell Diagram. A cell diagram is a shorthand way to show the movement of electrons in a REDOX reaction, specifically a galvanic cell. The single lines represent phase boundaries and the double line represents the two-phase boundary, which separates the two equations. To the left of the two-phase boundary is the anode, which as we read left to right, increases in charge due to loss of electrons, and therefore is an oxidation. To the right of the two-phase boundary is the cathode, which as we read left right, decreases in charge due to the gain of electrons, therefore it is a reduction.

The formula for finding \(E^o_{Cell} = E^o_{red,cathode} - E^o_{red, anode}\). You must first write out both half reactions of the redox reaction (including the number of electrons transferred), then find their corresponding standard reduction value (from the given standard reduction table), then plug the values into the formula. redox reaction (including the number of electrons transferred), then find their corresponding standard reduction value (from the given standard reduction table), then plug the values into the formula.

a.

Since Cu has a charge of 0, due to it being a lone element, and Cu+2 has a charge of +, 2 electrons were transferred; you put the electrons on the side that has a more positive charge in order to balance out the charges. Now both sides have a total charge of 0. Now that you have the half reaction, look up the reaction using the Standard Reduction Potential table.

Anode: \(Cu_{(s)}\rightarrow Cu^{2+}_{(aq)} + 2e^-_{}\); \(E^o_{Cell}\) = .340V

Using the same method as mentioned before, 1 electron is transferred.

Cathode: \(Ag^+_{(aq)} + e^-\rightarrow Ag_{(s)}\)); \(E^o_{Cell}\)

However, the anode and cathode must both have the same amount of electron transfer, therefo re you multiply the equation by 2. This does not affect the standard reduction potential at all, please do not multiply the given values by the number of moles.

Cathode: 2(\(Ag^+_{(aq)} + e^-\rightarrow Ag_{(s)}\)); \(E^o_{Cell}\) = .8V

Plug the values into the \(E^o_{Cell}\) equation.

\(E^o_{Cell}\)= .8V - .340V = .460V

b.

To solve this, use the method from the last problem.

Anode: \(Cu_{(s)}\rightarrow Cu^{2+}_{(aq)} + 2e^-_{}\); \(E^o_{Cell}\) = .340V

This one is a bit different, as there are two ions, as well as Platinum. Platinum serves as an inert electrode and therefore is not part of the reaction. Platinum serves as a conductor for electrons to move into the aqueous cathode solution. For these instances and for the purpose of the calculations, ignore it and just use the ions. There is 1 electron transferred, it is put on the left side to balance out the charge on both sides.

Cathode: \(Fe^{3+}_{(aq)} + e^-\rightarrow Fe^{2+}_{(aq)} {}\)); \(E^o_{Cell}\)

There also is only 1 electron transferred; this is a problem because the electrons transferred in the anode and cathode must be the same amount, as when they are put together to form one REDOX reaction the charges on both sides must be equal. Therefore, in order to get 2 electrons, we multiply the whole reaction by a factor of 2. KEEP IN MIND, THE STOICHIOMETRY OF THE EQUATION DOES NOT EFFECT THE STANDARD REDUCTION VALUE AT ALL.

Cathode: 2(\(Fe^{3+}_{(aq)} + e^-\rightarrow Fe^{2+}_{(aq)} {}\)); \(E^o_{Cell}\)) = .771V

Plug the values into the \(E^o_{Cell}\) equation.

\(E^o_{Cell}\) = .771V - .340V = .431V

c.

For both half reactions, you must balance them out by adding water and hydrogen atoms, as the way it is written in the cell diagram excludes these spectator ions. This is done by adding (a) water molecule(s) to the side without oxygen so that there is an equal amount of oxygen on both sides. Then on the other side, add (a) hydrogen atom(s) so that there is an equal amount of hydrogen on both sides.

anode:

These are both a bit different, as there is an element combined with oxygen. Remember that the charge of Oxygen is always -2, and the total charge is 0, therefore Pb must have a charge of 4, as 4 + 2(-2) = 0 (Multiplied due to there being two mols of oxygen).

\({PbO_2}_{(s)}\rightarrow Pb^{2+}_{(aq)} + 2e^-_{}\)

There is 2 mol of Oxygen on the left, therefore 2 mol of water are added to the left. Now 4 mol of Hydrogen exist on the right side, so you must add 4 hydrogens to the left side.

\({PbO_2}_{(s)} + 4H^+\rightarrow Pb^{2+}_{(aq)} +2{H_2O}_{(l)} +2e^-\); \(E^o_{Cell}\) = 1.455

cathode:

Use the same method above to find the number of electrons transferred.

\(Mn^{2+}_{(aq)} + 2e^-\rightarrow {MnO_2}_{(s)}{}\) =

There is 2 mol of oxygen on the left side, so 2 mol of water is added to the left. Now since there is 4 mol of hydrogen on the left, add 4 Hydrogen on the right.

\(Mn^{2+}_{(aq)} + 2e^- +2{H_2O}_{(l)}\rightarrow {MnO_2}_{(s)} + 4H^+{}\); \(E^o{Cell}\) = -1.23*

*This is negative because the equation is flipped (when comparing it to the standard reduction production table), this is only done if the cathode is flipped. You may keep the value as found on the standard reduction table, that is, without flipping the sign, if you use the formula \(E^o_{Cell} = E^o_{red,cathode} - E^o_{red, anode}\) as we did in the previous problems. You also use \(E^o_{Cell} = E^o_{red,cathode}+ E^o_{ox, anode}\) in order to account for this switch, as the prior method would give you a wrongly negative value.

\(E^o_{Cell}\) = (- 1.23V) + 1.455 = .225V

d.

2 electrons are transferred, add it to the left side to balance the charges on both sides.

anode: \(Cu_{(s)}\rightarrow Cu^{2+}_{(aq)} + 2e^-_{}\); \(E^o_{Cell}\)= .340V

*cathode:

\({O_2}_{(g)} + 2e^-\rightarrow {H_2O_2}_{(l)}\) =

Since there is already equal amounts of oxygen, you only need to add hydrogen; there are 4 hydrogen on the left side, therefore add 4 on the right.

\(4H^+ + {O_2}_{(g)} + 2e^-\rightarrow {H_2O_2}_{(l)}\); \(E^o_{Cell}\) = .695V

Plug the values into the \(E^o_{Cell}\) equation.

\(E^o_{Cell}\) = .695V - .340V = .355V

Q19.37C

Using the Nernst equation and a list of \(E^o_{Cell}\) values, calculate the \(E_{Cell}\) for the following cells:

a. \({Al(s)|Al^{3+}(0.36\:M)||Sn^{4+}(0.086\:M),\,Sn^{2+}(0.54\:M)|Pt(s)}\)

b. \({Sn(s)|Sn^{2+}(0.01\:M)||Pb^{2+}(0.700\:M)|Pb(s)}\)

S19.37C (original solution incorrect)

The Nernst equation is:

\[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}\] \[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{.0591}{n}\right)\ln Q \label{Eq4}\]

Since we are not given a temperature, we will assume standard conditions of 298K; therefore we can use this simplified version.

a)

first we will find \(E^o_{Cell}\) by solving for the half reactions and finding their standard reduction potential; the cell diagram gives both the anode and cathode reactants and products in order, therefore you simply just write them out as an equation, and balance them by adding the correct amount of electrons to the side with the higher charge. Notice that the two equations need to have an equal amount of electrons transferred, therefore they are multiplied by a number that will help them equal the same amount. Ignore the molarity for now.

There is a transfer of 3 electrons, they are placed on the left to balance out the 3+ charge of Al, making both sides equal in charge.

Anode: \(Al_{(s)}\rightarrow Al^{3+}_{(aq)} + 3e^-_{}\)

There is a transfer of 2 electrons (4-2 =2), they are placed on the right side to balance out the 4+ charge, making both sides equal in charge.

Cathode: \(Sn^{4+}_{(aq)} + 2e^-\rightarrow Sn^{2+}_{(aq)} {}\)

The number of electrons is not equal. Therefore you must make them equal by multiplying each equation by the number of electrons transferred in the opposite reaction; the anode will be multiplied by 2, and the cathode by 3. This gives both equations 6 electrons transferred. Use the Standard Reduction Potential table to find \(E^o{Cell}\) values (remember the moles of reaction do not affect these values), then plug them into the \(E^o_{Cell}\) equation.

Anode: 2(\(Al_{(s)}\rightarrow Al^{3+}_{(aq)} + 3e^-_{}\)); \(E^o_{Cell}\) = -1.676V

Cathode: 3(\(Sn^{4+}_{(aq)} + 2e^-\rightarrow Sn^{2+}_{(aq)} {}\)); \(E^o_{Cell}\) = .154V

Plug the values into the the equation:

\(E^o_{Cell} = E^o_{red,cathode} - E^o_{red, anode}\) = .154V - (-1.676V) = 1.83V

Next we plug our \(E^o{Cell}\) value into the Nernst equation. N = the number of electrons transferred in the redox reaction, which was 6 elections. Lastly we plug in values for Q, which is the molarity of the ions of the products over the molarity of the ions of the reactants, each raised to the power of their stoichiometry.

\[\log Q_{eq} = \dfrac{[Sn^{2+}]^{3} \times [Al^{3+}]^{2}}{[Sn^{4+}]^{3}}\]

\[\log Q_{eq} = \dfrac{.54^3 \times .36^2}{0.086^3}\]

\[\log Q_{eq} = 1.506\]

\[E_\textrm{cell}=1.83V -\left(\dfrac{.0592}{6}\right)(1.506) =1.815V\]

b)

Do what you did previously, and write out the half reactions and find their standard reduction potentials, in order to find \(E^o_{Cell}\).

Anode: \(Sn_{(s)}\rightarrow Sn^{2+}_{(aq)} + 2e^-_{}\); \(E^o_{Cell}\) = -.137V

Cathode: \(Pb^{2+}_{(aq)} + 2e^-\rightarrow Pb_{(s)}\); \(E^o_{Cell}\) = -.125V

Plug the values into the equation.

\(E^o_{Cell} = E^o_{red,cathode} - E^o_{red, anode}\) = -.125V - (-.137V) = .012V

We do the same as before, expect this time only 2 electrons were transferred

\[\log Q_{eq} = log\dfrac{[Sn^{2+}] }{[Pb^{2+}]}\]

\[\log Q_{eq} = log\dfrac{0.01}{0.700}\]

\[\log Q_{eq} = -1.845\]

\[E_\textrm{cell}=.012V -\left(\dfrac{.0592}{2}\right)(-1.845) =.0666V\]

Q20.57A

Predict the molecular formulas for the following metal carbonyls and explain how many electrons are contributed by the metal and how many are contributed by the carbon monoxide and which noble gas electron configuration does the molecule exhibit.

a. Titanium (T)

b. Manganese (Mn)

c. Tungsten (W)

S20.57A

A carbonyl is simply a carbon monoxide, which can act as a neutral ligand for transition metals. When bonding, it donates \(\mathrm{2e^{-}}\) through single bonding.

a.

Titanium has 22 electrons, yet it needs to reach 36 electrons to have the noble gas configuration of Krypton.

\({Ti(CO)_?}\)

Ligands form either tetrahedral (4 bonds) or octahedral (6 bonds) with metals, therefore it is either:

\({Ti(CO)_?}\) = \({Ti(CO)_4}\)

or

\({Ti(CO)_?}\) = \({Ti(CO)_6}\)

However, It cannot be a tetrahedral as 22 electrons + 4 moles Co (2 electrons donated) = 30 electrons (remember a carbonyl donates 2 electrons each); which six short of 36.

At first glance, it seems it cannot be octahedral either as 22 electrons + 6 mol Co (2 electrons donated) = 34 electrons total, is two short of 36; however, this is much closer to the goal of 36 electrons then it was if it was tetrahedral, so it cannot be anything else.

Therefore:

\({Ti(CO)_?}\) = \({Ti(CO)_6}\)

To get the 2 extra electrons, consider the possibility the compound is an ion with a charge of negative 2 (this means 2 extra electrons are added); therefore the number of electrons will be 36 and this compound has reached noble gas configuration of Krypton. This would allow a positively charged ion to come along and balance the charge out.

Answer:

\({Ti(CO)_6}^{2-}\)

*Notice, you cannot add 7 CO, as it is impossible for so many bonds to occur.

b.

Manganese is the same, as it can be either tetrahedral (make 4 bonds) or octahedral (make 6 bonds).

\({Mn(CO)_?}\) = \({Mn(CO)_4}\)

or

\({Mn(CO)_?}\) = \({Mn(CO)_6}\)

Manganese has 25 electrons, it needs a total of 36 to reach the configuration of Krypton.

Like the previous problem, it cannot be tetrahedral, as 25 electrons + 4mols CO (2 electrons donated) = 33 total electrons, this is too far away from the goal of 36.

The octahedral structure is a better choice, as 25 electrons + 6 mols CO (2 electrons donated) = 37; this is much closer to 36.

therefore:

\({Mn(CO)_?}\) = \({Mn(CO)_6}\)

However, this has 1 extra electron, therefore the compound will hav e a positive charge of plus one (this means one electron is lost).

Another possibility is that it could bond to a hydrogen instead of a sixth carbonyl, however, the former is the more correct answer in this context. With the positive charge, some negative ion will likely bond to the complex.

answer:

\({Mn(CO)_6}^+\) (or \({HMn(Co)_5}\))

c.

Once again, either a tetrahedral or octahedral can be formed.

\({W(CO)_?}\) = \({W(CO)_4}\)

or

\({W(CO)_?}\) = \({W(CO)_6}\)

This one is quite simple, W has 74 electrons, it needs to reach a total of 86 electrons to reach the configuration of the noble gas Radon. It will have a stable and relatively unreactive composition.

It cannot be tetrahedral as 74 electrons + 4 mol Co (2 electrons donated) = 84

Therefore it is octahedral, 74 electrons + 6 mol Co (2 electrons donated) = 86; this is the exact amount of electrons, therefore the answer is simply:

\({W(CO)_6}\)

Q24.3A

In the reaction \(\mathrm{A \rightarrow B}\), \(\mathrm{[A]}\) is found to be 0.750M at \(\mathrm{t = 61.2\,s}\) and 0.704M at \(\mathrm{t = 73.5\,s}\). Find the average rate of the reaction during this time interval.

S24.3A

To find the average rate of the reaction, you essentially find slope. The units for rate of the reaction is M/S, therefore it will be the change of concentration (final concentration - initial concentration) dived by the change in time (final time - initial time). The rate of the reaction measures how fast the reactants "disappear" (turn into products) therefore this equation has a negative sign in front of it. If we were measuring the rate of [B], it would be the same value that we had just calculated multiplied by -1. The rate of disappearance is equal to the opposite of the rate of production in this case.

\(\mathrm{Rate = -\left(\dfrac{\Delta[A]}{\Delta t}\right) = - \dfrac{0.704\,M-0.750\,M}{73.5\,s-61.2\,s} = 0.00374 = 3.74 \times 10^{-3}\, Ms^{-1}}\)

Q24.45C

Explain why

(a) A reaction rate cannot be calculated from the solely collision frequency.

(b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency rises a lot more slowly

(c) Introducing a catalyst to a reaction mixture can have such a significant impact on the rate of the reaction, even if the temperature is held constant.

S24.45C

(a) When considering a reaction rate, you must keep in mind the collision theory. When two reactants collide, they must collide with the right amount of energy to bind together (and break their bonds); however, this is not the only factor. Even if the reactants hit each other multiple times (which happens in correlation to a frequency increase), it does not mean they will react, as they need to collide in the right way; in the right orientation with enough energy. Therefore even if the collision frequency increased, there are still many other factors to consider, such as concentration and temperature. Therefore you could not accurately calculate reaction rate.

(b) When thinking about this question, you must consider all the factors that affect the rate of the reaction. This can be summarized nicely by this:

rate = (collision frequency)(steric factor)(fraction of collisions with E > Ea)

When increasing temperature, the molecules will increase in speed, and therefore increase in collision frequency. The fraction of collisions with E > Ea will also increase, as increasing temperature correlates with increasing energy, and the higher the rise in temperature, the more collisions with E > Ea will occur. Overall, temperature increases both of these factors simultaneously. The two factors themselves might not change as drastically as the rate of the reaction; however, since these two values are multiplied together the rate of the reaction will increase dramatically.

Another way to look at it is using the Arrhenius equation which is:

The A is a constant for each chemical reaction that defines the rate due to the frequency of collisions in the correct orientation. Looking at this equation, we can see that A is raised to a coefficient, therefore it can increase dramatically with temperature, even if the number of collisions does not increase dramatically.

(c) An enzyme is a kind of biological catalyst, which provides a "short cut" of a sort for the reaction by lowering the activation energy (the amount of energy needed to make a reaction "go"). The catalyst is not used up in a reaction, rather it forms an intermediate product with one of the reactants, this then reacts with the remaining reactant; the activation energy of both these steps is much lower than the original reaction without the catalyst. Therefore, since the activation energy is lower, the rate of the reaction will increase, as it is easier to attain this smaller amount of activation energy. Usually activation energy is reached by high heat, but since the activation energy is lower, it requires less heat, therefore the reaction's rate can increase with a catalyst even if held at a constant temperature.

Q25.21D

The disintegration rate of a sample with \(\ce{^{205}_{83}Bi}\) as the only radioactive nuclide is 3000 dis h^-1. The half-life of \(\ce{^{205}_{83}Bi}\) is 15.31 days. Estimate the number of atoms in the sample.

Relevant equations:
\(\mathrm{A=\lambda N}\), \(\mathrm{\lambda = \dfrac{0.693}{t_{1/2}}}\)

S25.21D

Since they already give you the equation, it is mostly up to one's understanding of the equations. The first equation is used to find activity, with A = activity (in dis/time, which is also known as becquerel), \({\lambda}\) = the decay constant, and n = the number of atoms in the reaction. The second reaction used to find the half-life; since nuclear reactions are 1st order reactions, you simply divide .693 (ln2) by the nuclear constant. Remember that first order reactions are independent of concentration or origonal quantity.

Since you do not have the decay constant to solve for the number of atoms in the same, you can use the half life equation to find it. Take note that the activity is dis per hours, so you must convert days into hours.

\(\mathrm{\lambda = \left(\dfrac{0.693}{15.31\: days}\right) \times \left(\dfrac{1\: day}{24\: hours}\right) = 0.00188602\: h^{-1}}\)

Now that you have the decay constant, you can solve for activity. You can arrange the equation for finding activity and solve for N.

\(\mathrm{N=\dfrac{A}{\lambda}}\)

\(\mathrm{N = \dfrac{3000\: dis\: h^{-1}}{0.00188602\: h^{-1}} = 1590649.357\: atoms\: \ce{^{205}_{83}Bi}}\)

Q19.6

What new element is formed when \(\textrm{K-40}\) decays by \(\beta\)-emission? Is the new element formed likely to be stable? Why or why not?

S19.6

\(\textrm{K-40}\) has 19 protons, as it is the 19th element on the periodic table; therefore it can be written as \(\ce{^{40}_{19}K}\). \(\beta\)- emission occurs when a heavy nucleus with too many protons or neutrons are transformed into each other, in this case since it is emission (also known as decay) a neutron decays into a proton, an electron, and an antineutrino (has neither mass nor charge). A \(beta\) particle is simply an electron, so you would place it on the right side since it is one of the products of the emission.

\(\ce{^{40}_{19}K} \rightarrow \ce{^{0}_{-1} e^{-}} + \ce{^{?}_{?}element}\ + antineutrino\)

Then you would evaluate the mass number and the atomic number of the left side, then make the right side have an equal total sum of mass and atomic numbers.

The mass on the left side is 40; on the right, since the mass number of an electron is 0, this unknown element has a mass of 40 (0 + 40 =40).

\(\ce{^{40}_{19}K} \rightarrow \ce{^{0}_{-1} e^{-}} + \ce{^{40}_{?}element}\ + antineutrino\)

The atomic number on the left side is 19; therefore the atomic number total on the right side must also be 19. Since an electron is -1, the element must have an atomic number of 20 (19 = -1 +20).

\(\ce{^{40}_{19}K} \rightarrow \ce{^{0}_{-1} e^{-}} + \ce{^{40}_{20}element}\ + antineutrino\)

This means that the element is Calcium, as it is the 20th element on the periodic table.

\(\ce{^{40}_{19}K} \rightarrow \ce{^{0}_{-1} e^{-}} + \ce{^{40}_{20}Ca}\ + antineutrino\)

\(\textrm{Ca-40}\) would be very stable, as the neuron (found by subtracting the atomic number from the mass number; 40-20=20) and the atomic number are both 20; therefore the ratio is 1:1. This ratio signifies that the element is stable, you can also note that both neutron and protons are even numbers as well, which also correlates to stability.

One could also instantly tell this is stable without looking at the ratio, as 20 is the "magic number" of stability for both neutrons and protons without exception.

Q21.3.5

A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer.

S21.3.5

This statement actually has merit, as stars act like nuclear reactors, as they convert elements into new elements, in the form of nuclear fusion. Starting with hydrogen, hydrogen will then make helium:

\[_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta
\\_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma
\\_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}\]

Much like this reaction, different types of elements are synthesized within a star's lifetime using fusion. These processes happen at extremely high temperatures and pressures within the core of a star.

Heavier elements that are Z > 28 are made due to supernovas. When a supernova explodes (due to the collapse of the core) it creates a nebula (space dust). This is where all the new elements that are made are distributed out of the dead star and into the universe. During this process there are so many neutrons that multiple proton-capture occur, leading to the creation of heavier elements that perhaps have less stable nuclides.

ex:

\[_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}\]

Therefore, since we ourselves are made of elements, we are made of stardust.