Extra Credit 25

Q19.8B

Cr₂O₂⁻⁷_(aq)+14H⁺_(aq)+6Ag_(s)→2Cr³⁺_(aq)+6Ag⁺_(aq)+7H₂O_(l)

Use the data table to determine:

1. E^∘cell
2. K
3. ΔG^∘
4. Whether the reaction goes substantially to completion when the reactants and products are initially in their standard state.

S19.8B

1. E^∘cell can be calculated using the standard reduction potentials given in Table 2. Using the reaction above, we can write the half reactions for each species. The half-reaction tells us which species is oxidized and which is reduced. Oxidation occurs when electrons are lost, and reduction occurs when electrons are gained. Oxidation occurs at the anode and reduction occurs at the cathode.

The half-reactions are as follows:

Cathode: 14H⁺_(aq) + Cr₂O₇^2-_(aq) +6e^- → 2Cr³⁺_(aq) +7H₂O_(l) (1.33V)

Anode: 6Ag_(s) → 6Ag⁺_(aq) +6e^- (0.800V)

From this we can use the equation: E^∘cell = (E^∘cathode) - (E^∘anode)

= (1.33V) - (0.800V)

=+0.53 V

2. In order to find K, we can use the equation E^∘cell=(-RT/nF)(ln K). R is the gas constant, T is temperature in Kelvin, n is the moles of electrons transferred, and F is Farraday's constant. K is the equilibrium constant we are looking to calculate.

We found E^∘cell in part 1 of the question - 0.53 V. From there we use the equation. We can algebraically rearrange the equation to solve for K.

This will result in the equation K=e^{(-nFE∘cell)/(RT)}

Plugging our values into the equation we get K=e^{(-6 mol x96485 C/mol e x 0.53 J/C)/(8.314 J/mol K x 298 K)}

Calculating this value yields 6.07x10⁵³.

3. We can use the following equation that describes the relationship between E^∘cell and ΔG^∘ under standard conditions where ΔG^∘ = Gibbs free energy, n=moles of electrons transferred, F=Farraday's constant, and E^∘cell = cell potential in Volts.

ΔG^∘ = -nFE^∘_cell

Using this equation, we can find ΔG^∘ easily. From the half reactions we found in part 1, we know that 6 mol e^- were transferred in the cell from the anode to the cathode. Farraday's constant is 96,485 Coulombs/mol e^-, and E^∘cell in volts can be changed to the equivalent unit of Joules/Coulomb.

ΔG^∘ = (6 mol e^-)(96,485 C/mol e^-)(0.53 J/C)

=-30682.23 J or -306.8 kJ.

From this, we can also tell that the reaction is spontaneous.

4. Yes, this reaction goes substantially to completion when the reactants and products are initially in their standard states. We know this because the equilibrium constant value (K) is so large. This means that the reaction readily goes from reactants to products. A large K value indicates that the equilibrium constant for the products is large and that the reaction will proceed to the right towards the products side, resulting in an increase of the concentration of the products.

Additionally, in part 3 we found that the ΔG^∘ is a very large negative value. This means that the reaction is extremely spontaneous, so the reaction can be assumed to go to completion from reactants to products in the standard state.

Q19.65B

Which of the following reactions are spontaneous under standard conditions? If the reaction is not spontaneous, determine its minimum voltage applied to undergo electrolysis.

1. Sn²⁺_(aq)+Zn²⁺_(aq)→Zn_(s)+Sn⁴⁺_(aq)Sn²⁺_(aq)+Zn²⁺_(aq)→Zn_(s)+Sn⁴⁺_(aq)
2. Li⁺_(aq)+Fe²⁺_(aq)→Li_(s)+Fe³⁺_(aq)Li⁺_(aq)+Fe²⁺_(aq)→Li_(s)+Fe³⁺_(aq)
3. 2H_2(g)+O_2(g)[in1MH⁺_(aq)]→2H₂O_(l)2H_2(g)+O_2(g)[in1MH⁺_(aq)]→2H₂O_(l)
4. 2Al³⁺_(aq)+3Pb_(s)→3Pb²⁺_(aq)+2Al_(s)2Al³⁺_(aq)+3Pb_(s)→3Pb²⁺_(aq)+2Al_(s)

S19.65B

Electrolysis is the use of an electric current to make a non-spontaneous reaction occur. It can be used to separate a substance into its original components. In electrolysis, an electric current is passed through an electrolyte and into aqueous solution in order to stimulate the flow of ions needed to make a non-spontaneous reaction occur.

Furthermore, in an electrolytic cell, a reaction must be induced and the anode and cathode are reversed (anode, left and cathode, right) - which is the opposite of a voltaic cell. In each of these questions, the process is the same - find the half reactions of both to determine which species is losing electrons and which species is gaining electrons. If the reaction is non-spontaneous, we flip the sign of the cell potential to determine the minimum voltage that needs to be applied for electrolysis to occur.

1. Sn²⁺_(aq)+Zn²⁺_(aq)→Zn_(s)+Sn⁴⁺_(aq)Sn²⁺_(aq)+Zn²⁺_(aq)→Zn_(s)+Sn⁴⁺_(aq)

The half reactions are as follows:

Cathode: Zn²⁺ + 2e^- --> Zn_(s) (-0.763 V)

Anode: Sn²⁺ --> Sn⁴⁺ + 2e^- (+0.154 V)

We use the equation:

E^∘cell = (E^∘cathode) - (E^∘anode)

And we plug in the values: (-0.763 V) -(+0.154 V) = -0.917 V.

From this data, we know that the reaction is non-spontaneous because the Ecell value is negative. So, we flip the sign of the Ecell value to determine the minimum voltage needed to make electrolysis occur and for the reaction to occur.

This value is +0.917 V.

2. Li⁺_(aq)+Fe²⁺_(aq)→Li_(s)+Fe³⁺_(aq)Li⁺_(aq)+Fe²⁺_(aq)→Li_(s)+Fe³⁺_(aq)

First we write the half-reactions to find out which species is losing electrons and which is gaining electrons. They are as follows:

Cathode: Li_(s) + e^- --> Li_(s) + Fe³⁺ (-3.040 V)

Anode: Fe²⁺ --> Fe³⁺ + e^- (+0.771 V)

We again use the equation:

E^∘cell = (E^∘cathode) - (E^∘anode)

And we plug in the values: (-3.040 V) - (+0.771 V) = -3.811 V.

From this data, we know that the reaction is non-spontaneous because the Ecell value is negative. So, again we flip the sign of Ecell value to determine the minimum voltage needed to undergo electrolysis.

This value is +3.811 V.

3. 2H_2(g)+O_2(g)[in1MH⁺_(aq)]→2H₂O_(l)2H_2(g)+O_2(g)[in1MH⁺_(aq)]→2H₂O_(l)

Again we write the half reactions to find out which species loses electrons and which species gains electrons.

The half-reactions are as follows:

Cathode: O_2(g) + 4H⁺ +4e^- --> 2H₂O_(l) (+1.229 V)

Anode: 2H_2(g) --> 4H⁺ + 4e^- (0 V)

We again use the equation: E^∘cell = (E^∘cathode) - (E^∘anode)

Then, we plug in the standard reduction potential values. This will be (+1.229 V) - (0 V) = +1.229 V.

No electrolysis is needed because we have a positive Ecell value, meaning the reaction is spontaneous and no electric current must be passed through solution in order for the reaction to occur.

4. 2Al³⁺_(aq)+3Pb_(s)→3Pb²⁺_(aq)+2Al_(s)2Al³⁺_(aq)+3Pb_(s)→3Pb²⁺_(aq)+2Al_(s)

We again find the half reactions to see which species gains electrons and which species loses electrons.

The half reactions are as follows:

Cathode: Al³⁺ + 3e^- --> Al_(s) (-1.676 V)

Anode: Pb_(s) --> 2e^- + Pb²⁺ (-0.126 V)

We again use the equation: E^∘cell = (E^∘cathode) - (E^∘anode).

Then we plug in the SRP values: (-1.676 V) - (-0.126 V) = -1.55 V.

Since the Ecell value is negative, we flip the sign to determine the minimum voltage needed to undergo electrolysis.

We find that +1.55V are needed to make electrolysis occur.

Q21.5B

1. Use crystal field theory to draw the most probable structure of [CoF₆]³⁻ in a weak field.
2. Determine whether [Cu(H₂O)₄]²⁺ is paramagnetic or diamagnetic

S21.5B

1. First, we must determine the charge of cobalt in [CoF₆]^3-. Since the fluorine ion has a -1 charge, we multiply that by 6 to get a -6 charge. This must be balanced out by the cobalt metal, but since the ligand has a 3- charge, we can determine that cobalt in this ligand is a Co³⁺ cation.

This will yield the electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁶.

Therefore there are 6 electrons in its d orbital.

Now, we can draw the most probable structure of [CoF₆]³⁻ in a weak field. A weak field means that there is a small crystal field splitting, which leads to a high spin. This means that the electrons will go to high spin before beginning to pair leftover electrons. In a weak field (∆_o˂P), meaning it takes less energy to go to high spin than pair electrons.

The diagram is shown here:

This shows that there are 4 unpaired electrons in its d orbital and is therefore paramagnetic.

2. To find out whether [Cu(H₂O)₄]²⁺ is paramagnetic or diamagnetic, we must first find the charge of Copper, as in part 1 above. Since H₂O is a neutral ligand, the charge of Cu is 2+.

Since the cation in the complex is Cu²⁺, the electron configuration is as follows:

1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁹. Therefore there are 9 electrons in the d orbital.

To draw the electron diagram, we must also consider that H₂O is higher up on the spectrochemical series. This means that crystal field splitting will be large and that the electrons will pair and go to low spin before filling up the top row on the diagram.

We can now draw the electron diagram for the d shell:

We see that there is one electron unpaired, meaning that [Cu(H₂O)₄]²⁺ is paramagnetic.

Q24.13B

The following rates of reaction were obtained in three experiments with the reaction 2NO(g) + Cl₂(g) --> 2NOCl(g).

What is the rate law of this reaction?

S24.13B

To find the rate law of this reaction, we must find it by observing the data and solving for the coefficients of the concentration for the rate law. The balanced chemical equation will not help in determining the rate law for the reaction.

First we begin by determining that in experiment 2, the concentration of [Cl₂] is doubled, while [NO] remains constant from experiment 1. We can set up the equation for the rate law and plug in our values:

2.23 x 10^-5 = k(0.0125)^x(0.0255)^y

4.46 x 10^-5 = k(0.0125)^x(0.0510)^y

Dividing these two equations, we obtain: 2=2^y and solving for y, y=1.

We can then set up our second set of equations.

We see that in experiment 3 the concentration of [NO] doubles and that [Cl₂] remains constant from experiment 1.

Now, we can set up our second set of equations.

2.23 x 10^-5 = k(0.0125)^x(0.0255)^y

8.92 x 10^-5 = k(0.0250)^x(0.0255)^y

Dividing these two equations, we obtain: 4=2^x and solving for x, x=2.

Now we can successfully set up our rate law to solve for k, the rate constant.

r = 2.0k[NO]²[Cl₂]

We can solve for the rate constant by plugging in this equation to any one of the equations above. For example, we can use 2.23 x 10^-5 = k(0.0125)^x(0.0255)^y and plug in our rate law. Plugging it in, we get 2.23 x 10^-5 = 2.00k(0.0125)²(0.0255) and we solve for k and obtain k=5.597.

Our final rate law is:

r=5.597[NO]²[Cl₂]

Q24.60A

The following substrate's concentration [S]versus time data were obtained during an enzyme-catalyzed reaction: t=0 min; [S]=1.00M[S]=1.00M; 30 min, 0.90M0.90M; 90 min, 0.70M0.70M; 120 min, 0.50M0.50M; 180 min, 0.20M0.20M. What order is this reaction with respect to S in the concentration.

S24.60A

In order to find the order of reaction when give time vs. concentration, we must plot the data and determine which reaction order equation will yield a straight line.

To plot the integrated rate law for a zero order reaction, we graph time vs. concentration.

This yields the following graph:

Since this integrated rate law produces a relatively straight line, we can determine that the reaction is a zero order reaction.

To further illustrate this point, we can plot the integrated rate forms of the first and second order reaction and see what the graph yields to check if our solution is correct.

Plotting the first order integrated rate form yields a downward curve, so we know that it is NOT a first order reaction.

Plotting a second order integrated rate form yields an upward curve, so we know it is also NOT a second order reaction.

Q25.29A

What should be the mass ratio of Po-210/U-238 in an object that is around 2.4X10⁹ years old? The half-life of U-238 is 4.47X10⁹ years. [Hint: one Po-210 atom is final decay product of one U-238 atom]

S25.29A

First, we must find the decay constant by using the equation:

λ=A/N, where A is a constant (ln 2) and N is the given radioactive half-life in yeras.

We plug in our values to find the decay constant.

λ=(0.693)/(4.47x10⁹ y) = 1.55x10^-10 1/y

We then find the ratio of Po atoms after 2.4x10⁹ years to the initial amount of Po atoms.

We use the equation:

ln(N_p/N_o)=−kt

Then, we plug in our values and solve for this ratio. K will be our decay constant found above, and t will be the "age" of the object/time elapsed.

ln(N_p/N_o) = -(1.55x10^-10)(2.4x10⁹)

ln(N_p/N_o) = -0.372

Raise each side by the power of e to eliminate the ln function.

e^-0.372= 0.689

N_p/N_o = 0.689

From this calculation, we know that for every mole of U-238 that is present initially, there will be 0.689 moles of U-238 and 0.310 moles of Po after 2.9x10⁹ years have elapsed.

To calculate the mass ratio, we need the molar mass of U-238 and Po-210.

(0.310mol ²¹⁰Po/0.689mol²³⁸U)×(1 mol ²³⁸U/238 g U)×(210 g Po/1 mol ²¹⁰Po)

= 65.1 g ²¹⁰Po / 166.12 g ²³⁸U

= 0.392 g ²¹⁰Po/1 g ²³⁸U

Q21.1.4

Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters.

S21.1.4

During the 20th century, the building of fallout shelters became popularized as nuclear weapons began to develop and nuclear power had been studied to use for energy purposes. During the Cold War, fallout shelters were prepared by governments and private citizens in their own homes in order to prepare for a nuclear crisis. Fear grew in the 1950s with the development of the Hydrogen bomb by both the USA and the Soviet Union. Many fallout shelters were built during both the Eisenhower and Kennedy administrations as it seemed that nuclear strikes were imminent.

Fallout shelters were built in order to protect people from radiation that follows a nuclear strike. People built them with concrete padding to block out gamma ray radiation. A basic shelter blocked gamma ray radiation by a factor of 1000 - meaning that concrete shelters needed to be at least 3.6 inches thick. Though the Cold War was an arms race and nuclear strike never occurred, these shelters were found to be pretty effective given that they were well-shielded by concrete that was thick enough, stocked with supplies, and had adequate air circulation.

Additionally, after the Chernobyl disaster in 1986, the USSR built a massive steel and concrete structure to cover the nuclear reactor #4 building of the Chernobyl Nuclear Power Plant to limit radioactive contamination of the environment. Though the structure was unstable immediately following this construction, further construction occurred to more stably and effectively block off the radioactive site and was completed in November of 2016.

Q24.6.1

Describe crystal field theory in terms of its

1. assumptions regarding metal–ligand interactions.
2. weaknesses and strengths compared with valence bond theory.

S24.6.1

1. The crystal field theory is a theory that describes the breaking of degeneracies of electron orbital states, usually for f or d orbitals, due to a static electric field that is created by surrounding charge distributions of anions. The basic concept of the crystal field theory is that ions are assumed to be "simple point charges."

The key assumption regarding metal-ligand interaction in the crystal field theory is that metal-ligand interactions are purely electrostatic in nature. This assumption however is not accurate for many complexes (complexes with neutral ligands like CO, for example). But, the crystal field theory lets chemists explain several properties of many transition metal complexes fairly accurately.

2. The crystal field theory explains many properties of transition metals that cannot be explained using valence bond theory alone. Complex formation is assumed to be the consequence of electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles around a metal ion.

Valence bond theory describes bonding using the hybridization of orbitals and electron pairs as an extension of Lewis structure methods that are commonly used for simpler molecules. Crystal field theory is an electrostatic approach that describes the split in d-orbital energies. It gives an approximate description of electronic energy levels that determine the ultraviolet and visible spectra, but does NOT describe bonding.

So to summarize the above, valence bond theory describes bond formation of hybridization while crystal field theory does not describe bonding. However, crystal field theory does explain how complex formation occurs between a central metal and negatively charged species to produce a complex structure. Using crystal field theory, we can determine bond strength, energy properties, electron distribution, and magnetic properties such as the color of the complex system.