Extra Credit 32

Q19.11C

All reactants and products are in their standard states, and use data from the standard electrode reduction potentials table to predict whether the reaction is spontaneous in the forward reaction:

a.) \(\mathrm{Cu(s) + 2Ag^+ (aq) \rightarrow Cu^{2+} + 2Ag (s)}\)

b.) \(\mathrm{2Al (s) + 3Zn^{2+} (aq) \rightarrow 2Al^{3+} (aq) + 3Zn (s)}\)

c.) \(\mathrm{Fe^{2+} (aq) + Ag^+ (aq) \rightarrow Fe^{3+} (aq) + Ag (s)}\)

d.) \(\mathrm{2Fe^{2+} (aq) + I_2 (s) \rightarrow 2Fe^{3+} (aq) + 2I^- (aq)}\)

S19.11C

a.) The cell reaction: \[\mathrm{Cu(s) + 2Ag^+ (aq) \rightarrow Cu^{2+} + 2Ag (s)}\] may be described as having both an oxidation half reaction and a reduction half reaction.

Oxidation: \(\mathrm{Cu (s) \rightarrow Cu^{2+} + 2e^-}\)

Reduction: \(\mathrm{(Ag^+ (aq) + e^- \rightarrow Ag (s)) x2}\)

The equation for standard cell potential between two species is given by the equation \[\mathrm{E°_{(cell)} = E°_{(reduction)} + E°_{(oxidation)}}\] To find \(\mathrm{E°_{(red)}}\), we look to the standard reduction potential table given and determine that \(\mathrm{E°_{(red)}}\) is equal to 0.7996 V.

Similarly, for the oxidation half reaction, we determine that the \(\mathrm{E°_{ox}}\) is equal to -0.3419 V. Recall that for the oxidation reaction, you must take the negative value of the provided reduction potential given on the chart, as we are approximating the reverse reaction for the copper species.

Thus, we may plug these values into the equation \[\mathrm{E°_{(cell)} = (0.7996) + (-0.3419)}\] \[\mathrm{E°_{(cell)} = 0.4577 V}\]

We see that the cell potential is a positive value. If you remember how standard cell potential is related to ΔG, you will recall that when standard cell potential is positive, ΔG is negative, and thus indicates that the reaction is spontaneous. The reaction will proceed in the forward direction.

b.) The cell reaction \[\mathrm{2Al (s) + 3Zn^{2+} (aq) \rightarrow 2Al^{3+} (aq) + 3Zn (s)}\] has the half reactions

Oxidation: \(\mathrm{(Al (s) \rightarrow Al^{3+} + 3e^-) x2}\)

Reduction: \(\mathrm{(Zn^{2+} +2e^- \rightarrow Zn (s)) x3}\)

As in the previous question, we use the same methodology to evaluate \(\mathrm{E°_{(red)}}\) as -0.7618 V, and \(\mathrm{E°_{ox}}\) as 1.676 V by looking at the standard reduction potential table. Note that even though the half reactions have some factor by which they are being multiplied, we do not see the same factor by which the standard cell potential. This is because \(\mathrm{E°}\) is measured in voltage, not in energy per reaction. In other words, it isn't a matter of how many reactions there are, it is a measure of how many reactions it takes to reach that charge. We may now solve for \(\mathrm{E°_{(cell)}}\).

\[\mathrm{E°_{(cell)} = E°_{(reduction)} + E°_{(oxidation)}}\] \[\mathrm{E°_{(cell)} = (-0.7618) + (1.676)}\] \[\mathrm{E°_{(cell)} = 0.9142 V}\]

Because the cell potential is positive, we know that it must correlate to a negative ΔG, and thus be a spontaneous reaction in the forward direction.

c.) The cell reaction: \[\mathrm{Fe^{2+} (aq) + Ag^+ (aq) \rightarrow Fe^{3+} (aq) + Ag (s)}\] has the half reactions

Oxidation: \(\mathrm{Fe^{2+} (aq) \rightarrow Fe^{3+} + e^-}\)

Reduction: \(\mathrm{Ag^+ (aq) + e^- \rightarrow Ag (s)}\)

As we have reviewed thus far, we may find the cell potential of this system by finding the standard potential of each species. In this case, it was found that \(\mathrm{E°_{(red)}}\) was 0.7996 V, and through negating the oxidation species, \(\mathrm{E°_{ox}}\) is -0.771. With these values, we can find that \[\mathrm{E°_{(cell)} = E°_{(reduction)} + E°_{(oxidation)}}\] \[\mathrm{E°_{(cell)} = (0.7996) + (-0.771)}\] \[\mathrm{E°_{(cell)} = 0.0286 V}\]

The \(\mathrm{E°_{(cell)}}\) value that we found is positive, which correlates to a negative ΔG, and is thus a spontaneous reaction in the forward direction.

d.) The cell reaction: \[\mathrm{2Fe^{2+} (aq) + I_{2} (s) \rightarrow 2Fe^{3+} (aq) + 2I^- (aq)}\] has the half reactions

Oxidation: \(\mathrm{(Fe^{2+} (aq) \rightarrow Fe^{3+} (aq) + e^-) x2}\)

Reduction:\(\mathrm{I_{2} (s) + 2e^- \rightarrow 2I^- (aq)}\)

As we have found before, we can find the standard reduction potential for each species. According to the table, \(\mathrm{E°_{(red)}}\) = 0.5355 V, and the negated oxidation species, \(\mathrm{E°_{ox}}\), is equal to -0.771. We can find \(\mathrm{E°_{(cell)}}\) with these values: \[\mathrm{E°_{(cell)} = E°_{(reduction)} + E°_{(oxidation)}}\] \[\mathrm{E°_{(cell)} = (0.5355) + (-0.771)}\] \[\mathrm{E°_{(cell)} = -0.2355 V}\]

Note that in this instance, the cell potential came out to be a negative value. In this case, ΔG came out to be positive, which indicates that the reaction is not spontaneous for the forward reaction. This reaction would require an outside source of energy in order to proceed in the direction that it was written. This cell would be called an electrolytic cell.

Q19.82C

The \(\mathrm{E°_{(cell)}}\) is 0.323 V for the following reaction under non-standard conditions \[\mathrm{Zn (s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)}\] Calculate a \(\mathrm{\Delta G}\) value for \(\ce{[Fe^2+]}\).

S19.82C

The question had stated that the system was under non-standard conditions, so we should know to use the Nernst equation. We can relate cell potential with Gibbs free energy, and equilibrium constant through the equation \[\Delta{G} = -nFE \tag{2}\] Notice how the equation suggests that a positive \(\Delta{G}\) correlates with a negative \(\mathrm{E°_{(cell)}}\), and vice versa. As for the variables of the equation, F stands for Faraday's constant (96500 Coulombs/mole), and n stands for the number of electrons transferred between the electrochemical species in the balanced equation. n may be found by analyzing the oxidation and reduction half reactions for the redox equation. \[\mathrm{Oxidation: Zn (s) \rightarrow Zn^{2+} (aq) + 2e^-}\] \[\mathrm{Reduction: Fe^{2+} (aq) + 2e^- \rightarrow Fe (s)}\] Notice how two electrons are transferred between the two species, so we may say that the value of n is 2. Now we can solve for \(\Delta{G}\): \[\Delta{G} = -(2)(96500Coulombs/mole)(0.323V)\] \[\Delta{G} = -62339 J or -62.339 KJ\]

When solving for \(\Delta{G}\)(\(\mathrm{[Fe^{2+}]}\), we may relate the free energy associated with it to the equation \[\mathrm{\Delta G = \Delta G[Zn^{2+}] -\Delta G[Fe^{2+}]}\] This equation essentially works by the principle of \(\mathrm{\Delta G = \Delta G[products] -\Delta G[reactants]}\), however not every chemical species in the parent equation is used because solids are not incorporated. Ultimately, the calculated value of \(\mathrm{[Fe^{2+}]}\), through the table of delta G values provided, will guide you to get the value -84.77 KJ.

Q21.6C

a.) \(\ce{[CoF6]^3-}\)is seen as green while \(\ce{[Co(H2O)6]^3+}\) is seen as blue.
Explain the color difference.

b.) One of the following complexes appears green and the other appears violet: \(\ce{[Cr(OH)6]^3-}\); \(\ce{[Cr(NH3)6]^3+}\). Indicate which is which, and explain.

S21.6C

a.) \(\ce{[CoF6]^3-}\) and \(\ce{[Co(H2O)6]^3+}\) are both octahedral transition metal complexes that have six monodentate ligands (containing one atom capable of binding to the central atom) attached to the central metal. The question had stated that the hexafluorocobaltate (III) ion appeared green, while the other hexaaquacobaltate (III) appeared blue. The frequency of the color blue is at a higher energy than the frequency associated with green, so why is it that one molecule exhibits one color over the other? The spectrochemical series associated with ligand connections explains. Ligands to the left of the spectrochemical series are considered to be weak field ligands, or in other words lack the field strength to pair electrons in d orbital before moving to the next energy level. The stronger the field strength of a particular ligand, the higher the frequency of the light emitted. Because \(\mathrm{H_{2}O}\) is a stronger field ligand than F, we see that hexaaquacobaltate (III) has a shorter wavelength than hexafluorinecobaltate (III): that \(\ce{[Co(H2O)6]^3+}\) is blue where \(\ce{[CoF6]^3-}\) is green.

b.) \(\ce{[Cr(OH)6]^3-}\) is lower on the spectrochemical series than \(\ce{[Cr(NH3)6]^3+}\). Because \(\mathrm{NH3}\) has a stronger field strength than \(\mathrm{OH^-}\), we should expect the stronger field ligand to absorb a higher energy wavelength than the weaker one. Because the opposite color of violet on the color wheel is yellow, which is higher than the opposite of green, which is red, we should expect \(\mathrm{NH3}\) to appear violet, and the \(\mathrm{OH^-}\) to appear green.

Q24.15B

One of the following statements is true and the other is false regarding the first-order reaction A \(\mathrm{\rightarrow}\) B + C. Identify the true statement and the false one, and explain your reasoning.

1. (a) The rate of the reaction decreases as more of B and C form.
2. (b) The time required for one-half of substance A to react is directly proportional to the quantity of A present initially.

S24.15B

(a) The rate of the reaction decreases as more of B and C form. This statement is true, and we may assess the validity of the statement by analyzing this reaction's first order rate law: \(\mathrm{Rate = k[A]^{1}}\). We know that, in order for more B and C to form, the reactant (A) must be used up simultaneously. Consequentially, according to the rate law equation, a decreased value of [A] times some rate constant value of k will always yield to a lower rate of reaction.

(b) The time required for one-half of substance A to react is directly proportional to the quantity of A present initially. This statement is false, and we may prove so due to the integrated form of first order reactions. With the insight of knowing that a reaction is first order, we may relate time and concentration by using the equation \[\mathrm{ln\dfrac{[A]}{[A]_{0}} = -kt}\] Notice how if we inserted the context of the question into the equation and made \(\mathrm{[A]_{0}}\) twice that of what \(\mathrm{[A]}\) would be, or in other words having \(\mathrm{[A]}\) half of what it's initial concentration is, we're left with the subsequent rate constant, which we should know based on the equation, and time, which we may call half-life or \(\mathrm{t_{1/2}}\). Now that we substituted these values into our skeleton equation, we may rearrange the equation to be shown as \[\mathrm{t_{1/2} = \dfrac{ln2}{k}}\] The equation shown above represents the relationship between the time needed to decay or use up half of the reactant's concentration for a first order reaction. Notice how there isn't any variable for concentration, which indicates that the time needed for half-life also remains constant for any concentration, [A]. In other words, the time required for one-half of substance A to react is not directly proportional to the quantity of A present initially.

Q24.78A

Find an expression to describe the units of rate constant, k, for a reaction in terms of order of the reaction (n) , concentration (M), and time (s). Then use this expression to find the units of the rate constant, k, for a zero, first, and second order reaction.

S24.78A

The equation for rate, controlling for the order of reaction, may be given by the equation: \[\mathrm{rate = k[A]^{n}}\] The units of concentration, [A], are \(\mathrm{M^{n}}\), while the units for rate will always be M/s. The n mentioned in the units of the concentration variable stands for the order of the reaction. This "n" may be measured through experimentation only. Thus, in order to solve for units of k, we can rearrange the equation to be \[\mathrm{k = \dfrac{rate}{[A]^{n}}}\] \[\mathrm{k = \dfrac{M·s^{-1}}{M^{n}}}\] \[\mathrm{k = M^{1-n} · s^{-1}}\] In order to find the units of k for the zero, first, or second order reaction, you must substitute n=0, 1, or 2.

When n=0 (zero order reaction) \[\mathrm{k = M^{1-0} / s}\] \[\mathrm{k = M/s}\]

When n=1 (first order reaction) \[\mathrm{k = M^{1-1} / s}\] \[\mathrm{k = 1/s}\]

When n=2 (second order reaction) \[\mathrm{k = M^{1-2} / s}\] \[\mathrm{k = 1/(M·s)}\]

Q25.33B

What is the energy (in joules) associated with the destruction of \(\mathrm{6 x 10^{-23}}\)g of matter?

S25.33B

Albert Einstein had theorized in the first half of the 20^th century that matter and energy are inexorably linked: specifically when it comes to nuclear processes. When looking at the mass of a specific nucleus, and comparing it to the theoretical mass of the sum of neutrons and protons, we find that it is smaller. This is because some of the mass that we put into the process was turned into energy, or rather, the binding energy. We can directly evaluate how much energy is in a sample mass of matter through Einstein's equation:\[\mathrm{Energy = (mass) · (speed of light)^{2}}\] Since we know the mass of the substance, and we know the speed of light, we can substitute our values \[\mathrm{Energy = (6 x 10^{-26}kg) (2.998 x 10^{8}m/s)^{2}}\] \[\mathrm{Energy = 5.393x10^{-9} \dfrac{kg·m^{2}}{s^{2}} = 5.393x10^{-9}J}\] Since the relationship between joules of energy and kg requires \(\mathrm{J = \dfrac{kg·m^{2}}{s^{2}}}\), we had to change the initial mass in grams and turn it into kilograms. This is the amount of energy that equates to exactly \(\mathrm{6 x 10^{-23}}\)g of matter.

Q21.1.11

There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 10⁶ decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why?

S21.1.11

In order for the boy scout to turn the thorium-232 into uranium-233, he would have had to have a source of neutrons to bombard with. The source of neutrons was taken from bombarding aluminum with alpha particles which came from decaying americium-241. The process may be written as \[\mathrm{Am^{241} \rightarrow He^{4} + Np^{237}}\] \[\mathrm{Al^{27} + He^{4} \rightarrow P^{30} + n^{1}}\] \[\mathrm{Th^{232} + n^{1} \rightarrow Th^{233} \rightarrow Pa^{233} + e^{0} + (antineutrino) \rightarrow U^{233} + e^{0} + (antineutrino)}\] The first step underwent alpha decay to produce an alpha particle, or helium atom, and the third step underwent two beta decays to become uranium-233. The boy was exposed to this process for 2 hours of every day, for two years. We can find the seconds that he spent exposed by \[\mathrm{2 hours · 2(365 days) · 60 minutes · 60 seconds = 5,256,000 seconds}\] We can find out how many decays happened in his presence \[\mathrm{3.5x10^{6}\dfrac{decays}{s} · 5,256,000s = 1.1038x10^{15}decays}\] Because each decay had only one particle thrown out, and we know that each particle had an energy of 5.24 MeV/particle, we may say that the total energy of the system would be \(\mathrm{1.1038x10^{15}decays}\) x 5.24 MeV, which is equal to \(\mathrm{5.78x10^{15}}\) MeV.

The question, however, asks for the exposure of the boy scout in rads, which is different from MeV in that rads is a measurement of energy/mass value. There are \(\mathrm{6.2415x10^{12}}\)MeV/kg for every 100 rads. So, in order to make the conversion, we must find how much energy was put into the 60kg boy scout through \[\mathrm{\dfrac{5.784x10^{13}MeV}{60kg} = 9.64x10^{13}\dfrac{MeV}{kg}}\] Now we have the proper units to convert to rads: \[\mathrm{9.64x10^{13}\dfrac{MeV}{kg} · \dfrac{kg}{6.2415x10^{12}MeV} = 15.44 rads}\] The boy likely had no ill effects from the experience, as he had only received radiation in the form of beta, alpha, and neutron emission, which doesn't do too much damage, cannot penetrate the surface level of the skin, or not ionizing enough to do any damage.

Q24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

(a) \(\mathrm{[Cu(NH_{3})_{4}]^{2+}}\)

(b) \(\mathrm{[Ni(CN)_{4}]^{2-}}\)

S24.6.8

(a) The ion, tetraamminecopper (II), has 4 strong field ammine ligands, and uses the copper (II) ion in the transition metal complex. Copper (II) has the electronic configuration of [Ar] \(\mathrm{3d_{9}}\). The shape of the molecule is square planar. The ammine ligand is typically considered to be strong field ligand, however in the d-orbital splitting diagram, the configuration will be the same. Ultimately there will be one unpaired electron.

(b) The ion, tetracyanonickelate (II), has 4 strong field cyanide ligands, and uses the nickel (II) ion in the transition metal complex. From this information, we may decipher the structure, spin, and number of unpaired electrons present. Nickel (II) has the electronic configuration [Ar] \(\mathrm{3d_{8}}\), and because the ligand associated with the central metal is strong field, we may say that the complex is low-spin. To determine the number of unpaired electrons present, we must draw the associated d-orbital splitting diagram related to the molecule's shape. Because the charge of the ion is 2+, and there are four ligands, we may deduce that the shape is square planar. Therefore, when we use the square planar d-orbital splitting diagram with a low spin approach, we observe that four out of five orbitals are completely filled, with no unpaired electrons left over.