Extra Credit 46

Q.17.6.5

Why would a sacrificial anode made of lithium metal be a bad choice despite its \(E_{Li+/Li}^∘=−3.04V\), which appears to be able to protect all the other metals listed in the standard reduction potential table?

S.17.6.5

Lithium appears at the top of the standard reduction table, meaning that it could protect all the metals below it in the standard reduction potential table. However, its chemical properties don't make it a good sacrificial anode. Since lithium is an akali metal, it is highly reactive and can even be explosive in water. Although it will react first to things that can oxidize the metal below it, it will also react to things that might not even oxidize the metal it is protecting. As a result, its high reactivity makes it a potential fire hazard and it would have to be replaced very often if it were to be used as a sacrificial anode.

Q.12.3.9

What is the instantaneous rate of production of N atoms Q12.3.8 in a sample with a carbon-14 content of 1.5 × 10⁻⁹ M?

S.12.3.9

From Q.12.3.8 we see that for the reaction:

\[^{14}_6C \rightarrow^{14}_6N+e^-\]

\[rate=k[_6^{14}C]\]

The rate constant for the radioactive decay of ¹⁴C is 1.21 × 10⁻⁴ year⁻¹.

Using the rate law provided, we can find the instantaneous rate of production of N atoms.

\[rate = (1.21 x 10^{-4} year^{-1})(1.5 x 10^{-9} M) = 1.8 x 10 ^{-13} M/year \]

Q.12.6.1

Why are elementary reactions involving three or more reactants very uncommon?

S.12.6.1

For a reaction involving three reactants to occur, all three reactants would have to collide at the same time with the right orientations and with enough energy. This is much more difficult to achieve with three or more reactants as opposed to a reaction between only two reacants. As a result, termolecular elementary reactions are very uncommon and can be very slow when they do occur.

Q.21.4.3

Define the term half-life and illustrate it with an example.

S.21.4.3

Half-life is the time it takes for the concentration of a reactant to decrease by half its previous concentration, or for half of a sample to decay. For example, carbon-14 has a half-life of 5730 years. A 50g sample of C-14 will have 25g left after 5730 years.

Q.20.3.1

Is \[2NaOH(aq)+H_2SO_4(aq)→Na_2SO_4(aq)+2H_2O(l)\] an oxidation–reduction reaction? Why or why not?

S.20.3.1

We can analyze whether a reaction is oxidation-reduction by assigning oxidation numbers to each element in the reaction and seeing if any element's number changes between the products and reactants. It is useful to refer to these rules for assigning oxidation numbers..

• First, look at the reactant side of the equation. \[2NaOH(aq)+H_2SO_4(aq)\]

In \(NaOH\), \(O\) has an oxidation of \(-2\), \(Na\) has an oxidation number of \(+1\) since it is an alkali metal, and \(H\) has an oxidation number of \(+1\).

In \(H_2SO_4\), \(O\) has an oxidation number of \(-2\), and \(H\) has an oxidation number of \(+1\). Since the compound is neutral, \(S\) will have to have an oxidation number of \(+6\).

• Now, look at the product side of the equation. \[Na_2SO_4(aq)+2H_2O(l)\]

In \(Na_2SO_4\), \(O\) has an oxidation number of \(-2\), and \(Na\) has an oxidation number of \(+1\). Since the compound is neutral, \(S\) has an oxidation number of \(+6\) to make the total state \(0\).

In \(H_2O\), \(O\) has an oxidation number of \(-2\) and \(H\) has an oxidation number of \(+1\).

Compare the oxidation numbers of the elements on both sides of the equations. We see that \(Na\) is \(+1\) on both sides, \(S\) is \(+6\) on both sides, \(O\) is \(-2\) on both sides, and \(H\) is \(+1\) on both sides. Since none of the oxidation numbers changed between the reactants and products, this is not a redox reaction.

Q.20.5.12

How many electrons are transferred during the reaction Pb(s) + Hg₂Cl₂(s) → PbCl₂(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg₂Cl₂ spontaneous? Calculate ΔG° for this reaction.

S.20.5.12

This question can be answered by analyzing the reaction as a redox reaction and splitting it into two half-reactions. Since Cl does not change oxidation states through the reaction, it can effectively be ignored in the half-reactions. The two unbalanced half-reactions can be written as:

\[oxidation: Pb\rightarrow Pb^{2+} \] \[reduction: Hg^{2+}\rightarrow 2Hg \]

We can initially balance the electrons in the oxidation reaction:

\[Pb\rightarrow Pb^{2+} + 2e^- \]

We can then balance the reduction reaction:

\[4e^- + 2Hg^{2+} \rightarrow 2Hg\]

Then multiply the oxidation reaction by 2 so the number of electrons transferred match:

\[2Pb \rightarrow 2Pb^{2+} + 4e^- \]

From balancing the half-reactions we see that 4 total electrons are transferred in the reaction.

The standard cell potential can be found by looking up the half-reactions in a standard reduction potential table. Recall:

\[E^∘_{cell} = E_{cathode} - E_{anode} \]

The metal being reduced (Hg) is the cathode and the metal being oxidized (Pb) is the anode. Here are entries that were found for the above half-reactions:

\[cathode: Hg^{2+}(aq)+2e^-\rightarrow Hg(l), E^ ∘ = 0.855V\]

\[anode: Pb^{2+}(aq)+2e^-\rightarrow Pb(s), E^ ∘ = -0.126V\]

Note that the stoichiometric coefficients of the reaction won't affect the cell potential.

\[E^ ∘{cell} = 0.855V - (-0.126V) = 0.981V \]

Finally, we can use the number of electrons transferred and the standard cell potential to find the ΔG° using

\[ΔG^∘ = -nFE_{cell},\]

where F is Faraday's constant (96,485 C/mol) and n is the number of electrons transferred.

Plug in the values you calculated:

\[ΔG^∘ = -(4 e^-)(96,485 C/mol)(0.981V) = -3.79x10^5 J/mol = -379 kJ/mol \]

Since the value for ΔG° is negative and the standard cell potential is positive, the reaction is spontaneous.

Q.24.6.8

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

1. [Cu(NH₃)₄]²⁺
2. [Ni(CN)₄]²⁻

S.24.6.8

We can predict the structure of the complexes using the spectrochemical series and the metal's oxidation number.

1. \([Cu(NH_3)_4]^{2+}\)

First we find the oxidation state of Cu. Since NH₃ is a neutral compound overall and the total complex has a charge of +2, Cu will have an oxidation state of +2. Cu²⁺ has 9 d-electrons.

Next, we use the spectrochemical series to determine the ligand strength. NH₃ is a strong-field ligand, so it will create a a higher splitting. Since Cu has a \(d9\) configuration and is bonded to 4 strong-field ligands, the structure will most likely be square planar.

Since there is only one possible arrangement of electrons for a \(d9\) configuration, the complex is technically neither high spin nor low spin. There is one unpaired electron.

2. \([Ni(CN)_4]^{2-}\)

CN^- has an overall oxidation state of -1. Since there are four, the total is -4 from the multiple CN^-. Since the total complex has a charge of -2, Ni has an oxidation state of +2. Ni²⁺ has a \(d8\) configuration.

From the spectrochemical series, we see that CN^- is a very strong ligand, creating higher splitting and a low-spin complex. Since Ni has a \(d8\) configuration and will be low spin, its structure is most likely square planar. There are no unpaired electrons in this configuration.

Q.14.7.12

A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.

S.14.7.12

In both cases, the production of Y will be favored.

\[Z\rightleftharpoons X \rightarrow Y\]

In the uncatalyzed reaction, pathway A is favored; therefore, its product Y will be favored naturally.

In the catalyzed reaction, pathway B is favored; however, it s a reversible reaction. Since it is reversible, any product Z that is created can be converted back to X, which could then be irreversibly converted to Y through pathway A. As a result, The production of Y will be favored over time.