Extra Credit 5

Solution 17.1.5

We are given the oxidation and reduction half reactions for a reaction. We can add the half reactions together to get the overall reaction, but we must make sure that electrons are not involved in the final reaction equation. Therefore, we must make sure that the amount of electrons on the right side of the oxidation equation equal the amount of electrons on the left side of the reduction equation. This ensures that the electrons will cancel out in the final equation.

What is the purpose behind balancing these half reactions? To balance or confirm that a chemical reaction is a redox reaction we typically split the reaction into its corresponding half reactions and use them as a guide; oxidation indicates a loss of an electron, and reduction indicates gaining an electron-- in terms of oxidation states, or numbers, oxidation is an increase ( for ex. +1 to +2) and reduction is a decrease ( +2 to +1). If a reaction has both, that is one element is increasing in its oxidation state, and another is being reduced, then you can confirm that the reaction being studied is a redox reaction.

More on balancing redox reactions: here

Problem 1

Oxidation half reaction: \({Ca} \rightarrow {Ca^{2+}}+{2e}^{−}\)

Reduction half reaction: \({F}_{2}+{2e}^{−} \rightarrow {2F}^{−}\)

The electron number in each half reaction is equal, so the electrons cancel and we get our final equation.

\[{F_2}+{Ca} \rightarrow {2F^−}+{Ca^{2+}}\]

Problem 2

Oxidation half reaction: \(2({Li} \rightarrow {Li^+}+{e^−})\)

\({2Li} \rightarrow {2Li^+}+{2e^−}\)

Reduction half reaction: \({Cl_2}+{2e^−} \rightarrow {2Cl^−}\)

We multiply the oxidation half reaction by 2 so that both half reactions involve 2 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

\[{Cl_2}+{2Li} \rightarrow {2Li^+}+{2Cl^−}\]

Problem 3

Oxidation half reaction: \(2({Fe} \rightarrow {Fe^{3+}}+{3e^−})\)

\({2Fe} \rightarrow {2Fe^{3+}}+{6e^−}\)

Reduction half reaction: \(3({Br_2}+{2e^−} \rightarrow {2Br^−})\)

\({3Br_2}+{6e^−} \rightarrow {6Br^−}\)

We multiply the oxidation half reaction by 2 and the reduction half reaction by 3 so that both half reactions involve 6 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

\[{3Br_2}+{2Fe} \rightarrow {2Fe^{3+}}+{6Br^−}\]

Problem 4

Oxidation half reaction: \(3({Ag} \rightarrow {Ag^+}+{e^−})\)

\({3Ag} \rightarrow {3Ag^+}+{3e^−})\)

Reduction half reaction: \({MnO^{−}_{4}}+{4H^+}+{3e^−} \rightarrow {MnO_2}+{2H_2O}\)

We multiply the oxidation half reaction by 3 so that both half reactions involve 3 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

\[{MnO^{−}_{4}}+{4H^+}+{3Ag} \rightarrow {3Ag^+}+{MnO_2}+{2H_2O}\]

Solution 19.1.3

Electron configurations are written models showing the distribution of electrons in the valance shells of an atom. In order to write the electron configuration, we begin by finding the element on the periodic table. Since Lanthanum, Samarium, and Lutetium are all a period below the noble gas Xenon, we know that the orbital configuration for Lanthanum will be the same as the orbital configuration for Xenon up to the first 54 electrons placed. Therefore, we can abbreviate \({1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}\) as [Xe] when writing the orbital configurations. We then find how many more electrons the element has than Xenon, and add these electrons to the orbital configurations using the Aufbau Principle. This gives us our neutral orbital configuration. To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital.

Lanthanum has 3 more protons than neutral Xenon, so it must have three more electrons as well. Therefore, these three more electrons will fill the 6s shell and place one electron in the 5d shell. In this specific case, the 5d shell is preferred by the lone electron over the 4f shell because the 5d shell is slightly lower in energy. When more than one electron needs to be located, the f shell is filled because electron pairing energy makes the 4f shell the lower energy shell. Its orbital configuration is:

\(La:\) [Xe] \({6{s}^2} {5{d}^1}\)

When Lanthanum is ionized to \({La^{3+}}\), it loses the two 6s electrons and a 5d electron, making its orbital configuration:

\({La^{3+}:}\) [Xe]

Samarium has 8 more electrons than Xenon, so the 6s orbital is filled and the 4f orbital receives 6 electrons. Its orbital configuration is:

\(Sm:\) [Xe] \({6{s}^2} {4{f}^6}\)

When Samarium is ionized to \({Sm^{3+}}\), it first loses the two 6s electrons, then one of its 4f electrons. Its orbital configuration is:

\({Sm^{3+}}:\) [Xe] \({4f^5}\)

Lutetium has 17 more electrons than Xenon. Two electrons go to the 6s orbital, 14 electrons go to the 4f orbital, and one electron goes to the d orbital. Its orbital configuration is:

\(Lu:\) [Xe] \({6s^2}{4f^{14}}{5d^1}\)

When Lutetium is ionized to \(Lu^{3+}\), it loses its two 6s electrons and its 5d electron, resulting in a stable and full 4f orbital. Its orbital configuration is:

\(Lu^{3+}:\) [Xe] \(4f^{14}\)

Some notable things about the electronic configurations

1) there are exceptions to the rules we have established for electron configuration-chromium and copper being the most common. In these notable cases the electron from 4s orbital rises to the 3 d orbital. So even though one may predict Cr to be [Ar]4s²3d⁴, it is actually [Ar]4s¹3d⁵. In these cases, a completely full or half full d sub-level is more stable than a partially filled d sub-level, so an electron from the 4s orbital is excited and rises to a 3d orbital.

2) As seen above the orbitals are filled in a specific order. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d...and the list continues. Because 4s is lower in energy than 3d, it is filled before the 4s. But when you remove the electrons, the one from the 4s is removed first.

Here is a diagram that illustrates the different energy levels:

Image taken from: chem.libretext

Solution 19.2.5

We are instructed to draw all geometric isomers and optical isomers for the specified compound. Optical isomers exist when an isomer configuration is not superimposable on its mirror image. This means there are two distinct molecular shapes. Often a left and right hand are cited as an example; if you were to take your right hand and place it upon your left, you cannot make the major parts of your hand align on top of one another. The basic idea when deciding whether something is optically active is to look for a plane of symmetry--if you are able to bisect a compound in a manner that establishes symmetry, then the compound does not have an optical isomer.

Cis isomers exist when there are 2 ligands of the same species placed at 90 degree angles from each other. Trans isomers exist when there are 2 ligands of the same species placed at 180 degree angles from each other.

Problem 1

This compound is an octahedral molecule, so the six ligands (atoms in the complex that are not the central transition metal) are placed around the central atom at 90 degree angles. Two optical isomers exist for [Co(en)₂(NO₂)Cl]⁺. The second isomer is drawn by taking the mirror image of the first.

Problem 2

This compound is also an octahedral molecule. Two cis (optical) isomers and one trans isomer exist for [Co(en)₂Cl₂]⁺. The trans isomer can be drawn by placing the chlorine ligands in positions where they form a 180 degree angle with the central atom. The first cis isomer can be drawn by placing the chlorine ligands in positions where they form a 90 degree angle with the central atom. The second cis isomer can be found by mirroring the first cis isomer, like we did in problem 1.

Problem 3

This compound is also an octahedral molecule. One trans isomer and one cis isomer of [Pt(NH₃)₂Cl₄] exist. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle with the central atom.

Problem 4

This compound is also an octahedral molecule. Two optical isomers for [Cr(en)₃]³⁺ exist. The second optical isomer can be drawn by taking the mirror image of the first optical isomer.

Problem 5

This compound is a square planar complex, so the ligands are placed around the central atom in a plane, at 90 angles. A trans isomer and a cis isomer exist for the complex [Pt(NH₃)₂Cl₂]. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle in the plane with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle in the plane with the central atom.

Solution 12.3.18

Information about reaction rates can be found here.

Problem 1

The rate equation for an \(n\) order reaction is given as \(\frac{dr}{dt}={k}{[A]^n}\). For each set of data points, we can plug concentration and rate values into this equation for \([A]\) and \(\frac{dr}{dt}\). We can then use two of these equations to solve for \(n\). The reason we use these equations is because we do not know k,the rate constant, and we also do not know (n), the rate of the reaction; but we do know rate constant is just that--a constant, in that manner we can cross out the k and solve for n

Rate equation 1: \(4.17 \times {10}^{-4}={k}{[0.230]^n}\)

Rate equation 2: \(9.99 \times {10}^{-4}={k}{[0.356]^n}\)

\[{\frac{4.17 \times {10}^{-4}}{9.99 \times {10}^{-4}}} = {\frac{k[0.230]^n}{k[0.356]^n}} \]

\[{0.417}={0.646^n}\]

\[ln{0.417}={n \cdot ln{0.646}}\]

\[\frac{ln{0.417}}{ln{0.646}}=n=2\]

Therefore, the rate equation is second order in A and is written as \(\frac{dr}{dt}={k}{[A]^2}\).

Problem 2

Since \(\frac{dr}{dt}={k}{[A]^2}\) is our rate equation, we can solve for \(k\) by plugging in our first data points \( [A]=0.230 \frac{mol}{L}\) and \( \frac{dr}{dt} = 4.17 \times {10}^{-4} \frac{mol}{L \cdot s}\). We get:

\[4.17 \times {10}^{-4} \frac{mol}{L \cdot s}={k}{[0.230 \frac{mol}{L}]^2}\]

\[k=7.88 \times {10}^{-3} \frac{L}{mol \cdot s}\]

By plugging in values to our rate equation, we find the value of k as well as its proper units.

Solution 12.6.9

Answer: Option 2: \({CH_3}+{HCl} \rightarrow {CH_4}+{Cl}\)

Explanation:

Chain reactions involve reactions that create products necessary for more reactions to occur. In this case, a reaction step will continue the chain reaction if a radical is generated. Radicals are highly reactive particles, so more reactions in the chain will take place as long as they are present. The chlorine is considered a free radical as it has an unpaired electron; for this reason it is very reactive and propagates a chain reaction. It does so by taking an electron from a stable molecule and making that molecule reactive, and that molecule goes on to react with stable species, and in that manner a long series of "chain" reactions are initiated. A chlorine radical will continue the chain by completing the following reaction:

\({Cl \cdot}+{CH_4} \rightarrow {CH_3 \cdot}+{HCl} \)

The \({CH_3}\) generated by this reaction can then react with other species, continuing to propagate the chain reaction.

Option 1 is incorrect because the only species it produces is \({CH_3Cl}\), a product in the overall reaction that is unreactive. This terminates the chain reaction because it fails to produce any \(Cl\) or \(CH_3\) radicals that are necessary for further propagating the overall reaction.

Option 2 is the correct answer because it produces a \(Cl\) radical. This \(Cl\) radical can continue the chain by colliding with \(CH_4\) molecules.

Option 3 is incorrect because it fails to produce a radical capable of continuing the chain.

Option 4 is incorrect because it produces \(Cl_2\), a molecule that does not react unless additional light is supplied. Therefore, this step breaks the chain.

Solution 21.4.21

Problem 1

For this problem, we must assume that the rock sample contained no strontium when it was formed. Otherwise, our calculations will be inaccurate (see solution to problem 2). Through beta decay, some of the rubidium in the rock decayed to strontium. The beta decay process is:

\[{^{87}_{37}Rb} \rightarrow {^0_{-1}\beta}+{^{87}_{38}Sr}\]

The original amount of rubidium in the rock is therefore the sum of the current amounts of rubidium and strontium (rubidium-87 and strontium-87 have the same numbers of nucleons, so their molar masses are the same and their current weights can be added to obtain the original weight of rubidium).

\[{8.23mg}+{0.47mg} = {8.70mg}\]

Because the half-life of the decay of rubidium by \( \beta \) emission is 4.7 × 10¹⁰ years, we can calculate \( \lambda \), the decay constant. If you recognize this is very similar to first order chemical kinetics; the initial concentration does not impact the half life.

\[ {t_{1/2}} = \frac{ln{(2)}}{\lambda} \]

\[ {4.7 \times 10^{10}} = \frac{ln{(2)}}{\lambda} \]

\[ \lambda = 1.47 \times 10^{-11} \]

We can use \( \lambda \) to calculate how old the rock is using the radioactive decay equation:

\[N=N_o{e}^{- \lambda {t}} \]

\[8.23=8.70{e}^{(-1.47 \times 10^{-11})t}\]

\[\ln{\frac{8.23}{8.70}}=-1.47 \times 10^{-11}t \]

\[t=3.77 \times 10^9 \]

The rock sample is 3.8 billion years old.

Problem 2

The rock would be younger than the age calculated in question 1. If Strontium was originally in the rock, less than the 0.47mg of strontium used to calculate part (a) would have been generated by the beta decay of Rubidium. As this amount would be smaller and the age is proportional to the amount of Strontium created by beta decay, the rock would be younger.

Solution 20.3.9

Problem 1

We need to write both the oxidation and reduction half reactions. From the main reaction, we see that lead is reduced (it gains electrons), and Vanadium is oxidized (it loses electrons). The unbalanced half reactions are:

Reduction: \({Pb(s)} \rightarrow {Pb^{2+}(aq)}\)

Oxidation: \({VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}\)

To balance the reactions, we first add \(H_2O\) to balance the oxygen, then add \(H^+\) to balance the hydrogen. We then add electrons to balance the charge.

Reduction: \({Pb(s)} \rightarrow {Pb^{2+}(aq)}\)

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

Oxidation: \({VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}\)

\({VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}+{H_2O(l)}\)

\({VO^{2+}(aq)}+{2H^+} \rightarrow {V^{3+}(aq)}+{H_2O(l)}\)

Problem 2

The cathode is defined as the electrode where reduction takes place. This means the reduction reaction occurs at the cathode.

The anode is defined as the electrode where oxidation takes place. This means the oxidation reduction occurs at the anode.

Problem 3

The cathode is positively charged because electrons flow across the wire to this electrode.

The anode is negatively charged because electrons flow across the wire from this electrode.

It helps to remember this; regardless of whether you cell is a galvanic cell or an electrolytic cell, reduction always happens at the cathode, and oxidation at the anode. Most student attempt to memorize it using signs that; cathode is positively charged and anode is negatively charged. But that is only the case in a galvanic cell, because in an electrolytic cell you are running a nonspontaneous reaction.

Solution 20.5.20

We are given the reduction half reaction:

\(Ni^{2+}(aq)+{2e^-} \rightarrow {Ni(s)} \)

The question states that hydrogen gas is the reductant and that the pH of the solution is involved. This means our second half reaction is the oxidation reaction between hydrogen gas and hydronium ions, so the oxidation half reaction must be:

\({H_2(g)} \rightarrow {2H^+(aq)}+{2e^-} \)

The same amount of electrons are in each half reaction, so they cancel when we combine the two half reaction equations.

\({Ni^{2+}(aq)}+{H_2} \rightarrow {Ni(s)}+{2H^+}\)

The standard reduction potential for the reduction (cathode) half reaction is \(−0.257 V \). By looking up the standard reduction potential for our oxidation (anode) half reaction in a chart of standard reduction potentials, we get a value of \( 0.00 V \). We can find our standard cell potential by subtracting the standard cell potential of the anode from the standard cell potential of the cathode:

\[E^o_{cell}=E^o_{cathode}−E^o_{anode} \]

\[E^o_{cell}=−0.257 V − 0.00 V= −0.257 V\]

The question asks "What pH is needed for this reaction to take place...?", which means our reaction will not be under standard conditions. We must use the Nernst equation to find the concentration of hydronium ions required for the reaction to be spontaneous. Since the reaction occurs at 298K, we can use the simplified version of the Nernst equation where \(R\), \(T\) and \(F\) are simplified to \(0.0592 V \) and \(\ln{Q}\) becomes \(\log{Q}\). A reaction is spontaneous when \(E_{cell} > 0\).

\[E_{cell}={E^o_{cell}} − \frac{0.0592 V}{n} \log{Q} \]

\[0<{−0.257} − \frac{0.0592}{n} \log{Q} \]

In our reaction, \(n=2\) because there are two mols of electrons transferred for every mol of a product consumed (see half reactions). Based on the law of mass action, \(Q\) for our reaction will be:

\[Q=\frac{[H^+]^2}{(1 atm) [1 M]} \]

\[Q={[H^+]^2}\]

The Nernst equation becomes:

\[0<{−0.257} − \frac{0.0592}{2} \log{[H^+]^2} \]

\[0<{−0.257} − 0.0296 \log{[H^+]^2}\]

We can now solve for the concentration of hydronium ions:

\[\frac{−0.257}{0.0296} > \log{[H^+]^2}\]

\[exp(-8.68) > {[H^+]^2}\]

\[\sqrt{exp(-8.68)} > [H^+]\]

\[ {[H^+]} < 4.56 \times 10^{-5} \]

We now can find an inequality for the pH:

\[pH=−\log{[H^+]} \]

\[ −\log{[H^+]} > −\log{4.56 \times 10^{-5}} \]

\[pH > 4.34\]

A pH greater than 4.34 is needed for the reaction to take place at 25°C. This means the reaction favors basic conditions. We can check this answer by plugging in pH values greater than 4.34 to the Nernst equation. If our answer is right, the Nernst equation will yield a positive cell potential.

As a student, it is always in your best interest to check your answers; let us take a hypothetical pH value of 6 and see if the above answer is correct.

-0.257-(0.0592/2)log((10^-6)² = 0.0982. A positive E value indicated that the reaction is spontaneous.