Extra Credit 30

Q17.4.3

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

1. \[Hg(l)+S{^{2−}}(aq,0.10M)+2Ag^{+}(aq,0.25M)⟶2Ag(s)+HgS(s)\]
2. The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br₂(l) is the same as that of Br₂(aq).

S17.4.3

1.) This problem requires us to find standard cell potential and then cell potential under stated conditions.

Standard cell potential: Write out the half reactions for the redox reaction to see what is oxidized and reduced: \[Ag{^+}(aq)+e{^-}⟶Ag(s)\ \ \ \ \ \ \ S{^{2-}}(aq)⟶ S(s)+2e{^-}.\] Here, we made sure to balance charges by adding the appropriate amount of electrons. Note that we did not consider Hg in these half reactions since it is a liquid. From these equations, we can tell that the silver is being reduced (gaining electrons) and the sulfur is being oxidized (losing electrons). And, oxidation always occurs at the anode so the sulfur equation is occurring at the anode of the cell and the silver one is occurring at the cathode.

We can model such a battery by using a cell diagram of the standard conditions (standard conditions are at concentrations of 1.0M for concentration:

\[ \ HgS{_{(s)}} \ | \ Hg{_{(l)}} \ | \ S{^{2-}}(aq, 1.0M) \ || Ag{^+}(aq, 1.0M) \ | \ Ag{_{(s)}}\]

or without the liquid:

\[ \ S{_{(s)}} \ | \ S{^{2-}}(aq, 1.0M) \ || \ Ag{^+}(aq, 1.0M) \ | \ Ag{_{(s)}}\] Here, the single line depicts a new cell (phase change) and a double line depicts a salt bridge.

To find the standard cell potential we then write each half reaction in terms of reduction (flip the anode equation):

\[Ag{^+}(aq)+e{^-}⟶Ag(s)\ \ \ \ \ \ \ S(s)+2e{^-}⟶ S{^{2-}}(aq)\]

Next, we use a table of standard reduction potentials (for more info see 20.2.1) and find the overall standard cell potential via the equation:

\[E^{\circ}=E^{\circ}(cathode)-E^{\circ}(anode)\].

From the table we see for the cathode we have \[E^{\circ}(cathode)={\textbf{.80V}} \ \ \ \ \ \ \ E^{\circ}(anode)={\textbf{-.48V}}\] and thus we can do: \[0.80V-{-0.48V}=1.28V\] Since our cell potential is positive, the reaction is spontaneous. This is because the more positive (\({\Delta}E\)) the more negative the (\({\Delta}G.\))

Now we must consider the cell at the stated conditions: The nonstandard condition considers the concentrations of the ions at values other then 1.0M. The cell diagram here would be: \[ S{_{(s)}} \ | \ S{^{2-}}(aq, .1M) \ || \ Ag{^+}(aq, .25M) \ | \ Ag{_{(s)}} \ |\]

To calculate the cell potential at these nonstandard conditions we use the Nernst Equation. Since the problem is given at 298.15 K (standard temperature) we can use the simplified form: \[E=E^{^\circ}-\frac{0.0592V}{n}logQ\] In the equation Q is the concentration of the products over the reactants (seen from overall reaction). The activities for liquids and solids are 1: \[ \frac{1}{[Ag^{+}]^{2}[S^{2-}]}\]

We use the standard cell potential calculated above (1.28V)

And n is the number of electrons transferred in the total reaction. This involves balancing both the half reactions:

\[Ag{^+}(aq)+e{^-}⟶Ag(s)\] \[S{^{2-}}(aq)⟶ S(s)+2e{^-}.\]

We see that we must multiply the top equation by 2 to cancel out the electrons. So, we see here that n=2. So now, we just need to plug in numbers into the equation:

\[1.28V-\frac{0.0592V}{2}log\frac{1}{[.25M]^{2}[.1M]}=1.21V\]

Since this value is positive, the conditions are spontaneous.

We can follow the same procedure for the next two problems:

2.) The problem says that this is for a galvanic cell so we assume that the standard potential will be spontaneous with a positive standard E. So we write the half reactions in terms of reduction (we disregard counter ions):

\[Ni{^{2+}}(aq)+2e{^-}⟶Ni(s)\ \ \ \ \ \ \ Al{^{3+}}(aq)+3e{^-}⟶ Al(s)\] Where the table gives the SRP's for each to be - 0.26V and - 1.66V respectively. We know we want a spontaneous reaction (positive magnitude) so we can assume that the reaction with nickel is at the cathode and the one with aluminum is at the anode since that is what results in a positive value when plugged in and since aluminum has the lower reduction potential: \[E^{\circ}=E^{\circ}(cathode)-E^{\circ}(anode)\]we get \[- 0.26V-{- 1.66V}=1.40V.\]

This is positive and thus spontaneous.

Now we plug this into the Nernst equation. We must consider the concentrations given. Looking at the half reactions above we can see we are going to need to multiply by 2 for the aluminum equation and 3 for the nickel, making n = 6. We must also sum up the half reactions (oxidation and reduction) to get Q: \[Ni{^{2+}}(aq)+2e{^-}⟶Ni(s)\] \[Al(s)⟶Al{^{3+}}(aq)+3e{^-} \] \[3Ni{^{2+}}(aq)+2Al(s)⟶3Ni(s)+2Al{^{3+}}(aq)\] This gives a Q of: \[ \frac{[Al]^{2}}{[Ni]^{3}}\] Now, we can plug into the Nernst equation: \[1.40V-\frac{0.0592V}{6}log\frac{[.015M]^{2}}{[.25M]^{3}}=1.42V\] This positive value indicates spontaneity.

3.) Now we can apply the same concepts here and just analyze what is given. We are told that the aluminum ion is being reduced so that is the cathode. From this one can say that the bromine is the anode. We will write in terms of reduction now, though:

\[Al{^{3+}}(aq)+3e{^-}⟶Al(s)\ \ \ \ \ \ \ Br_{2}+2e{^-}⟶ 2Br^{-}(aq)\]

Then, we plug in their SRP's and do cathode minus anode: \[-1.66V-1.07V=-2.73V.\] This is negative and nonspontaneous.

Like before, after balancing the equation: \[6Br{^{-}}(aq)+2Al{^{3+}}(aq)⟶2Al(s)+3Br_{2}(aq)\]

we plug in the values of conditions given into the Nernst equation:

\[-2.73V-\frac{0.0592V}{6}log\frac{1}{[.023M]^{6}[.11M]^{2}}=-2.82V\] This negative value shows this is not spontaneous.

Q12.1.3

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction \[Cl_2(g)+3F_2(g)⟶2ClF_3(g).\]

Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl₂ and F₂ and the formation of ClF₃

S12.1.3

In this problem we are asked to write the equation that relates rate expressions in terms of disappearance of the reactants of the equation and in terms of the formation of the product. A reaction rate gives insight to how rate is affected as a function of concentration of the substances in the equation. Rates can often be expressed on graphs of concentration vs time expressed in change (\({\Delta}\)) of concentration and time and in a short enough time interval, the instantaneous rate can be approximated. If we were to analyze the reaction given, the graph would demonstrate that Cl₂ decreases, that F₂ decreases 3 times as quickly, and then ClF₃ increases at a rate doubles. The reactants are being used and converted to product so they decrease while products increase.

For this problem, we can apply the general formula of a rate to the specific aspects of a problem where the general form follows: \[aA+bB⟶cC+dD\].

And the rate can then be written as \(rate=-\frac {1}{a}\frac{{\Delta}[A]}{{\Delta}t}\) \(=-\frac {1}{b}\frac{{\Delta}[B]}{{\Delta}t}\) \(=\frac {1}{c}\frac{{\Delta}[C]}{{\Delta}t}\) \(=\frac {1}{d}\frac{{\Delta}[D]}{{\Delta}t}.\) Here the negative signs are used to keep the convention of expressing rates as positive numbers.

In this specific case we use the stoichiometry to get the specific rates of disappearance and formation (back to what was said in the first paragraph). So, the problem just involves referring the to the equation and its balanced coefficients. Based upon the equation we see that Cl₂ is a reactant and has no coefficient, F₂ has a coefficient of 3 and is also used up, and then ClF₃ is a product that increases two-fold with a coefficient of 2. So, the rate here can be written as: \[rate=-\frac{{\Delta}[Cl_2]}{{\Delta}t}=-\frac {1}{3}\frac{{\Delta}[F_2]}{{\Delta}t}=\frac {1}{2}\frac{{\Delta}[ClF_3]}{{\Delta}t}\]

An extra note: if one wanted to find the rate of appearance or disappearance of a specific substance (say F₂) they could simply isolate to get the rate without the coefficient.

Q12.5.1

Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

S12.5.1

The basis of this question is the collision theory. The collision theory describes what needs to occur for a reaction to happen and what can affect the rate of a reaction. This theory operates under the basis that even if two particles do collide (which they must for a reaction to occur), they do not always react. If that were the fact, reactions would be complete quite quickly. The collision theory describes that for a reaction to occur the molecules must collide with sufficient energy and the proper orientation (steric factor). These are the explanations for why this is:

Sufficient Energy: If two particles were to collide without sufficient kinetic energy, it is quite unlikely that they would react (they might bounce off each other). So, the particles must collide with sufficient energy, where the energy minimum is called the activation energy. So the activation energy is a barrier to the reaction and if the particles have an energy below the activation energy, the reaction will not occur. This has many applications to biology. In biology it is common that many processes are regulated by the fact that they have to meet a certain activation energy and then catalysts can help in regulation by lowering the activation energy. The concept of activation energy was explained by Svante Arrhenius. He described that adding heat can help in meeting the activation energy.

Proper Orientation (Steric Factor): Having the proper orientation ensures that the correct atoms are aligned and that correct bonds are broken and reformed. When particles react in a way that does not align the proper atoms, the reaction does not occur.

This figure illustrates the need for a certain amount of energy (top row) and the proper orientation (steric factor).

Ultimately, particles need to meet a minimal energy in order to react and they must react with the proper orientation. These factors (and also frequency) can be manipulated to affect the rate of the reaction as well.

Q21.3.5

Write a balanced equation for each of the following nuclear reactions:

1. the production of ¹⁷O from ¹⁴N by α particle bombardment
2. the production of ¹⁴C from ¹⁴N by neutron bombardment
3. the production of ²³³Th from ²³²Th by neutron bombardment
4. the production of ²³⁹U from ²³⁸U by \({^2_1}H\) bombardment

S21.3.5

The question revolves around nuclear chemistry, which is a form of chemistry depending on changes in the nucleus (as opposed to the electron like in redox chemistry). This involves radioactivity by which instability causes a change in the nuclei–specfically via the emission of particles. Ernest Rutherfold identified three types of radiation which are distinguished by their differences in reactions when an external magnetic/electric field is applied:

Alpha Radiation/Decay involves the emission of an alpha particle: \({^4_2}He\) and tends to occur in heavy nuclei off of the belt of stability. In these reactions, mass number is decreased by 4 and atomic number is decreased by 2. Beta Radiation involves either the emission of an electron \({\beta}^-\) along with an anti-neutrino or the emission of a positron \({\beta}^+\) with a neutrino. Another is gamma decay where a gamma ray is emitted and involves no change in mass.

In nuclear chemistry, it is important to write equations where mass number and atomic number are balanced on both sides of the equation: \[{^A_Z}Element Symbol\]

In general, Z denotes atomic number (proton count)vand A denotes the number neutrons and protons in total.

1.) This reaction describes the equation as the reactant ¹⁴N colliding with the alpha particle to yield ¹⁷O. At first we can write this as stated: \[^{14}N+{^4_2}He⟶^{17}O.\]

However, the mass numbers here do not balance (18 on the left and 17 on the right) and neither do the atomic numbers (9 on the left and 8 on the right). Since the right side of the equation is missing 1 for each unit on the right, this gives insight that we need to account for the emission of a proton (hydrogen): \({^1_1}H\). So, we must balance the equation by adding a proton to the right side: \[^{14}N+{^4_2}He⟶^{17}O+{^1_1}H.\] If we check this we can see that the masses and atomic numbers are equal on both sides.

2.) Now, we are asked to depict what happens when a neutron collides with ¹⁴N and yields ¹⁴C. A neutron has no charge and a mass of about 1 amu and can be depicted as such: \({^1_0}n.\) So, we check to make sure things are balanced:\[^{14}N+{^1_0}n⟶^{14}C.\] Once more, we see we need to balance the numbers by adding units of 1 to the right side again. So, the neutron bombardment knocked off a proton so that the solution looks as such: \[^{14}N+{^1_0}n⟶^{14}C+{^1_1}H.\]

3.) The same concept occurs here by which ²³²T is bombarded with a neutron and yields ²³³Th. So we can follow the same procedure as before: \[^{232}Th+{^1_0}n⟶^{233}Th.\] When looking at this equation, we see that the numbers on each side are balanced (we just added one to the mass) so this is the solution.

4.) In this last one, ²³⁹U is produced when \({^2_1}H\) bombards ²³⁸U. So, following the instructions we write: \[^{238}U+{^2_1}H⟶^{239}U.\] Since our units are off by one here, we need to add a proton to the right side of the equation. So the solution is: \[^{238}U+{^2_1}H⟶^{239}U+{^1_1}H.\]

Q20.2.1

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

S20.2.1

The answer to this question relies on redox chemistry. Redox reactions must always have a substance that is being oxidized and another that is being reduced. Oxidants are oxidizing agents which mean they cause other substances to be oxidized and are themselves reduced. A reductant, reducing agent, causes the oxidant to and is oxidized. Oxidation involves the loss of electrons and reduction involves the gain of electrons. An acronym that can be used to remember this is OIL RIG where OIL= oxidation is loss and RIG= reduction is gain. The best oxidants are those with higher electronegativity: which is the affinity for an electron. Since the atoms that "want" electrons more gain the electrons, they are reduced and thus be the best oxidants. Electronegativity increases moving to the right and up the periodic table (making F the most electronegative). So the best oxidants are those that are the most electronegative (fluorine, oxygen, chlorine, halogens, etc). The best reductants are likely to donate an electron to the oxidant. They are the elements with lower electronegativity. Common ones are the s block metals since they have unfilled valence electron shells and are highly reactive (and can easily give an electron to a good oxidant (Na+Cl⟶NaCl):

Another way to look at this is via a table of standard reduction potentials. The higher reduction potential, means a higher probability of being reduced:

So, F₂ is at the top and has a better chance of being reduced and is better at being an oxidant. In contrast, Lithium has a low, negative reduction potential and so is a better reductant. This concept has applications in electrochemistry.

Q20.4.20

Will each reaction occur spontaneously under standard conditions?

1. Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
2. Zn²⁺(aq) + Pb(s) → Zn(s) + Pb²⁺(aq)

S20.4.20

We will treat this problem like a cell. The first step in tackling these problems is to split the equation into half reactions of oxidation (electron loss) and reduction (electron gain): \[Cu(s)→Cu^{2+}(aq)+2e^-\] \[2H{^+}(aq)+2e{^-}→H{_2}(g)\] Note that we balanced charges with electrons.

Here we see that the H⁺ is gaining an electron and is thus reduced and then Cu(s) is losing an electron and is thus oxidized. In a cell, oxidation always occurs at the anode and reduction always occurs at the cathode. Thus, we can use standard cell potential to evaluate spontaneity of the reaction. The more positive the cell potential, the more negative the \({\Delta}G\) is, and so the more spontaneous the reaction is. This is because cell potential measures how much a cell can force electrons through a circut.

Next, we use a table of standard reduction potentials to evaluate whether the equation is spontaneous or not. But first we must write each equation in terms of a reduction reaction (simply flip the anode equation): \[Cu^{2+}(aq)+2e^-→Cu(s)\] \[2H{^+}(aq)+2e{^-}→H{_2}(g).\]

Now we can locate each equation on the table (for more insight on the table see problem 20.2.1):

We see that the equation for the reduction of Cu²⁺ has a \(E^{\circ}={\textbf{0.34V}}\) and for the reduction of 2H⁺ the standard cell potential is \(E^{\circ}={\textbf{0V}}\) (Standard Hydrogen Electrode). The overall standard cell potential is given by the equation \[E^{\circ}=E^{\circ}(cathode)-E^{\circ}(anode)\].

So we simply subtract the standard reduction potential of the anode from the cathode: \[0V-{0.340}=-0.34V\] Since the overall standard cell potential the reaction is NOT spontaneous and this is likely an electrolytic cell.

We can follow the same process for the next problem:

Step one: determine anode and cathode:\[Zn^{2+}(aq)+2e^-→Zn(s)\] \[Pb(s)→Pb^{2+}(aq)+2e^-.\] Oxidation occurs with the lead so that is the anode and then the zinc is the cathode.

Next, we write in terms of reduction: \[Zn^{2+}(aq)+2e^-→Zn(s)\] \[Pb^{2+}(aq)+2e^-→Pb(s).\]

Now we look at the table and see: \[E^{\circ}(cathode)={\textbf{-0.76V}}\] and \[E^{\circ}(anode)={\textbf{-0.13V}}\]

Last, we subtract cathode minus anode: \[-0.76V-{-0.13V}=-0.63V\] So, the overall cell potential is negative and the reaction is not spontaneous.

Q20.9.5

The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?

1. AlCl₃
2. MgCl₂
3. FeCl₃

S20.9.5

1.) The problem asks for how much metal is plated. So we are looking at what happens when solid Al is formed (we do not consider chlorine because it is not a solid; but we do use to to find the charge on the Al ion as 3+). So we are looking at what happens when aluminum is plated which is given by this reaction \[Al^{3+}(aq)+3{e^-}→Al(s)\] All we are doing here is converting moles to grams per electron via how many moles are given in the balanced half reaction. So here for every one mole of aluminum plated, there are three moles of electrons. We can then use the molar mass of aluminum to find grams of metal: Our work follows: \[\frac{1mol \ Al}{3 mol \ e{^-}} \ \frac{26.981539g \ Al}{1mol \ Al}= \ 8.99g \ per \ mol \ e{^-}\]

2.) We use the same concept here but the reaction this time is: \[Mg^{2+}(aq)+2{e^-}→Mg(s).\] And we have two electrons per mole Magnesium. SO we do:

\[\frac{1mol \ Mg}{2 mol \ e{^-}} \ \frac{24.305g \ Mg}{1mol \ Mg}= \ 12.2g \ per \ mol \ e{^-}\]

3.) For this problem we only have three moles of electrons per iron: \[Fe^{3+}(aq)+3{e^-}→Fe(s).\] So, we follow the procedure for one but for iron this time: \[\frac{1mol \ Fe}{3 mol \ e{^-}} \ \frac{55.845g \ Fe}{1mol \ Fe}= \ 18.6g \ per \ mol \ e{^-}\]

Here is a general example of electroplating that is occurring:

Q14.5.1

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates.

a.) How does the activation energy relate to the chemical kinetics of a reaction?

b.) Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

S14.5.1

a.) This problem relates to the collision theory. One aspect of the collision theory is that a reaction cannot occur (and rate depends on) if the particles do not collide with enough kinetic energy. The minimum energy requirement that must be met is called the activation energy.

So, for a reaction to occur the molecules must collide with enough energy to meet this activation energy. In addition, the energy of the particles be manipulated to affect the rate of a reaction. An example of this is a catalyst. A catalyst often lowers the activation energy and allows for the reaction to proceed quickly (without changing the heat of the reaction):

The height of the activation energy and how quickly such can be reached affects the activation energy. Usually, the lower the activation energy, the quicker the rate is. This was recognized by Arrhenius.

b.) The increase in the temperature increases the reaction rate because it increases the kinetic energy of a reaction. An increase in particle movement increases the probability the molecules will collide and thus increases the number of tries the molecules have to collide in a productive way (see 12.5.1 on collision theory), making the probability of reaction increase and thus increasing rate. This still occurs even if the average kinetic energy of the particles is below the activation energy. Though the average rate of kinetic energy is below the activation energy, there are likely some molecules that collide with enough energy. An increase in heat still increases the probability of effective collisions (especially of those higher than the energy of activation).