Extra Credit 24

Q17.3.3

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

\({\mathrm{Cu}(s)│{Cu^2+}(aq)║{Au^3+}(aq)│{Au}(s)}\)

S17.3.3

\({\mathrm{3Cu(s)+2Au^{3+}(aq)} \rightarrow 3Cu^{2+}(aq)+2Au(s); +1.16\:V}\) ; spontaneous(s

We are asked to determine the overall reaction. From the cell notation we can determine the oxidized element to be \({\mathrm{3Cu}}\) because the elements on the left of the double bars partakes in the oxidation half reaction and on the right of the double bars are the elements that partake in the reduced half reaction. From that we can determine the oxidation half reaction to be :

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

and the reduced half reaction is :

\({\mathrm{ Au}^{3+}(aq)+{3e}^- \rightarrow {Au}(s)}\)

Now that we have determined the half reactions, we now have to combine the two together to form a whole reaction. We do that by multiplying each half reaction so that they have equal electrons on each side of the equation. In the context of this problem, we multiply the oxidation half reaction by 3 to have 6 electrons and the reduction half reaction by 2 to also have 6 electrons. After, we combine the two half reactions and cancel out the electrons on each side of the reaction to find the overall reaction to be

\(\mathrm{3Cu}(s)+{2Au^{3+}}(aq)\rightarrow{3Cu^{2+}}(aq)+{2Au}(s)\)

Now to determine the cell potential at 25 °C, we need to look at a reduction potential table.

If we look on the table we see that Cu half reaction has a reduction potential of .34 and Au has a standard reduction potential of 1.50. To calculate to reduction potential of the whole reaction we need to subtract the oxidation reaction from the reduction reaction using the equation E_cell = E_cathode - E_anode. By doing this we get

1.50 - .34= 1.16V

Since the reduction potential is positive we can therefore determine that the reaction is spontaneous because the spontaneity of a reaction is negatively related to the reduction potential. In other words, if we have a negative reduction potential, then the reaction isn't spontaneous and if we have a positive reduction potential then the reaction is spontaneous.

Correct Solution

Q19.1.22

Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.

\(({\mathrm{Co(NO_3)_2}(s)\rightarrow{Co_2O_3}(s)+{NO_2}(g)+{O_2}(g)}\)

In order to balance this reaction we first need to determine what the oxidation and reduction half reactions are. If we look at the reaction, we find that Cobalt is oxidized from +2 to +3, Oxygen is oxidized from -2 to 0, and Nitrogen is reduced from +5 to +4 so the oxidized half reactions are

\(\mathrm(Co(NO_3)_2(s)\rightarrow Co_2O_3(s)\)

\(\mathrm H_2O(l)\rightarrow O_2(g)\)

and the reduced half reaction is

\(\mathrm(Co(NO_3)_2(s)\rightarrow 2NO_2(g)\)

Since there are 3 half reactions we need to combine 2 of them to so that we will only have 2 half reactions. Since \(\mathrm NO_2(g)\) and \(\mathrm(Co(NO_3)_2(s)\) have the same reactant we can combine the two together. Now our two half reactions are

R: \(\mathrm 2(Co(NO_3)_2(s)\rightarrow 4NO_2(g)+ Co_2O_3(s)\)

O: \(\mathrm H_2O(l)\rightarrow O_2(g)\)

First we will work on the reduced half reaction.This equation isn't balanced yet so we now need to do that. First we need to balance the oxygen. Since there are 6 oxygen on the left and 5 on the right we need to add 1 more oxygen. We add oxygen by adding water, so our new equation is

\(\mathrm 2(Co(NO_3)_2(s)\rightarrow 4NO_2(g)+ Co_2O_3(s)+H_2O(l)\)

Now we have an excess amount of Hydrogen on the right so we need to balance the number of Hydrogen. Our new equation is

\(\mathrm 2(Co(NO_3)_2(s)+2H^+(aq)\rightarrow 4NO_2(g)+ Co_2O_3(s)+H_2O(l)\)

Finally we need to balance the overall charge, so we add electrons. The charge of the reactant side is +2 more than the product's side. So we balance the charges by adding two electrons to the reactant side. Our new equation is

\(\mathrm 2(Co(NO_3)_2(s)+2H^+(aq)+2e^-\rightarrow 4NO_2(g)+ Co_2O_3(s)+H_2O(l)\)

Now we need to repeat the process for the oxidation half reaction which is

\(\mathrm 2H_2O(l) \rightarrow O_2(g)\)

The Oxygen is already balanced, so we now add Hydrogen. The new equation is

\(\mathrm 2H_2O(l) \rightarrow O_2(g)+4H^+(aq)\)

And finally the electrons

\(\mathrm 2H_2O(l) \rightarrow O_2(g)+4H^+(aq)+4e^-\)

Now we have both half reactions, so all we need to do now is balance the overall charge of the molecules. Since the reduced reaction has only 2 electrons on the left and the reduced has 4 on the right, we need to multiply the reduced reaction by a factor of 2. Our reactions are now

O: \(\mathrm 2H_2O(l) \rightarrow O_2(g)+4H^+(aq)+4e^-\)

R: \(\mathrm 4(Co(NO_3)_2(s)+4H^+(aq)+4e^-\rightarrow 8NO_2(g)+ 2Co_2O_3(s)+2H_2O(l)\)

Now we can add the equations together and cancel out the electrons and protons.

\(\mathrm 2H_2O(l)+4(Co(NO_3)_2(s)+4H^+(aq)+4e^-\rightarrow O_2(g)+4H^+(aq)+4e^-+8NO_2(g)+ 2Co_2O_3(s)+2H_2O(l)\)

Our final equation is

\(\mathrm 4(Co(NO_3)_2(s)\rightarrow O_2(g)+8NO_2(g)+ 2Co_2O_3(s)\)

Correct Solution

Q12.4.14

There are two molecules with the formula C₃H₆. Propene, CH3CH=CH2CH3CH=CH2, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:

When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of 5.95 × 10⁻⁴ s⁻¹. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?

To find the half life of this reaction, we first need to know the formula of half life, which is :

\({\mathrm {t_\frac12}=\frac{\ln2}{k}}\)

where k is the rate constant

and t is the time in seconds.

Using this equation, we can then determine the half life of the reaction by using 5.95 × 10⁻⁴ s⁻¹ as k and solving for t. When we solve for t we should get that the half life is 1164.95 seconds.

Now we need to find the fraction of cycloproane that remains after .75 hrs at 499 °C. In order to do this we need to use the equation:

\({\mathrm[A]=[A_\circ]{e^{kt}}}\)

where

\({\mathrm[A] }\)is the current/final concentration

\({\mathrm[A_\circ]}\) is the initial concentration

k is the rate constant

t is the time in seconds that the reaction takes to complete.

First we need to convert \(\mathrm{t}\) from hours into seconds. We can convert by

\(\mathrm{t}=\frac{3hrs}{4} ×\frac{60min}{1hr}×\frac{60sec}{1min}=2700 sec\)

Using the equation, \({\mathrm[A]=[A_\circ]{e^{kt}}}\), we can then solve for the ratio of the final concentration to the initial concentration by using \(\mathrm{t}=2700sec\), and k = 5.95 × 10⁻⁴ s⁻¹ . After doing the calculations, we find that \(\mathrm{\frac{[A]}{[A_\circ]}}= .20\). From that, we find that the 20% remains.

Correct Solution

S12.4.14

1164.95 seconds; 20% remains

Q21.2.9

Which of the following nuclei lie within the band of stability

a)chlorine-37

b)calcium-40

c) ²⁰⁴Bi

d)⁵⁶Fe

e)²⁰⁶Pb

f)²¹¹Pb

g)²²²Rn

h)

First in order to solve these problems we need to calculate the number of protons and neutrons in each element. We can do this by finding the atomic number of the element on the periodic table. We then subtract this number from the mass number given to get the number of neutrons.

a) Chlorine-37 has 20 neutrons and 17 protons. By looking on the graph, we find that the band of Chlorine-37 lies on the band of stability.

b)Calcium-40 has 20 neutrons and 20 protons, so it lies on the band of stability.

c)Bismuth-204 has 83 protons and 121 neutrons, so it lies on the band of stability.

d) Iron-56 has 26 protons and 30 neutrons, so it lies on the band of stability.

e) Lead-206 has 82 protons and 124 neutrons, so it lies on the band of stability.

f) Lead-211 has 82 protons and 129 neutrons, so it doesn't lie on the band of stability.

g) Radon- 226 had 86 protons and 140 neutrons, so it doesn't lie on the band of stability.

h) Carbon-14 has 6 protons and 8 neutrons so it doesn't lie on the band of stability

Correct Solution

S21.2.9

(a), (b), (c), (d), and (e)

Q21.7.1

If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:

1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)

1. Since cobalt-60 is a strong emitter of gamma rays, the hospital would need to use lead, iron, or another thick metal plate container to prevent the gamma rays from being emitted because they can not pass through these substances.
2. molybdenum-99 is a beta emitter, so the hospital would need to use thin plates of wood or aluminum to prevent the emission of beta.

Correct Solution.

Q20.4.10

For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?

1. measuring the potential of a Cl⁻/Cl₂ couple
2. measuring the pH of a solution
3. measuring the potential of a MnO₄⁻/Mn²⁺ couple

1. To measure the potential of a Cl⁻/Cl₂ couple, we should use an inert electrode because we only want there to be a transfer of electrons, nothing else. The measured potential would also be less than the corresponding E° because there are always experimental errors when conducting an experiment.
2. We should use an inert electrode for measuring the pH because we want to measure the electron transfer without any other ions entering the solution. The measured potential would also be less than the corresponding E° because there are always experimental errors when conducting an experiment.
3. We should use an inert electrode for this couple because we want to measure the potential of electron transfer since it is the same element. So, we should use either graphite or platinum. The measured potential would also be less than the corresponding E° because there are always experimental errors when conducting an experiment.

Correct Solution

S20.4.11

\(\mathrm Pt(s)∣H_2(g, 1 atm) | H^+(aq, 1M)∥Cu^{2+}(aq)∣Cu(s)\)

Q20.4.11

Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction:

\(\mathrm H_2(g)+Cu^{2+}(aq)→2H^+(aq)+Cu(s)\)

When drawing a cell diagram, we need to first understand what each symbol means.

| means a phase change between the product and reactant of an element.

|| means a salt bridge in the reaction

An SHE electrode can be either a platinum or graphite electrode. In this solution we will use Platinum. It is used because they are considered to be inert electrodes.

When constructing a cell diagram we put the oxidized reactant, then the oxidized product, salt bridge, then reduced reactant, and reduced product. Finally if we use an inert electrode, we put it either in the very front of the cell diagram if it for the oxidized element or at the very end if it is for the reduced element.

So, our equation should look like this:

\(\mathrm Pt(s)∣H_2(g, 1 atm) | H^+(aq, 1M)∥Cu^{2+}(aq)∣Cu(s)\)

Correct Solution

Q20.8.2

What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?

When a metal is being coated with a sacrificial layer, the metal is being covered in another metal that is more easily corroded or more easily oxidized. The sacrificial layer is supposed to act as an anode to prevent the corrosion of the metal underneath it. By doing this, we are able to protect the metal underneath the sacrificial layer from corroding. This is different from galvanic protection because galvanic protection doesn't completely cover the metal in a sacrificial layer. It only has blocks of the sacrificial anode attached to the metal.

Correct Solution