Extra Credit 28

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Q17.4.1

For the standard cell potentials given here, determine the ΔG° for the cell in kJ.

0.000 V, n = 2
+0.434 V, n = 2
−2.439 V, n = 1

S17.4.1

Important notes to remember if reaction is spontaneous:
	Spontaneous	Nonspontaneous	Equilibrium
Cell Potential	$\varepsilon >0$	$\varepsilon <0$	$\varepsilon =0$
Gibbs Free Energy	$\Delta G<0$	$\Delta G>0$	$\Delta G=0$
			$K=Q$

Here, we will use the following equation in order to solve for the Gibbs free energy, where n is the number of electrons transferred, F is Faradays constant 96,485 C/mol e-, and ε° is cell potential in volts.

$\Delta G=-nF\varepsilon$

In this question, we are trying to find the Gibbs free energy of the cell from the standard cell potential. The number of electrons transferred is 2, and the cell potential in 0.000V, and we will also use the Faraday constant. Now just plug everything into the equation.

$\Delta G=-(2mol)(96,485\frac{C}{mol})(0.000V)=0J$

Then you must convert the answer into kJ, so just divide by 1000 J

$0J(\frac{1kJ}{1000J})=0kJ$

Since $\Delta G=0$ , the reaction is at equilibrium.

$\Delta G=-nF\varepsilon$

In this question, we are trying to calculate the Gibbs free energy of the cell from the standard cell potential. The number of electrons transferred is 2, and the cell potential in 0.434V, and we will also use the Faraday constant. Now just plug everything into the equation.

$\Delta G=-(2mol)(96,485\frac{C}{mol})(0.434V)=-83,748.98J$

Then you must convert the answer into kJ, so just divide by 1000 J

$-83,748.98J(\frac{1kJ}{1000J})=-83.7kJ$

EDIT: -83,748 J *(1 KJ/ 1000 J) = -83.7 kJ

Since $\Delta G<0$ , the reaction is spontaneous.

$\Delta G=-nF\varepsilon$

In this question, we are trying to calculate the Gibbs free energy of the cell from the standard cell potential. The number of electrons transferred is 2, and the cell potential in -2.439V, and we will also use the Faraday constant. Now just plug everything into the equation.

$\Delta G=-(2mol)(96,485\frac{C}{mol})(-2.439V)=235,326.915J$

Then you must convert the answer into kJ, so just divide by 1000J

$235,326.915J(\frac{1kJ}{1000J})=235.3kJ$

Since $\Delta G>0$ , the reaction is nonspontaneous.

Q12.1.1

What is the difference between average rate, initial rate, and instantaneous rate?

S12.1.1

The instantaneous rate is the rate of change at any given time or moment. It can be measured by the tangent line. The initial rate is the rate at the beginning of a reaction, as it is starting to occur. The average rate is the average of all the rates in a time period, so it is the average of all the instantaneous rates. It can be measured by finding the slope of the secant line in order to find the reaction rate of an interval.

To find the average, we generally use the general rate of reaction (avg rate) for the following reaction $aA+bB\rightarrow cC+dD$ :

$rate=-\frac{1\Delta [A]}{a\Delta t}$ $=-\frac{1\Delta [B]}{b\Delta t}$ $=\frac{1\Delta [C]}{c\Delta t}$ $=\frac{1\Delta [D]}{d\Delta t}$

Where the negative indicates rate of disappearance while the positive represent rate of appearance.

EDIT:

To further clarify what the average rate of the reaction is, it is the change in concentration over a set amount of time (ex. 3 seconds, 4 minutes, etc...). The initial rate is the change in concentration over the first set of recorded time, which is usually 0 units of time to another point. The instantaneous rate of reaction is the change in concentration over a time frame that is almost zero (or in other words, as the limit of the change in time as it approaches zero).

Q12.4.19

Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:

Initial [C₃H₅N₃O₉] (M)	4.88	3.52	2.29	1.81	5.33	4.05	2.95	1.72
t (s)	300	300	300	300	180	180	180	180
% Decomposed	52.0	52.9	53.2	53.9	34.6	35.9	36.0	35.4

S12.4.19

In order to solve these equation, we must realize that these are first order reactions because the samples underwent decomposition. So, we must use the first order integrated rate law:

$ln[A]=-kt+ln[A]_{0}$

1) The initial concentration is 4.88M, the time of decomposition is 300s, and the percent of the sample decomposed is 52%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 4.88M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$4.88M(.52)=2.538M$

$4.88M-2.538M=2.342M$

Now just plug in your numbers and solve for k.

$ln(2.342M)=-k300s+ln(4.88M)$

Take the ln of the molarities.

$0.8512=-k300s+1.585$

Subtract 1.585 from 0.8512.

$-0.734=-k300s$

Divide by 300s.

$-2.45*10^{-3}s^{-1}=-k$

Divide by -1 in order to solve for k.

$2.45*10^{-3}s^{-1}=k$

2)The initial concentration is 3.52M, the time of decomposition is 300s, and the percent of the sample decomposed is 52.9%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 3.52M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$3.52M(0.529)=1.862M$

$3.52M-1.862M=1.658M$

Now just plug in your numbers and solve for k.

$ln(1.658M)=-k300s+ln(3.52M)$

Take the ln of the molarities.

$0.5056=-k300s+1.258$

Subtract 1.258 from 0.5056.

$-0.753=-k300s$

Divide by 300s.

$-2.51\cdot 10^{-3}s^{^{-1}}=-k$

Divide by -1 in order to isolate k.

$2.51\cdot 10^{-3}s^{^{-1}}=k$

3) The initial concentration is 2.29M, the time of decomposition is 300s, and the percent of the sample decomposed is 53.2%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 2.29M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$2.29M(0.532)=1.072M$

$2.29M-1.2183M=1.072M$

Now just plug in your numbers and solve for k.

$ln(1.072M)=-k300s+ln(2.29M)$

Take the ln of the molarities.

$0.0693=-k300s+0.829$

Subtract 0.829 from 0.0693.

$-0.759=-k300s$

Divide by 300s.

$-2.54\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.54\cdot 10^{-3}s^{-1}=k$

4) The initial concentration is 1.81M, the time of decomposition is 300s, and the percent of the sample decomposed is 53.9%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 1.81M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$1.81M(0.539)=0.9756$

$1.81M-0.9756M=-0.834M$

Now just plug in your numbers and solve for k.

$ln(0.834M)=-k300s+ln(1.81M)$

Take the ln of the molarities.

$-0.181=-k300s+0.593$

Subtract 0.593 from -0.181.

$-0.774=-k300s$

Divide by 300s.

$-2.58\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.58\cdot 10^{-3}s^{-1}=k$

5)The initial concentration is 5.33M, the time of decomposition is 180s, and the percent of the sample decomposed is 34.6%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 5.33M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$5.33M(0.346)=1.844M$

$5.33M-1.844M=3.486M$

Now just plug in your numbers and solve for k.

$ln(3.486M)=-k180s+ln(5.33M)$

Take the ln of the molarities.

$1.249=-k180s+1.673$

Subtract 1.673 from 1.249

$-0.424=-k180s$

Divide by 180s.

$-2.35\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.35\cdot 10^{-3}s^{-1}=k$

6)The initial concentration is 4.05M, the time of decomposition is 180s, and the percent of the sample decomposed is 35.9%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 5.33M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$4.05M(0.359)=1.454M$

$4.05M-1.454M=2.596M$

Now just plug in your numbers and solve for k.

$ln(2.596M)=-k180s+ln(4.05M)$

Take the ln of the molarities.

$0.954=-k180s+1.399$

Subtract 1.399 from 0.954

$-0.445=-k180s$

Divide by 180s.

$-2.47\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.47\cdot 10^{-3}s^{-1}=k$

7)The initial concentration is 2.95M, the time of decomposition is 180s, and the percent of the sample decomposed is 36%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 2.95M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$2.95M(0.36)=1.062M$

$2.95M-1.062M=1.888M$

Now just plug in your numbers and solve for k.

$ln(1.888M)=-k180s+ln(2.95M)$

Take the ln of the molarities.

$0.636=-k180s+1.082$

Subtract 1.082 from 0.636

$-0.446=-k180s$

Divide by 180s.

$-2.47\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.47\cdot 10^{-3}s^{-1}=k$

8) The initial concentration is 1.72M, the time of decomposition is 180s, and the percent of the sample decomposed is 35.4%. $[A]_{0}$ is the initial concentration of the sample and [A] is the concentration of the sample after decomposition. The initial concentration is 1.72M. In order to find [A], we need to find the amount of concentration that has decomposed then subtract that from the original concentration.

$1.72M(0.354)=0.6089M$

$1.72M-0.6089M=1.1111M$

Now just plug in your numbers and solve for k.

$ln(1.1111M)=-k180s+ln(1.72M)$

Take the ln of the molarities.

$0.1053=-k180s+0.5423$

Subtract 0.5423 from 0.1053

$-0.437=-k180s$

Divide by 180s.

$-2.43\cdot 10^{-3}s^{-1}=-k$

Divide both sides by -1 in order to isolate k.

$2.43\cdot 10^{-3}s^{-1}=k$

EDIT:

Everything looks correct.

Q21.3.3

Complete each of the following equations by adding the missing species:

$_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{He}\rightarrow ? + _{0}^{1}\textrm{n}$
$_{94}^{239}\textrm{Pu}+? \rightarrow _{96}^{242}\textrm{Cm}+_{0}^{1}\textrm{n}$
$_{7}^{14}\textrm{N}+_{2}^{4}\textrm{He}\rightarrow ? + _{1}^{1}\textrm{H}$
$_{92}^{235}\textrm{U}\rightarrow ?+_{55}^{135}\textrm{Cs}+4_{0}^{1}\textrm{n}$

S21.3.3

It is important to remember this general rule:

$_{A}^{Z}\textrm{X}$

Where Z is the mass number, A is the atomic number, and X is the element. A is also the number of protons, so to find the number of neutrons, it is Z-A.

Particle	Nuclear Symbol
Alpha	$_{2}^{4}\textrm{He}$ or $\alpha$
Beta (positron)	$_{1}^{0}\textrm{e}$ or $\beta ^{+}$
Beta (electron)	$_{-1}^{0}\textrm{e}$ or $\beta ^{-}$
Proton	$_{1}^{1}\textrm{p}$
Neutron	$_{0}^{1}\textrm{n}$
Gamma	$_{0}^{}\textrm{0}$ $\gamma$

1. $_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{He}\rightarrow ? + _{0}^{1}\textrm{n}$

(1) In this reaction, an aluminum isotope is reacting with an alpha particle to produce a neutron and an elemental isotope.

(2) The total mass number in reactants must equal the total mass number in the products (the number at the top left is corner of the element is the mass number), and the total atomic number in reactants must equal the total atomic number in the products (the number at the bottom left corner of the element is the atomic number).

(3) 27+4=31 in the reactants, so for products, x + 1 = 31, solve for x, and x =30. So the mass number of the unknown element is 30.

$27+4=31$

$x+1=31\therefore x=30$

(4) 13+2=15 in reactants, so for products, x+0=15, so x=15. So the atomic number of the element is 15.

$13+2=15$

$x+0=15\therefore x=15$

(5) Phosphorus is the unknown element because its atomic number is 15.

$_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{He}\rightarrow _{15}^{30}\textrm{P}+_{0}^{1}\textrm{n}$

EDIT: We know phosphorous-30 is the answer because phosphorus has the atomic number 15 and is represented by the shorthand "P".

2. $_{94}^{239}\textrm{Pu}+? \rightarrow _{96}^{242}\textrm{Cm}+_{0}^{1}\textrm{n}$

(1) In this reaction, a plutonium isotope and another elemental isotope reacts to produce curium and a neutron.

(3) 242+1=243 in the products, so for reactants, 239+x=243, solve for x, and x=4. So the mass number of the unknown element is 4.

$242+1=243$

$239+x=243\therefore x=4$

(4) 96+0=96 in the products, so for reactants, 94+x=96, solve for x, so x=2. So the atomic number of the element is 2.

$96+0=96$

$94+x=96\therefore x=2$

(5) Helium is the unknown element because its atomic number is 2 and it is also called an alpha particle.

$_{94}^{239}\textrm{Pu}+_{2}^{4}\textrm{He}\rightarrow _{96}^{242}\textrm{Cm}+_{0}^{1}\textrm{n}$

EDIT: We know that an alpha particle is the answer because it is similar to helium which has the atomic number 2 with the mass of 4 and is represented by the shorthand "He".

3. $_{7}^{14}\textrm{N}+_{2}^{4}\textrm{He}\rightarrow ? + _{1}^{1}\textrm{H}$

(1) In this reaction, a nitrogen isotope reacts with a helium isotope to produce an unknown elemental isotope and hydrogen.

(2) The total mass number in the reactants must equal the total mass number in the products (the number at the top left is corner of the element is the mass number), and the total atomic number in reactants must equal the total atomic number in the products (the number at the bottom left corner of the element is the atomic number).

(3) 14+4=18 in the reactants, so for products, x+1=18, solve for x and you get x=17. So the mass number of the unknown element is 17.

$14+4=18$

$x+1=18\therefore x=17$

(4) 7+2=9 in the reactants, so for the products, x+1=9, solve for x and you get x=8. So the atomic number of the element is 8.

$7+2=9$

$x+1=9\therefore x=8$

(5) Oxygen is the unknown element because its atomic number is 8.

$_{7}^{14}\textrm{N}+_{2}^{4}\textrm{He}\rightarrow _{8}^{17}\textrm{O}+_{1}^{1}\textrm{H}$

EDIT: We know oxygen-17 is the answer because oxygen has the atomic number 8 and is represented by the shorthand "O".

4. $_{92}^{235}\textrm{U}\rightarrow ?+_{55}^{135}\textrm{Cs}+4_{0}^{1}\textrm{n}$

(1) In this reaction, uranium decays into an elemental isotope, cesium , and 4 neutrons.

(3) 253 in the reactants, so for products, x+135+4(1) (because there are four neutrons) =235. Solve for x and you get x=96. So the mass number of the unknown element is 96.

$x+135+4(1)=235\therefore x=96$

EDIT: THERE IS A TYPO (This should say 235 =x +135 + 4(1) where x = 96)

(4) 92 in the reactants, so for the products, x+55+4(0) (because there are four neutrons)=92. Solve for x and you get x=37. So the atomic number of the unknown element is 37.

$x+55+4(0)=92\therefore x=37$

EDIT : THERE IS A TYPO (This should say x + 55 + 4(0) = 92 where x = 37)

(5) Rubidium is the unknown element because its atomic number is 37.

$_{92}^{235}\textrm{U}\rightarrow _{37}^{96}\textrm{Rb}+_{55}^{135}\textrm{Cs}+4_{0}^{1}\textrm{n}$

EDIT: We know rubidium-96 is the answer because it has the atomic number 37 and is represented by the shorthand "Rb".

Q21.7.5

Given specimens neon-24 ( $t_{1/2}=3.38min$ ) and bismuth-211 ( $t_{1/2}=2.14min$ ) of equal mass, which one would have greater activity and why?

S21.7.5

Neon-24 is most likely stable because since it has a mass number of 24, the atomic number is 10 (which is also the number of protons), so the number of neutrons is 24-10, which is 14. Thus, it is most likely stable because it has an even number of protons and neutrons. Bismuth-211 is most likely radioactive because it has a mass number of 211, the atomic number is 83 (which is also the number of protons), so the number of neutrons is 211-83, which is 128. Thus, it is probably radioactive because it has an odd number of neutrons and an even number of protons. Therefore, based on this, Bismuth-211 would have greater activity because it is radioactive and unstable, so it would undergo more decay.

EDIT: Stability is not always dependent on an element having an even number of protons and neutrons. It has more to do with the stability curve which can be viewed HERE. In this case, neon-24 lies on the line of stability, while bismuth-211 is off the line and the graph indicates that it undergoes radioactive decay.

Another way to approach this would be to refer to the half life nuclear equation:

$t_{1/2}=$ $\frac{ln2}{\lambda }$ or $t_{1/2=}\frac{0.693}{\lambda }$

Since we know the half lives of the two specimens, we can just plug it in and solve for $\lambda$ .

EDIT: We actually need to find activity which is usually calculated by Activity (A) = decay constant (lambda) x number of undecayed nuclei (N). Because we are only given half-lives, we need to use the above equation to find the decay constant (or lambda) first. Then from there, the relative activity between neon-24 and bismuth-211 can be found.

For Neon-24:

$3.38min=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{3.38min}$

$\lambda =0.205\frac{1}{min}$ --- This would actually be in s^-1 not Bq because it is a decay constant

EDIT: So far, the decay constant is found but the number of undecayed nuclei must be found to do the calculation. Neon-24 means it has 24 g/mol.

N = mass of sample/ (24 g/mol) x (6.02 E23 nuclei/mol) -- this calculation would normally find the amount of nuclei in a sample

The result is A = 0.205 s^-1 x (mass of sample x (6.02E23 nuclei/24 g))

= 0.205s^-1 x (mass of sample x 6.69 E21)

= (mass of sample)(5.14 E21) Bq

For Bismuth-211:

$2.14min=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{2.14min }$

$\lambda =0.325\frac{1}{min}$ --- This would actually be in s^-1 not Bq because it is a decay constant

EDIT: Do the same calculations as in neon-24.

N = mass of sample/ (211 g/mol)x (6.02 E23 nuclei/mol) -- this calculation would normally find the amount of nuclei in a sample

A = 0.325 s^-1 x (mass of sample/ ((6.02 E23 nuclei/211 g))

= 0.325s^-1 x (mass of sample x 2.85 E21)

= (mass of sample)(9.27 E20) Bq

~~Therefore, since Bismuth-211 has a larger constant, it will be more active. This shows the relationship that the larger the half life is, the less active it will be because it will result in a smaller $\lambda$ .~~

EDIT:

From the calculations, it seems that neon-24 would be more active than bismuth-211 because it has a greater activity of 5.14E21 Bq compared to the same mass of bismuth which has 9.27E20 Bq.

This method would only work if the question states that the two samples have equal mass and it only specifies to find which has greater activity.

Q20.4.18

You have built a galvanic cell using an iron nail, a solution of FeCl₂, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is E_cell?

S20.4.18

In addition to the corrosion of the iron nail due to oxidation, other observations would include the solution to bubble because of the standard hydrogen electrode, where $H_{2}^{+}+2e^{-}\rightarrow H_{}2$ . Thus, the solution would form hydrogen gas, which causes the solution to bubble. SHE has an reduction potential of 0.0V and so all the other half reactions have standard hydrogen reduction potentials relative to hydrogen. Also, the $FeCl_{2}$ would disassociate in the solution forming $Cl^{-}$ and $Fe^{2+}$ ions. Therefore another observation would be that the solution would turn green because $Fe^{2+}$ ions produce a green color in an aqueous solution.

Image result for reduction potential table

EDIT:

The E°_cell of this reaction would be calculated using the formula E°_cell =E°cathode - E°_anode

The cathode side is the reduction: $H_{2}^{+}+2e^{-}\rightarrow H_{}2$ E°cathode = 0.00 V

The anode side is the oxidation: Fe(s) → Fe²⁺ + 2e^-E°_anode = -0.44 V

E°_cell = 0.00 V + 0.44 V = 0.44V

Q20.9.3

Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?

S20.9.3

Pure salts inhibit the ability of ions to move because it is in a fixed state, such as in the form of a solid. This would not be beneficial in electrolysis because ions to to be able to move around. Therefore, this would result in incorrect voltages or the current would not be able to flow through the solution. However, using mixtures of molten salts means that its ions are more capable of moving freely between the aqueous solutions in an electrochemical reaction because it is not in a fixed state, such as in a liquid phase. In addition, the bonds of the ions in molten salts are weaker than the bonds of ions in pure salts, so when molten salts are placed in an aqueous solution, it is easier to break the bonds of the ions compared to if a pure salt was placed in an aqueous solution. Therefore, since the ions are capable to move around, currents can pass through. All in all, molten salts have a greater ability to conduct electricity compared to pure salts.

EDIT: Some parts of this explanation is a little confusing, so clarification is needed. The paragraph is essentially saying the following:

Electrolysis is when an electric current it sent through an electrolyte and into solution in order to create a flow of ions needed to run an otherwise non-spontaneous reaction.
Pure salts are usually in a solid form, so ions and electrons cannot freely move about. As a result, nothing will happen in the electrolysis because nothing will be able to flow between the electrodes.
Molten salts break the barrier faced by pure salts because they are a liquid, which allow for the flow of both ions and electrons. This free flow is necessary because there must be a way for an electron current to form to do an electrolysis.

Q20.9.1

Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?

S20.9.1

Electrolysis is the process of turning nonspontaneous reactions into spontaneous reactions by electricity. In order for an electrochemical reaction to be thermodynamically favored, a high electrical potential must be supplied by an external power supply, specifically higher than a certain voltage in the reverse direction, in order for the reaction to be spontaneous and to turn from a galvanic to electrolytic cell. Therefore, there are 3 ways in order to determine if an electrochemical reaction is thermodynamically favorable. In addition, an overvoltage can help overcome the activation energy of a reaction so that the reaction can proceed and be spontaneous.

Favorable	Nonfavorable	Equilibrium
$\varepsilon >0$	$\varepsilon <0$	$\varepsilon =0$
spontaneous	nonspontaneous	equilibrium
shift to products	shift to reactants	no shift

EDIT: Let's define the next part of the question. Overvoltage is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system (definition from Bodner Research Web). This can boil down to needing extra charge to overcome the activation energy.

Thus, an electrochemical reaction that may be thermodynamically favored may require an overvoltage because the activation energy is too high for the reaction on its own to overcome. Even the most exothermic reaction that produces stoichiometrically more products than reactants have an activation energy that must be overcome for any reaction to take place.

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