Extra Credit 12
- Page ID
- 82870
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17.2.1
Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.
- Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)
- 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
- Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Au(s)
- 3CuNO3(aq)+Au(NO3)3(aq)⟶3Cu(NO3)2(aq)+Au(s)
Answer:
(a) Mg(s)│Mg2+(aq)║Ni+(aq)│Ni(s)
(b) Cu(s)│Cu2+(aq)║Ag+(aq)│Ag(s)
(c) Mn(s)│Mn2+(aq)║Sn2+(aq)│Sn(s)
(d) Pt(s)│Cu+(aq),Cu2+(aq)║Au3+(aq)│Au(s)
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The format which the cell notation is written is the anode electrode, aqueous anode, salt bridge, aqueous cathode, then cathode electrode from left to right.
This is the format: anode electrode| aqueous anode || aqueous cathode | cathode electrode. Single straight lines are used to denote a phase change and the double straight line is used to denote the salt bridge and the separator. Notice that it is written to represent electron flow from left to right as electrons flow from the anode to the cathode.
The cathode is the oxidizing agent or the species that is reduced. The anode is the reducing agent or the species that is oxidized.
- Mg(s) + Ni2+(aq) --> Mg2+(aq) + Ni(s)
-Determine which is reduced and which is oxidized.
-Mg is oxidized because it loses two electrons Mg(s) -> Mg2+(aq)
-Ni is reduced because it gains two electrons Ni2+(aq) -> Ni(s)
-Therefore, Mg is the anode and Ni is the cathode.
anode electrode|anode||cathode||cathode electrode
- Answer: Mg(s)│Mg2+(aq)║Ni+(aq)│Ni(s)
I feel like doing the tutorial in bullet points makes it hard for the student to really follow along, but I like how you make the formula so that it is easy to see what you are trying to do. The same is applied for all of these types of problems.
- 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
-Determine which is reduced and which is oxidized
-Ag is oxidized because it loses one mole of electron per mole of Ag
-Cu is reduced because it gains two electrons
-Therefore, Ag is the anode and Cu is the cathode
- Answer: Cu(s)│Cu2+(aq)║Ag+(aq)│Ag(s)
c. Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s) (the Au is probably a typo)
-Mg is oxidized because it loses two electrons as shown in the half reaction: Mn(s)+(NO3)2 --> Mn(NO3)2 + 2e-
-Sn is reduced because it gains two electrons as shown in the half reaction: Sn(NO3)2 +2e --> Sn(s) + 2(NO3)-
-Therefore Mn is the anode and Sn is the cathode
- Answer: Mn(s)│Mn2+(aq)║Sn2+(aq)│Sn(s)
d. 3CuNO(aq)+Au(NO3)3(aq)⟶3Cu(NO3)2(aq)+Au(s)
-Determine which is reduced and which is oxidized by determine the oxidation states
-Au is reduced because it loses 3 moles of electrons per mole of Au. Determine the oxidation state of Au and we see that it goes from -3 --> 0. Because nitrate(NO3) has a -1 charge, and it takes 3 nitrates per gold (Au), we can determine that gold has an initial oxidation state of +3. At the end of the experiment, it is a standalone solid, therefore having a charge of +0. Therefore it is reduced because Au goes from +3 --> 0 oxidation state.
-Cu is oxidized because it loses one mole of electrons per mole. We know that the charge on nitrate(NO3) is -1 so from that we can see that one nitrate is bound to one Copper(Cu). Therefore we can determine the oxidation state of copper as it goes from +1 --> +2 because Cu(NO3) -->Cu(NO3)2
-We need a platinum electrode in this situation because the copper is not a solid anywhere in the redox reaction, therefore it cannot act as an electrode and we will need an inert electrode. In this case it is platinum.
-Therefore Cu is the anode and Au is the cathode
-Cu+ comes from the Cu(NO3) and Cu2+ comes from the Cu(NO3)2
- Answer: Pt(s)│Cu+(aq),Cu2+(aq)║Au3+(aq)│Au(s)
19.1.10
Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.
Answer: In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases.
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-To answer this question, we need to use the Lewis Acid Base theory. The theory states that acids and bases react with each other to exchange electrons. The Lewis Acid is any substance that can accept a pair of electrons. It can be an H3O+, but the definition allows any substance be an acid. Similarly, a Lewis Base is any substance that can donate a pair of electrons. It can be an OH-, but not necessarily. Any lone pair of electrons ready to be donated is a Lewis Base.
-Here is an example of a Hydronium Ion. The hydrogen ion pairs with a lone pair of electrons on the oxygen atom in water to form H3O+. Each line between the Oxygen and Hydrogen represent two electrons. The Hydrogen ion is the Lewis Acid and the H2O is the Lewis Base.
-The auto ionization of water means that water is constantly breaking into H+ and OH-
-Aqueous means that it was placed into water. Placing it into water means that it is subjected to H2O molecule as well as H+ and OH- molecules
-Manganese Oxide can accept lone pairs. Therefore it would act as a Lewis Acid
Again, I am not crazy about the bullet points, but your explanation and diagram is a representation of what you are trying to tell the student.
19.3.2
Draw the crystal field diagrams for [Fe(NO2)6]4− and [FeF6]2−. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.
I would describe what you are doing here with the diagram, meaning telling the student what the arrows are supposed to mean. Do not just assume that the student would be able to tell that the arrows mean electrons. Explain the mechanics of what you are doing, down to the smallest feature. However, I do like the visuals!
Answer:
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-For the element iron, its bonding electrons are in the d orbital, therefore it would have an octahedral ligand field.
-The octahedral ligand field has the following orbital format
.
-Each d orbital can hold 2 electrons. One electron as a spin up and the other as a spin down.
-According to Hund's Rule, electrons will prefer to fill orbitals with the same energy first before they pair with each other.
-According to Aufbau's principle, electrons will prefer the lower energy orbital first before filling the higher energy orbitial.
-The spin of the electron depends on the strength of the ligand on it. Weak field ligands tend to allow the electrons to settle at the higher spin orbitals without paring with the lower energy electrons. And strong field ligands tend to cause the electrons to pair with the lower energy orbitals first before rising to the higher energy orbital.
-For [Fe(NO2)6]4− , we see that (NO2) is a strong ligand. Because of this, we can see that the electrons will occupy the lower orbitals first and pair. It is a low spin because all of the electrons are on the lower orbital. The number of electrons is determined by the oxidation state of the element. [Fe(NO2)6]4− . We know that (NO2) has a -1 charge and the overall coordination compound has a -4 charge. Because there is 6 (NO2) it will be a -6 charge and because it is an overall -4 charge, the iron has to be a +2 oxidation state. Using the electron configuration of iron, [Ar] 3d^6 4s^2 , up to two s orbital electrons are removed first. therefore there will be 6 electrons in the d orbital, hence 6 arrows. This is a low spin.
-For [FeF6]2− , we see that Fluoride is a weak field ligand. Because of this, we see that the electrons will occupy all of the available orbitals before forming pairs with the other electrons. It is a high spin because there are electrons on the higher orbitals. Iron has an oxidation number of +4. The electron configuration of iron is [Ar] 3d6 4s2, two of the s orbital electrons are removed, then two of the d orbital electrons are removed. Leaving 4 d orbital electrons. It will still occupy the lower energy first before jumping to the higher energy. This is a high spin.
12.4.2
Use the data provided to graphically determine the order and rate constant of the following reaction: SO2Cl2⟶SO2+Cl2
I would explain more on how you know that it is a first order reaction. Describe to the student how you are able to tell that it is a first order, do not just assume that he or she would be able to tell.
Answer:
Plotting a graph of ln[SO2Cl2] versus t reveals a linear trend; therefore we know this is a first-order reaction:
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-Create three graphs using concentration vs time. The first graph represents zero order and is [SO2Cl2] vs time. The second graph represents first order and is Ln(SO2Cl2) vs time, and the third graph represents second order and is 1/[SO2Cl2] vs time.
![Ln([SO2Cl2]).png](https://chem.libretexts.org/@api/deki/files/112434/Ln(%255BSO2Cl2%255D).png?revision=1&size=bestfit&width=537&height=286)
![1[SO2Cl2].png](https://chem.libretexts.org/@api/deki/files/112433/1%25EF%2580%25A2%255BSO2Cl2%255D.png?revision=1&size=bestfit&width=526&height=289)
-To determine the graph that determines which order the reaction is, we draw a line between each point and determine the one that forms a straight line. If [SO2Cl2] formed a straight line, then it is zero order as concentration does not affect rate. If Ln([SO2Cl2]) forms a straight line, then it is first order as concentration does affect rate. If 1/[SO2Cl2] forms a straight line, then it is second order as concentration does affect rate at a exponentially double rate.
-In this example, the straightest line is formed by the first order graph, therefore this reaction is first order.
12.7.5
For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:
Answer: The lowering of the transition state energy indicates the effect of a catalyst. (a) A; (b) B
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-A catalyst lowers the activation energy of a reaction. To find a catalyst, compare the amount of energy required by the non catalyzed reaction. In a non catalyzed reaction, the activation energy in a transition state will be higher.
-In (a), reaction is an endothermic reaction because the products start with lower internal energy and end with higher internal energy. This reaction has intermediates and we can see that the intermediate is affected by a catalyst at the second peak. Because the peak on (b) is lower than the peak on (a), therefore, (b) will be a catalyzed reaction.
-In (b), this reaction is an exothermic reaction because the products start with higher internal energy and end with lower internal energy. This reaction has intermediates as there are two peaks, which represent the activation energy of that intermediate. We can see that (b) is the catalyzed reaction because it has a lower activation energy.
21.4.28
Write a balanced equation for each of the following nuclear reactions:
a. mercury-180 decays into platinum-176
b. zirconium-90 and an electron are produced by the decay of an unstable nucleus
c. thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium-228 by beta decay
d. neon-19 decays into fluorine-19
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I think that if you were to do this in a step-wise method of teaching, it would be more effective for the student. The abc method can be hard for a student to follow along.
-Atomic mass is the total mass of the atom: the proton, neutron, and electron.
-A proton and neutron are approximately 1 amu and an electron is approximately 0 amu.
-There is 4 types of nuclear decay. Alpha decay, beta decay, positron decay, and gamma decay.
-Alpha decay, the nucleus loses 2 protons and 2 neutrons. It looses a total of 4 amu per alpha.
-Beta decay, the nucleus loses one electron and one neutron and gains a proton. The neutron decays an electron and becomes a proton. the atomic mass doesn't change, but the atomic number increases by one per beta decay.
-Positron decay, where a proton transforms into a neutron, and emits an anti-electron (positive charge electron). The atomic mass stays the same, but the atomic number decreases by one per positron decay.
-Gamma decay, a type of decay where the nucleus of an atom emits a photon which is a gamma ray. The atomic mass and the number of protons and neutrons do not change. This occurs because the nucleus of an atom is relatively unstable after alpha and beta decays.
a) Mercury 's atomic mass is 180 and it has an atomic number of 80 which is how many protons it has. Determine the number of neutrons. 180 - 80 = 100 neutrons. This element decays into platinum 176. According to the periodic table, platinum has an atomic number of 78, meaning it has 78 protons.
-This is alpha decay because alpha decay results in -4 amu and -2 protons. Therefore
- 180Hg --> 176Pt + 4α where Hg has 80 protons, Pt has 78 protons
b) Zirconium has 40 protons, and an atomic mass of 90. Because an electron is produced, this is beta decay.
- So ??? --> 90Zr + 0-1β
-During beta decay, the nucleus gains one proton, but atomic mass isn't changed. Therefore
??? = 90Nb
- 90Nb --> 90Zr + 0-1β Where Nb has 41 protons, Zr has 40 protons
c) This is a two step equation. There are two decays.
-First determine the first decay, which is alpha decay so
232Th --> 4α + 228Ra Where Th has 90 protons and Ra has 88 protons.
-Now determine the second decay, which is a beta decay
228Ra --> 0-1β + 228Ac Where Ra has 88 protons and Ac has 89 protons
- 232Th --> 4α + 228Ra --> 0-1β + 228Ac
d) Determine what type of decay this is.
-Neon turns into Fluorine, so the atomic number is changing from 10 to 9, but atomic mass does not change. Therefore this is a positron decay.
- 19Ne --> 19F + 01β Where Ne has 10 protons and Fluorine has 9 protons
20.3.14
For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.
- Ag(s) + Fe3+(aq) → Ag+(aq) + Fe2+(aq)
- Fe3+(aq) + 1/2H2(g) → Fe2+(aq) + H+(aq)
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-To determine which direction the reaction happens spontaneously, use a Standard Reduction Potential Table.
-We will use this formula to determine the cell potential of the two elements in the redox
-E°cell gives us the theoretical voltage that a cell will produce if the elements are combined.
E°cell = E°(cathode) - E°(anode)
ΔG° = -nFE°cell Where n is number of electrons exchanged, F is Faraday's constant = 96486C/(mole e-) and E°cell is cell potential.
-Using ΔG°, we can determine the spontaneity of the reaction. If ΔG° is negative, then it is spontaneous, and like wise, if it is positive, then it is non spontaneous.
a)
-First write out the appropriate half reactions.
Ag(s) --> Ag+(aq)
Fe3+(aq) --> Fe2+(aq)
-Balance the electrons
Ag(s) --> Ag+(aq) +e-
Fe3+(aq) +e- --> Fe2+(aq) Note that one electron is being exchanged
-Write out the E°voltage using the standard reduction potentials table
Ag(s) --> Ag+(aq) 0.80V
Fe3+(aq) --> Fe2+(aq) 0.771V
-Determine E°cell
E°cell = 0.8V - 0.771V = 0.029V if Ag is the cathode and Fe is the anode
E°cell = 0.771V - 0.8V = -0.029V if Ag is the anode and Fe is the cathode
-Determine if spontaneous
ΔG° = -nFE°cell
n=1 because one electron is being exchanged
F = 96486C/(mole e-)
ΔG° = -1*96486*0.029 = -2800 if Ag is the cathode and Fe is the anode
ΔG° = -1*96486*-0.029 = 2800 if Ag is the anode and Fe is the cathode
-Therefore the reaction is spontaneous if Ag is the cathode and Fe is the anode
- Pt | Fe3+(aq) , Fe2+(aq) || Ag+(aq) | Ag(s)
b)
-First write out the appropriate half reactions.
Fe3+(aq) --> Fe2+(aq)
1/2H2(g) --> H+(aq)
-Balance the electrons
Fe3+(aq) + e- --> Fe2+(aq)
H2(g) --> 2H+(aq) + 2e- Note that two electrons are being exchanged
-Write out the E°voltage using the standard reduction potentials table
Fe3+(aq) --> Fe2+(aq) 0.771V
H2(g) --> 2H+(aq) + 2e- 0.00V
-Determine E°cell
E°cell = 0.00V - 0.771V = -0.771V if H2 is the cathode and Fe is the anode
E°cell = 0.771V - 0.00V = 0.771V if H2 is the anode and Fe is the cathode
-Determine if spontaneous
ΔG° = -nFE°cell
n=2 because two electron is being exchanged
F = 96486C/(mole e-)
ΔG° = -2*96486*0.771 = -149000 if H2 is the cathode and Fe is the anode
ΔG° = -2*96486*-0.771 = 149000 if H2 is the anode and Fe is the cathode
-Therefore the reaction is spontaneous if H2 is the anode and Fe is the cathode.
-Because H2 is a gas, we need an electrode to help facilitate electron movement.
- Pt | Fe3+(aq) , Fe2+(aq) || H+(aq) | C(s)
20.5.27
Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction O2(g) + 4H+(aq) + 4e− → 2H2O(l).
- Will aqueous acidic KMnO4 evolve oxygen with the formation of MnO2?
- At pH 14.00, what is E° for the oxidation of water by aqueous KMnO4 (1 M) with the formation of MnO2?
- At pH 14.00, will water be oxidized if you are trying to form MnO2 from MnO42− via the reaction 2MnO42−(aq) + 2H2O(l) → 2MnO2(s) + O2(g) + 4OH−(aq)?
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1:
Begin by writing the two half reactions
O2(g) + 4H+(aq) + 4e− → 2H2O(l)
MnO4- + 4H+ + 3e--->MnO2 + 2H2O(l)
Combine the two equations.
-We want the MnO4 to be the reactant
2H2O(l) --> O2(g) + 4H+(aq) + 4e− Flipped Around
MnO4- + 4H+ + 3e--->MnO2 + 2H2O
-Sum the two equations, making sure to multiply the equations such that the number of electrons are the same
3(2H2O(l) --> O2(g) + 4H+(aq) + 4e−)
2(MnO4- + 4H+ + 3e--->MnO2 + 2H2O)
6H2O + 4MnO2 + 16H+ --> 3O2 + 12H+ + 4MnO2 + 8H2O
-Simplify
4MnO2 + 4H+ --> 3O2 + 4MnO2 + 2H2O
- Because oxygen is a product of this reaction, acidic KMnO4 will evolve oxygen.
2:
Begin by writing the two half reactions
MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V
O2 + 4H+ +4e- --> 2H2O 1.23V
Determine which will be oxidized and which will be reduced
-Use the standard cell potential, E°cell = E°(cathode) - E°(anode) , to determine which is cathode and which is anode
-A spontaneous reaction would have an E°cell >0
1.68V - (-1.23V) = 2.91V
1.23V - (1.68V) = -0.45V
-Therefore MnO4- will be the cathode, the species that will be reduced and O2 will be the anode, the species that is oxidized.
Combine the two equations so that MnO4- is reduced and O2 is oxidized
MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V
O2 + 4H+ +4e- --> 2H2O 1.23V
-Flip the O2 equation
MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V
2H2O --> O2 + 4H+ +4e- -1.23V
-Manipulate both equations so that they have the same number of electrons
4(MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l)) 1.68V
3(2H2O --> O2 + 4H+ +4e-) -1.23V
-Combine them and simplify
4MnO4- + 4H+ --> 4MnO2 + 2H2O + 3O2
Calculate the E°cell potential
-Use the standard cell potential, E°cell = E°(cathode) - E°(anode).
E°cell = 1.68V - (-1.23V) = 2.91V
Use the Nernst Equation to calculate Ecell @ pH 14
Q = product concentration over reactant concentration
\[E =\frac{8.3145*298}{12*96485.33}*ln\frac{1}{10^{-14}}\]
E = 2.63V
- Because E is positive, therefore it is spontaneous at pH 14. Therefore at E°, the oxidation of water is at -1.23V.
3:
Begin by writing the three half reactions and potentials at standard.
MnO4- + 4H+ + 3e--->MnO2 + 2H2O 1.68V
MnO42- --> MnO4- + e- -0.56V
2H2O --> O2 + 4H+ +4e- -1.23V
Sum the first two equations without eliminating the electrons
MnO42- --> 4H+ + 2e- --> MnO2 + 2H2O 2.252V
Sum the third equation
MnO42- --> 4H+ + 2e- --> MnO2 + 2H2O 2.252V
2H2O --> O2 + 4H+ +4e- -1.23V
2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V
Compensate for basic reaction
2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V
-Add OH- to both sides of the equation and simplify
2MnO42- + 2H2O --> O2 + 2MnO2 + 4OH- 3.482V
Use Nernst equation to determine E @ pH 14
-Where E is the final cell potential, E° is the cell potential at standard conditions, R is the ideal gas constant, T is the temperature in K, n is the number of electrons exchanged, F is Faraday's constant and Q is concentrations in:
2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V
\[E =\3.482 - frac{8.3145*298}{4*96485.33}*ln\frac{1}{(10^-14)}\]
E = 3.482
- Therefore water will be oxidized because in the equation, it is already shown to donate its electrons, and the reaction is spontaneous at pH 14.
I like how you posted the formulas in bigger font to make it easier for the student to comprehend what they are supposed to use!