Extra Credit 7
- Page ID
- 83286
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Q17.1.6
Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem. (seen below)
a. H2O2 + Sn2+⟶ H2O + Sn4+
b. PbO2 + Hg ⟶ Hg22++ Pb2+
c. Al + Cr2O72−⟶ Al3+ + Cr3+
S17.1.6
To answer this problem we must first understand what Oxidation and Reduction agents look like, see this page for an in depth look at identifying the different elements of Redox reactions. As a summary;
Oxidation: the net loss of electrons (the specific element increases in oxidation number, becoming more positive) ie. Zn(s) → Zn2++ 2e− (0 charge becomes +2 charge, gain of electrons)
Reduction: the net gain of electrons (the specific element decreases in oxidation number, becoming more negative) ie. Cu2+ + 2e− → Cu(s) (+2 charge becomes 0 charge, loss of electrons)
Oxidizing agent: the compound or element in a reaction that oxidizes the other compounds it bonds with, making them lose electrons and become more positive. The oxidizing agent in turn becomes reduced by the reducing agent.
Reducing agent: the compound or element in a reaction that reduces the other compounds it bonds with, making them gain electrons and become more negative. The reducing agent in turn it gets oxidized by the oxidizing agent.
PROBLEM SOLUTION:
a. Sn2+ is the reducing agent because it undergoes oxidation (Sn2+ becomes Sn4+ losing 2 electrons and becoming more positive)
H2O2 is oxidizing agent because it undergoes reduction (H2O2 becomes H2O gaining 2 electrons and becoming more negative)
b. Hg is reducing agent because it undergoes oxidation (Hg becomes Hg22+ losing 2 electrons and becoming more positive)
PbO2 is oxidizing agent because it undergoes reduction (PbO2 becomes Pb2+ gaining two electrons and becoming more negative)
c. Al is reducing agent because it undergoes oxidation (Al becomes Al3+ losing 3 electrons and becoming more positive)
Cr2O72− is oxidizing agent because undergoes reduction (Cr2O72− becomes Cr3+ gaining 3 electrons becoming more negative)
Q19.1.5
Which of the following elements is most likely to be used to prepare La by the reduction of La2O3: Al, C, or Fe? Why?
S19.1.5
In order to effectively reduce a material, the chosen compound must have enough force to initiate the reaction. According to the table of Standard Reduction Potentials, the larger the the E° value is for a given half reaction, the more energy it takes to reduce (or add electrons) to that molecule. Compounds that are difficult to reduce make good Reducing agents, because they have the power to Oxidize other materials they react with. The same rule works in reverse too, half reactions with low E° values are easy to reduce, and make good Oxidizing agents.
Therefore, a compound on the table with a high E° value, has the power to reduce anything with a lower E° value than itself, and anything on the table with a lower E° value will Oxidize anything else with an E° value higher than itself.
Looking at the example given, if we were to try to reduce a molecule of La2O3 (which is essentially La3+) into La, we would want to use a reducing agent with a low enough E° value to make it happen. According to the table, we see that:
La3+ + 3e−v⇌ La(s) has an E° value of –2.38v
We can then find the three potential reducing agents and compare their E° values.
Al3+(aq) + 3e- ⇌ Al(s) E° = –1.676
Fe2+(aq) + 2e- ⇌⇌ Fe(s) E° = -0.44
CO2(g) + 2H+ + 2e−⇌⇌ CO(g) + H2O(l) E° = –0.106
Looking at all of three options, we can clearly see that Al has the lowest E° value, and would be the best Reducing agent for this reaction.
Note: Be careful though, because sometimes the table provided records Standard Oxidation Potentials rather than Reduction Potentials, in which case you would need to reverse the hierarchy of all values provided. (strong Reducers are weak Oxidizers, and vice versa)
Correction: Standard Oxidation potentials are the opposite of standard reduction potentials. It originally said that SOP were opposites of SOP.
Q19.2.7
Name each of the compounds or ions given in Exercise Q19.2.5. (seen below)
a. [Co(en)2(NO2)Cl]+
b. [Co(en)2Cl2]+
c. [Pt(NH3)2Cl4]
d. [Cr(en)3]3+
e. [Pt(NH3)2Cl2]
S19.2.7
See the Nomenclature of Coordination Complexes page for more information and specific naming exceptions.
Transition Metal Nomenclature:
Coordination complexes are made up of a single transition metal surrounded by multiple ligands (usually 4 or 6 atoms or molecules) covalently bonded to the central atom. When written, the coordination complex is written in square brackets [] usually expressing the transition metal first, followed by the ligands. If the complex is an ion, then a charge will appear outside the brackets, and if the coordination complex is bonded with a counter ion, then that ion will appear in parenthesis (). Whichever of the two ions is positively charged is the one that appears first.
examples:
1. [Fe(OH2)6]+ has an Iron central transition metal atom, 6 OH2 ligands, and an overall charge of +1
2. (NH4)2[CoCl4] has a Cobalt central transition metal atom, 4 Cl ligands, and is bonded to two positively charged ammonium ions
Naming Conventions:
Reading the expression from left to right, naming the positive complex first followed by the negative. The ion is named according to normal ion nomenclature laws. When writing the coordination complex name however, write the ligand names first (appearing in alphabetical order if there are more than one different kind of atom or molecule bonded to the central atom) including any prefixes to express number as needed (ie. mono-, di-, tri-, tetra-, penta-, and hexa-). See the above link for common ligand root names. Then name the transition metal. If the coordination complex has an overall negative charge, then add the ending -ate to the transition metal (ie. Nickel becomes Nickelate), followed by the charge of the transition metal in roman numerals in parentheses ().
examples:
1. k4[Ni(CN)4] would be potassium tetracyanonickelate (0)
2. [Fe(ox)3]-3 would be trioxalateferate (III) ion
3. [Pt(NH3)2]Cl2 would be diamineplatnum (II) Chloride
Solutions:
a. [Co(en)2(NO2)Cl]+ chlorobisethylenediaminenitrocobalt (III) ion
b. [Co(en)2Cl2]+ dichlorobisethylenediaminecobalt (III)
c. [Pt(NH3)2Cl4] diamminetetrachloroplatinum (IV)
d. [Cr(en)3]3+ trisethalinediaminechromium (III)
e. [Pt(NH3)2Cl2] diamminedichloroplatinum (II)
Phase 2: I agree with all of their solutions. I liked how they added examples in beginning to show how to name the compounds.
Q12.3.20
The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6.2 × 10−4 min−1
2N2O5 ⟶ 4NO2 + O2
What is the rate of the reaction when [N2O5] = 0.40 M?
S12.3.20
With the knowledge that all decomposition reactions are first order, we can complete this calculation rather quickly. Using the First Order rate law equation:
\(\mathrm rate = k[A]^1 \)
where the rate of the reaction is equal to the rate constant (k) multiplied by the molarity of the substance to the first power, we can plug in our numbers.
rate = 6.2 × 10−4 min−1[0.40M]
rate = 2.48 x 10-4 mol/L/min
This tells us that the decomposition reaction of dinitrogen pentoxide in chloroform takes place at a rate of 2.48 x 10-4 mol/L every minute.
Q12.6.11
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:
- Cl2(g) ⇌ 2Cl(g) (fast, k1 represents the forward rate constant, k−1 the reverse rate constant)
- CO(g) + Cl(g) ⟶ COCl(g) (slow, k2 the rate constant)
- COCl(g) + Cl(g) ⟶ COCl2(g) (fast, k3 the rate constant)
- Write the overall reaction.
- Identify all intermediates.
- Write the rate law for each elementary reaction.
- Write the overall rate law expression.
S12.6.11
Scientists make reaction mechanisms as forms of hypotheses to explain the order and details of how a specific reaction occurs. The rates of these reactions can be determined in the lab, but there are sometimes intermediate steps that can slow down the reaction that must be accounted for.
1. In order to write the overall reaction when given the individual steps, we must combine them all and cancel any intermediates that appear on both the reactant and product sides.
Reaction 1 Cl2(g) ⇌ 2Cl(g)
Reaction 2 CO(g) + Cl(g) ⟶ COCl(g)
Reaction 3 COCl(g) + Cl(g) ⟶ COCl2(g)
Overall reaction: Cl2(g) + 2Cl(g) + CO(g) + COCl(g) ⟶ 2Cl(g) + COCl(g) + COCl2(g)
Simplified overall: Cl2(g) +CO(g) ⟶ COCl2(g)
2. Intermediates are any molecules in the reactions that appear in the first three elementary reactions but do not appear in the final overall reaction because they become canceled out. In this example, the intermediates are; Cl(g) and COCl(g)
3. In order to write the rate laws of each individual elementary reaction, multiply the specific step's rate constant, by the multiples of the Molarities of each reactant according to the rules of determining rate laws. Note that there are multiple different classifications of elementary reactions depending on the number of reactants;
Unimolecular: reactions with only one reactant
Bimolecular: reactions with two reactants
Termolecular: reactions with three different reactants
Note: it is highly improbable that there will be any more than three separate atoms naturally colliding at the same time, therefore the rate constants are not calculated for 4+ different molecule collisions.
For this problem, we can write the rate laws by multiplying constant k by the molarities. Notice that reaction 1 is Unimolecular while reactions 2 and 3 are Bimolecular.
|
Elementary Reaction 1 Cl2(g) ⇌ 2Cl(g) rate = k1[Cl2] rate = k-1[Cl]2 |
Elementary Reaction 2 CO(g) + Cl(g) ⟶ COCl(g) rate = k2[CO][Cl] |
Elementary Reaction 3 COCl(g) + Cl(g) ⟶ COCl2(g) rate = k2[COCl][Cl] |
4. To write the overall rate law expression for these three consecutive elementary reactions, we must combine their individual rate laws, but without the intermediates.
In order to do this, we need to solve for the intermediate, [Cl], using other rate law expressions
k1 [Cl2] = k-1 [Cl]2
= k1 [Cl2] = k-1 [Cl]2
= k1 [Cl2]/k-1 = [Cl]2
= ( k1 [Cl2]/k-1 )1/2 = [Cl]
then substitute into the second rate equation
k2 [CO][Cl]
= k2 ( k1 [Cl2]/k-1 )1/2 [CO]
= k [Cl2]1/2 [CO]
Q21.4.23
Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 109 years ago?
S21.4.23
This problem focuses on our knowledge of half lives. Many naturally forming substances on Earth will slowly decay away through nuclear radiation, forming other more stable elements over time. This is the logic behind the science of Carbon dating to determine the age of certain collectables based on the quantity of C-14 remaining in the item.
This question, however, is asking about the decay of U-238 into Pu-239. With the knowledge that the half life of Pu-239 is 24,100 years, we can plug values into the floowing equation.
[A]t = [A]0e-kt
By plugging in arbitrary numbers such as 100 and 50 for the quantity of pu-239 sample remaining after the first half life (24100) years has elapsed, we can determine the k constant.
[50] = [100]e-k(24100)
Then through simple algebra, we can determine the variable k;
k = ln(2)/24100
which we can now plug back into the original equation to get;
[A]t = [A]0e-(ln(2)/24100)t
By plugging in a sample size and the age of the solar system... we can determine if it is possible for there to still be trace amounts of Pu-239 remaining on earth from the beginning of the solar system.
[A]4.7 × 10^9 years = [100]e-(ln(2)/24100) x (4.7 × 10^9 years)
We can then calculate that the remaining quantity of Pu-239 would be...
[A]4.7 × 10^9 years = 0
Meaning, that it would be impossible for any Pu-239 created from the formation of the Earth to be remaining for the scientists to discover in the 1940's. Concequently, any plutonium now present could not have been formed with the uranium.
Phase 2: correct
Q20.3.11
Sulfate is reduced to HS− in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?
S20.3.11
We know that Sulfate (SO42−) is reduced, which means that it becomes more negatively charged, in this case, through the process of creating water. Follow the rules of Writing Equations for Redox Reactions, adding first water to balance oxygen, then H+ to balance Hydrogen, then electrons (e-) to balance charge for the first half reaction.
reduction half reaction: SO42−(aq) + 9H+(aq) + 8e− → HS−(aq) + 4H2O(l)
Now, complete the same process for the oxidation of glucose (C6H12O6) into bicarbonate (HCO3−) to form the oxidation half reaction.
oxidation half reaction: C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq) + 24e−
In order to combine these two half reactions into a single overall reaction, the electrons must be able to evenly cancel. In this case, the reduction reaction has 8 electrons and the oxidation reaction has 24, so multiply the first reaction by the common multiple of 4.
4(SO42−(aq) + 9H+(aq) + 8e− → HS−(aq) + 4H2O(l)
4SO42−(aq) + 36H+(aq) + 24e− → 4HS−(aq) + 16H2O(l)
Now we can line up both equations and cancel anything that appears on both sides of the equation to achieve the final overall equaiton.
reduction: 4SO42−(aq) + 36H+(aq) + 24e− → 4HS−(aq) + 16H2O(l)
Oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq) + 24e−
Overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3−(g) + 3H+(aq) + 3HS−(aq)
Phase 2 correct
Q20.5.22
Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:
Mn3+(aq) + e− → Mn2+(aq) E°=1.51 V
Mn3+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + e− E°=0.95 V
- What is E° for the disproportionation reaction?
- Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
- How could you prevent the disproportionation reaction from occurring?
S20.5.22
Disproportionation reactions are a sub category of Oxidation-Reduction Reactions, where a single element from the reaction undergoes both Oxidation and Reduction to form two completely different products.
1. E°cell = E°cathode - E°anode
= 1.51 - 0.95 = 0.56v
2. Disproportionation is more thermodynamically favored at a pH less than 7 because as the pH increases the forward reaction will slow down from the increase in H+ ions
3. One way to prevent disproportionation from occurring is by removing water from the reaction.