Extra Credit 46
- Page ID
- 82960
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Q17.6.5
Why would a sacrificial anode made of lithium metal be a bad choice despite its E∘Li+/Li=−3.04VELi+/Li∘=−3.04V, which appears to be able to protect all the other metals listed in the standard reduction potential table?
S17.6.5
Lithium is an alkali metal and alkali metals are very reactive. This means that lithium would have a reaction with many different substances and the lithium anode would be depleted very quickly. These seems inefficient because while lithium will attempt to protect a substance it will end up protecting other substances not involved in oxidation. This will lead to having to replace the lithium anode frequently which seems like a waste of lithium. Another short reason is if water happens to react with lithium a small fire can form from the reaction.
Part II Agreement (Vincent Raggio)
Sacrificial Anodes are created from a metal alloy with a more negative electrochemical potential than the other metal it will be used to protect meaning that if a more negative electrochemcial metal was introduced into a galvenic cell, it would begin to undergo oxidation rather than the less electrochemical metal. Because oxidation leads to the disintegration of a metal, the more negative electrochemical metal "sacrifices" itself to protect the other metal. However, the reason lithium would not be a good sacrificial anode because of its' reactivity. Being a alkali metal means that it reacts with a large number of molecules and it could react with the solution of the galvenic cell.
Q12.3.9
What is the instantaneous rate of production of N atoms Q12.3.8 in a sample with a carbon-14 content of 1.5 × 10−9 M?
S12.3.9
K=1.21 x 10-4 year-1
Rate =1.21 x 10-4 year-1 [6.5 × 10-9 M]
Rate=7.9 × 10-13 mol/L/year
Part II Revision (Vincent Raggio)
In Q 12.3.8, the rate law and the rate constant is provided for the emission of a beta particle from \(_{6}^{14}C\). The rate law is \({rate} = k[_{14}^6C]\) and the rate constant, k, equals \(1.21x10^{-4} year^{-1}\).
From there you can plug in your value for k and the provided concentration of 1.5x10-9 M
\[{Rate} = 1.21x10^{-4} * [1.5x10^{-9}] = 1.82x10^{-13} \]
Also make sure to have proper units which in this case would be M/year.
Q12.6.1
Why are elementary reactions involving three or more reactants very uncommon?
S12.6.1
An elementary reaction involving three or more reactants would be termed as a Termolecular reaction. In these reactions three reactants must collide, however; the reason theses reactions are so rare is because when these reactants collide they must do so with enough energy and proper orientation to make a reaction happen.
Part II Agreement (Vincent Raggio)
Elementary reactions are broken down steps of an equation which must add up to the original reaction. Elementary reactions exhibit molecularity which is defined as the definite molecular encounter during the course of a reaction or more simply, the number of molecules colliding in one step. More reactions involve one or two molecules, termed unimolecular or bimolecular, but an elementary step with three reactant molecules is known as termolecular and is exceeding rare. In order for a reaction to take place the molecules must collide with an sufficient amount of energy (activation energy) and must have the correct orientation to each other. These two components are expressed in Collisional Theory which dissects the rate law into 3 distant categories:
[Fraction of molecules with required orientation]x[Fraction of collisions with required energy]x[Collision frequency (number of collisions per sec)]= Rate.
Because of these intense and multiple requirements for a successful reaction, termolecular reaction which have more variables to address, occur in extremely rare circumstances.
Q21.4.13
Define the term half-life and illustrate it with an example.
S21.4.13
Half life is the time needed for the atoms in a radioactive sample to decay to half of the original amount. An example would be if you take the half life of krypton (Kr) which is 10,756 years and started with 3.00-g sample of Kr after 10,756 years have passed there would only be 1.50g left from the original sample due to radioactive decay.
Part II Agreement (Vincent Raggio)
A Half-Life is the time required for the concentration of a reactant to decrease to one-half its initial value. All half lives are 1st order reactions meaning that the half life can be defined as \({t_{1/2}} =\dfrac{.693}{k}\) where k is the rate constant. An example of a half life would be if a sample of 5.00 grams of \(_{92}^{238}U\) decayed for 4.468x109 years, only 2.50 grams of it would remain.
Q20.3.1
Is 2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l) an oxidation–reduction reaction? Why or why not?
S20.3.1
The reaction 2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l) is not an oxidation-reduction reaction. This reaction is instead an acid-base neutralization reaction.
Part II Agreement (Vincent Raggio)
\[2NaOH_{(aq)} + H_{2}SO_{4(aq)} \rightarrow Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)}\]
This reaction is not an oxidation-reduction reaction reaction because there is no change in oxidation states due to the lack of transferring of electrons. This reaction exhibits the characteristics of a acid-base neutralization reaction because of the substitution of ions from the base to the acid and the formation of water.
Q20.5.12
How many electrons are transferred during the reaction Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg2Cl2spontaneous? Calculate ΔG° for this reaction.
S20.5.12
Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)
Oxidation: Pb2++2e-→Pb
Reduction: Hg2Cl2+2e-→2Hg+2Cl-
Eocell=Eo(Cathode)-Eo(Anode)
Eocell=.27-(-.13)=.4
ΔG°=-nFEocell
ΔG°=-2(96486)(.4)
ΔG°=-77.189kJ/mol
Reaction is spontaneous. 2 electrons are transferred.
Part II Agreement (Vincent Raggio)
\[Pb_{(s)} + Hg_{2(s)}Cl_{2} \rightarrow PbCl_{2(aq)} + 2Hg_{(l)}\]
The oxidation reaction is \(Pb_{(s)} \rightarrow Pb^{2+} + 2e^- \) because when looking at the oxidation states \(Pb_{(s)}\) goes from neutral to 2+ meaning it lost electrons.
The reduction reaction is \(Hg_{2}Cl_{2(s)} + 2e^- \rightarrow 2Hg_{(l)} + 2Cl^{-} \) because when looking at the oxidation states of \(Hg_{2}Cl_{2(s)}\) the Hg falls from a 2+ to neutral meaning it gained electrons.
To find the standard cell potential, you find the cell potential from the Cell Potential Table and then plug them into the reaction \(E°_{cell} = E°_{cathode} − E°_{anode} \). (Reduction happens at the cathode and Oxidation occurs at the Anode).
When looking at the corresponding table, we find that for \(Pb_{(s)} \rightarrow Pb^{2+} + 2e^- \) V=-.13 and for \(Hg_{2}Cl_{2(s)} + 2e^- \rightarrow 2Hg_{(l)} + 2Cl^{-} \) V=.27.
Next plug them into the equation to find Eocell
\[.27 + .13 = .4 = E^o_{cell}\]
After we have Eocell, we plug the value into the Nernst Equation under standard conditions which is \(ΔG°=-nFE_{cell}\) where (\n\) is the number of electrons transferred in the full reaction and F is Faraday's constant:
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214\times10^{23}}{\textrm{1 mol e}^-}\right) \\&=9.64855\times10^4\textrm{ C/mol e}^-\simeq 96,486/(\mathrm{V\cdot mol\;e^-})\end{align}
To find the number of electrons transferred, combine the two half reaction and cancel out the electrons.
\[Pb_{(s)} \rightarrow Pb^{2+} + 2e^- \]
\[Hg_{2}Cl_{2(s)} + 2e^- \rightarrow 2Hg_{(l)} + 2Cl^{-} \]
In this equation, we can see that 2 electrons are on each side of the half reactions meaning they can cancel
\[Pb_{(s)} + Hg_{2(s)}Cl_{2} \rightarrow PbCl_{2(aq)} + 2Hg_{(l)}\]
And we have our \(n\) value which equals 2.
Finally plug into the Nernst Equation under standard conditions \(ΔG°=-nFE_{cell}\)
\[ΔG°=-(2)*(96,485)*(.4) = -77.189\]
Again DO NOT FORGET PROPER UNITS. In this case they would be kJ/mol.
Finally because ΔG° is a negative value we know that the equation of \(Pb_{(s)} + Hg_{2(s)}Cl_{2} \rightarrow PbCl_{2(aq)} + 2Hg_{(l)}\) is spontaneous.
Q24.6.8
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
- [Cu(NH3)4]2+
- [Ni(CN)4]2−
S24.6.8
a. [Cu(NH3)4]2+
b. [Ni(CN)4]2−
a. Cu2+→ 27 Electrons→ [Ar]3d9
One unpaired electron. Square planar. Neither low spin nor high spin.
b. Ni2+→ 26 electrons→ [Ar]3d8
No unpaired electrons. Square planar. Low spin.
Part II Revision (Vincent Raggio)
To figure out the number of unpaired electrons you first must look at the charge of the ligands in the complex molecules. In this molecule, ammine or NH3 is a neutral ligand meaning the charge solely belongs to Cu making it Cu2+. Next, we must look at the Periodic Table and find the electron configuration of Cu. Cu is a unique molecule with an unfilled 4s shell and full 3d shell making it [Ar] 4s13d10. When it loses 2 electrons, it becomes [Ar]3d9. Because it has 4 NH3 which is a monodentate ligand, the structure of the molecule is either tetrahedral or square planar. To find the correct structure we must draw the Crystal Field Splitting Diagram for each and see which one has the lower energy:
vs.
Because the square planar has lower energy, the square planar structure makes the most sense.
Finally, to predict which it is high or low spin, we must look at the strength of the ligands. However, square planar cannot be high or low spin and once we fill up the Crystal Field Diagram, we see that there is only one unpaired electron.
[Ni(CN)4]2−
To figure out the number of unpaired electrons you first must look at the charge of the ligands in the complex molecules. In this molecule, CN- is a negatively charged ligand meaning the charges interact and form the charge of the molecule. Because there is a 2- for the molecule and 4 negative CN- we see that Ni has a 2+ charge. Next, we must look at the Periodic Table and find the electron configuration of Ni. Ni is [Ar] 4s23d8. When it loses 2 electrons, it becomes [Ar]3d8.
To find the structure, we do the same procedure as the first molecule (CN- is also monodentate) and we find that it is also square planar because of the lower net energy.
There are no unpaired electron because CN- is a strong field ligand meaning it have a high Δsp which means that the electrons will fill the lower orbitals first which makes it a low spin structure.
Q14.7.12
A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.
S14.7.12
With or without a catalyst the product of A will always be favored. Since B is reversible all the product Z that is formed will eventually be turned into X and this will be converted to Y in the first reaction.
Part II Agreement (Vincent Raggio)
In both cases, the product A will be produced. This is the case because Z will eventually be converted to X if it is produced with the catalyzed reversible pathway B. When it converts back to X, it will then convert to A through the uncatalyzed reaction.