Extra Credit 44
- Page ID
- 82753
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Q17.6.3
If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.
S17.6.3
The standard reduction potential (SRP) of metals indicates how likely it is to be reduced. The more positive the SRP in comparison to another, the more likely that the metal will be reduced and serve as the cathode in a reaction. Similarly, the more negative the SRP in comparison to another, the more likely that the metal will be oxidized and serve as the anode in a reaction. Corrosion is a form of oxidation. Both of these cases involve a cathodic protection and sacrificial anode, where the metal with the more negative SRP will sacrifice itself and corrode first. In the first case, iron has an SRP of −0.447 V and zinc has an SRP of −0.7618 V. Therefore, zinc serves as the sacrificial anode and corrodes, protecting the iron. In the second case, iron has an SRP of −0.447 V and copper has an SRP of 0.34 V. Therefore, iron serves as the sacrificial anode and corrodes, protecting the copper.
*just reworded some things to made it easier to understand
Q12.3.7:
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:
\[_{15}^{32}P⟶^{32}_{16}S+e^{−1}\]
\[Rate = 4.85×10^{-2} day^{−1}[^{32}P]\]
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?
S12.3.7:
The rate given indicates the rate constant of the reaction. Since the units are in \(day^{-1}\), this indicates that the rate is a first order reaction.
The rate law is given: \(4.85×10^{-2} day^{−1}[^{32}P]\).
Plugging in the given concentration yields \(4.85×10^{-2} day^{−1}[0.0033]\), which simplifies to rate = \(1.601×10^{-4}\).
The instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M is \(1.601×10^{-4}\).
* separated the steps to make it easier to read
*fixed a few of the equations
Q12.5.16
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A+BC⟶AB+C reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
S12.5.16:
As the amount of energy that the A atom is shot with varies, the success of the reaction varies. When the Total Energy line at launch is below the transition state of the Potential Energy line, the reaction cannot proceed because the A atom does not hit the BC molecule with enough energy to overcome the activation energy and reach the transition state. When the when the Total Energy line at launch is above the transition state of the Potential Energy line, the A atom hits the BC molecule with enough energy to overcome the activation energy and reach the transition state, and therefore the reaction proceeds. Furthermore, the reverse reaction also proceeds since the Total Energy of the corresponding system is still high enough to overcome the activation energy and reach the transition state.
*no corrections needed
Q21.4.11
Write a nuclear reaction for each step in the formation of \(_{84}^{218}Po\) from \(_{92}^{238}U\), which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order.
S21.4.11
Alpha particles are a nucleus with 2 protons and an atomic mass of 4, written as \(_2^4He\). Beta particles are either electrons or positrons. Since the question does not specify, it is assumed to be an electron, written as \(_{-1}^0e\). It is important to make sure that the mass numbers (the number in the top left) and the atomic numbers (the number in the bottom left) are equal on both sides of the reaction. The mass number and atomic number of the missing element are determined by subtracting the numbers of the emitted particle from the initial element, and the atomic number that results designates which element is produced.
In the first reaction, an alpha particle is emitted. Therefore, the first step of the overall reaction is as follows: \(_{92}^{238}U⟶_{90}^{234}Th+_2^4He\).
In the second reaction, a beta particle is emitted. Therefore, the second step of the overall reaction is as follows: \(_{90}^{234}Th⟶_{91}^{234}Pa+_{-1}^0e\).
In the third reaction, a beta particle is emitted. Therefore, the third step of the overall reaction is as follows: \(_{91}^{234}Pa⟶_{92}^{234}U+_{-1}^0e\).
In the fourth reaction, an alpha particle is emitted. Therefore, the fourth step of the overall reaction is as follows: \(_{92}^{234}U⟶_{90}^{230}Th+_2^4He\).
In the fifth reaction, an alpha particle is emitted. Therefore, the fifth step of the overall reaction is as follows: \(_{90}^{230}Th⟶_{88}^{226}Ra+_2^4He\).
In the sixth reaction, an alpha particle is emitted. Therefore, the sixth step of the overall reaction is as follows: \(_{88}^{226}Ra⟶_{86}^{222}Rn+_2^4He\).
In the seventh reaction, an alpha particle is emitted. Therefore, the seventh step of the overall reaction is as follows: \(_{86}^{222}Ra⟶_{84}^{218}Po+_2^4He\).
*no corrections needed
Q20.2.15
Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:
S20.2.15
- \(Pt^{2+}_{(aq)} + Ag_{(s)} → \): No Reaction
- \(HCN_{(aq)} + NaOH_{(aq)} → NaCN_{(aq)}+ H_2O_{(l)}\): Acid-Base Reaction
The second reaction is an acid base reaction because HCN is a strong acid an NaOH is a strong base. NaCN is soluble in pH 7, and since the reactants are strong acids and bases, the overall pH will be 7.
3. \(Fe(NO_3)_{3_{(aq)}} + 3NaOH_{(aq)} → Fe(OH)_{3_{(s)}} + 3NaNO_{3_{(l)}}\): Precipitate Reaction
The third reaction is a precipitate reaction because \(Fe(NO_3)_3\) is a weak acid and NaOH is a strong base. \(Fe(OH)_3\) is not soluble, so it precipitates out of the solution.
4. \(CH_{4_{(g)}} + O_{2_{(g)}} → CO_{2_{(g)}} + 2H_{2_{(g)}}\): Redox Reaction
The fourth reaction is a redox reaction because the oxidation numbers change. The initial oxidation numbers are as follows: For \(CH_4\), H has an oxidation number of +1, and since the entire molecule is neutral overall, C must have an oxidation number of -4. For \(O_2\), the oxidation number of both oxygen atoms are 0. \(CO_2\) has overall charge of 0 and O typically has an oxidation number of -2, so C must have an overall charge of +4. Finally, each H in \(H_2\) has an overall charge of 0. Since the oxidation numbers of each element changed, it is a redox reaction.
*rearranged the set up to make a little more sense and look cleaner.
Q20.5.10
Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.
S20.5.10
Given the Nernst Equation \(E^o_{cell} = {RT\over nF}lnK\), a cell at equilibrium that contains a sparingly soluble salt as an anode and cathode can be used to find the \(E^o_{cell}\), which can then be converted to \(K_{sp}\).
For example, consider a cell in which the cathode is a silver wire in a 1.0M \(Cl^-\) solution that is saturated with AgCl and the anode is a silver wire in a 1.0M \(Ag^+\) solution. The cell potential will be the difference in the concentration of \(Ag^+\), which can be substituted into the Nernst Equation to find the \(K_{sp}\).
*fine as is
Q24.6.6
Do strong-field ligands favor a tetrahedral or a square planar structure? Why?
S24.6.6
Strong field ligands have a greater \(∆_o\), meaning that the energy required to fill electrons on the next energy level is more than that of a weaker field ligand. Electrons will seek to fill the position where the cost of energy is lowest. For weak field ligands, the \(∆_o\) < P, the spin pairing energy, and therefore electrons will structure themselves in a high spin configuration. However, for strong field ligands, the \(∆_o\) > P, and therefore electrons will structure themselves in a low spin configuration. Since tetrahedral has two orbitals at low energy and three at medium energy, the strong fields will favor a tetrahedral structure over square planar, which has two orbitals at low energy, one at medium energy, one at high energy, and one at really high energy. Therefore, as orbitals fill up, the \(∆_o\) for the square planar structure will be greater than that of tetrahedral structure, and as a result, strong-field ligands will favor a tetrahedral structure.
*no corrections needed
Q14.7.11
A particular reaction was found to proceed via the following mechanism:
- A + B → C + D (slow)
- 2C → E (fast)
- E + A → B + F (fast)
What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates.
S14.7.11
The overall reaction is given by 2A + C → D + F. This is found by adding together all the steps in the mechanism:
\[A + \cancel{B} + 2C + \cancel{E} + A → C + D + \cancel{E} + \cancel{B} + F\]
The B and E cancel out, one of the Cs cancel out, and the two As add up to produce 2A + C → D + F as the overall reaction.
The intermediates for this reaction are C and E because they are produced in one of the initial reactions and are consumed in a later reaction.
This reaction is catalytic, with B as the catalyst. This is because it is a reactant in the first step and a product in the last step, meaning it remains unchanged throughout the reaction.
*fixed one of the equations, otherwise no additional corrections needed