Extra Credit 44
- Page ID
- 82855
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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.
S12.3.7
Iron has a higher reduction potential (-o.447) than zinc (-0.7618), therefore it is a better oxidizing agent and zinc is a better reducing agent. Another way to think of it is that iron is the cathode because it has a more positive reduction potential, and zinc is the anode because it has a more negative reduction potential. The cathode, iron, reduces and the anode, zinc, oxidizes. As a result, zinc is the sacrificial anode that oxidizes faster than iron. Similarly, iron has a more negative reduction potential than copper, making it the sacrificial anode and corroding before copper.
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
1. The units of a 1st order reaction is \(sec^{-1}\) (or any time period, such as minutes, hours, or seconds). This means the units given in the problem are also 1st order because our units are \(day^{-1}\).
2. Plug in the given concentration, [0.0033], into the given rate equation: \(Rate = 4.85x10^{-2} day^{-1}[^{32}P]\). The exponent is 1 because it is a first order reaction.
3. Calculate. \(Rate = 4.85x10^{-2} day^{-1}[0.0033]\) = \(1.601x10^{-4}\).
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
The success of the reaction varies in relation to the potential power that the "A" atom is shot with. The reaction is unable to continue if the Total Energy line is below the transition state of the Potential Energy line. This is because the "A" atom doesn't hit the "BC" molecule with enough energy to surpass the activation energy to successfully reach the transition state. However, when the Total Energy line is above the transition state of the Potential Energy Line at launch, the "A" atom is able to collide with the "BC" molecule, allowing the reaction to continue. The reverse reaction also continues because the Total Energy is high enough to surpass the activation energy, reaching the transition state.
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
1. Subtract mass # and atomic # of the \(\beta\) particle (written as \(_{-1}^0e\)) or \(\alpha\) particle (written as \(_2^4He\)) emitted. In the first step, an \(\alpha\) particle is emitted, so subtract 4 from the mass # and atomic # of the original element.
New mass #: mass# of original element - mass # of \(\alpha\) particle \(\rightarrow\) 238-4=234
New atomic #: atomic # of original element - atomic # of \(\alpha\) particle \(\rightarrow\) 92-2=90
2. Look at the new atomic # (in this case 90) and find the corresponding element that correlates to that number.
atomic #:90 \(\rightarrow\) Th
3. Rewrite the product with new mass # and atomic #.
\(_{92}^{238}U⟶_{90}^{234}Th+_2^4He\)
4. The next nuclear reaction is with a beta particle, \(_{-1}^0e\). Use the same steps as above.
New mass #: mass# of original element - mass # of \(\beta\) particle \(\rightarrow\) 234-(0)=234
New atomic #: atomic # of original element - atomic # of \(\beta\) particle \(\rightarrow\) 90-(-1)=91
The new element (the daughter cell/new formed element) that corresponds to the new atomic number of 91 is Pa. Therefore rewrite the next equation with the new parent cell as \(_{91}^{234}Pa\)
The same rules as above can be applied to the rest of the problems:
\(_{90}^{234}Th⟶_{91}^{234}Pa+_{-1}^0e\)
\(_{91}^{234}Pa⟶_{92}^{234}U+_{-1}^0e\)
\(_{92}^{234}U⟶_{90}^{230}Th+_2^4He\)
\(_{90}^{230}Th⟶_{88}^{226}Ra+_2^4He\)
\(_{88}^{226}Ra⟶_{86}^{222}Rn+_2^4He\)
\(_{86}^{222}Ra⟶_{84}^{218}Po+_2^4He\)
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
A. \(Pt^{2+}_{(aq)} + Ag_{(s)} → \): No Reaction
\(Pt^{2+}\) and \(Ag_{(s)}\) are 2 transition metals, therefore, they cannot react to form anything.
B. \(HCN_{(aq)} + NaOH_{(aq)} → NaCN_{(aq)}+ H_2O_{(l)}\): Acid-Base Reaction
When a weak acid and strong base react together, that forms an Acid-Base Reaction. In this case, \(HCN_{(aq)}\) is a weak acid and \(NaOH_{(aq)}\) is a strong base. In addition, water is a product which is one of the characterisitics of an acid base reaction.
C. \(Fe(NO_3)_{3_{(aq)}} + 3NaOH_{(aq)} → Fe(OH)_{3_{(s)}} + 3NaNO_{3_{(l)}}\): Precipitate Reaction
This reaction is a double replacement reaction. After finding the products, we know that \(3NaOH_{(aq)}\) is aqueous and \(Fe(OH)_{3_{(s)}}\) is insoluble due to the solubilty rules.
D. \(CH_{4_{(g)}} + O_{2_{(g)}} → CO_{2_{(g)}} + 2H_{2_{(g)}}\): Redox Reaction
This reaction is a redox reaction because it has the same pattern as a displacement redox reaction \(AB + C ⟶ A + CB\). Additionally, the the oxygen is being reduced because its oxidation state is becoming more negative, whereas the carbon is being oxidized because its oxidation state is increasing.
Q20.5.10
Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.
S20.5.10
A common procedure to find the solubility of a sparingly soluble salt, set up a cell in which one of the electrode reactions involves the insoluble salt and whose net cell reaction corresponds to the dissolution of the salt.
ex) Find the Ksp of silver iodide, AgI.
\(AgI_{(aq)} ⟶ Ag^+_{(aq)} + I^-_{(aq)}\)
\(Ksp = [Ag^+][I^-]\)
\(AgI_{(aq)} + e^- ⟶ Ag_{(s)}+I^-_{(aq)}\) \(E^o = -0.15\)
\(Ag_{(s)} ⟶ Ag^+_{(aq)}+e^-\) \(E^o = -0.80\)
\(Final {E^o} = -0.95 V\)
The Nerst Equation can then be used because the final \(E^o\) can be plugged in to solve for K using the equation: \(E^o_{cell} = {RT\over nF}lnK\).
Q24.6.6
Do strong-field ligands favor a tetrahedral or a square planar structure? Why?
S24.6.6
Strong field ligands have a large \(∆_o\) (splitting energy), therefore they prefer low spin configurations, meaning that electrons prefer to be paired. Square planar complexes have high \(∆_o\) and they tend to have low spin configurations (electrons are unpaired) whereas tetrahedral structures have small \(∆_o\) and therefore are high spin (electrons are paired). As a result, most strong field ligands are generally square planar.
See link below for more information:
https://chem.libretexts.org/LibreTexts/University_of_California_Davis/UCD_Chem_002C/UCD_Chem_2C%3A_Larsen/Chapters/Unit_2%3A_Coordination_Chemistry/21.08%3A_Bonding_in_Non-Octahedral_Complex_Ions%3A_Crystal_Field_Theory
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
1. Write out all the reactants and products in one equation:
\(A + B + 2C + E + A ⟶ C + E + D + B + F\)
2. Cancel out all spectator ions (duplicates that appear on both the products and the reactants):
\(A + \require{cancel} \cancel{B} + \cancel{2C} + \cancel{E} + A ⟶ \cancel{C} + \cancel{E} + D + \cancel{B} + F\)
3. Write final equation without spectator ions:
\(2A + C ⟶ D + F\)
Intermediates: C, E because they are produced, then consumed.
Catalyst: B because it is consumed, then produced.