Extra Credit 44

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

This process can be explained by using the reduction potential. Corrosion occurs when a metal is oxidized (losing electrons). If we look at the chart, we see that copper has a potential of 0.34V, iron with a potential of -0.45V, and zinc with a potential of -0.76V. This means that out of the three, copper will least likely be oxidized while zinc will most likely be oxidized.

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:

\(^{32}_{15}P\rightarrow^{32}_{16}S+e^{-}\)

\(Rate ={4.85\times10^{-2}}day^{-1}\) \([^{32}P]\)

What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?

S12.3.7

This reaction is the first order because the units of the rate constant (k) are \(day^{-1}\). This means we can calculate the instantaneous rate by plugging in 0.0033 M for \([^{32}P]\). We get \(1.601\times10^-4 \scriptsize\frac{M}{day} \) as our answer.

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

If the total energy is below the transition state, the A atom does not have enough force to pass through the activation energy bump. However, if the total energy is above the transition state, the A atom will have enough force to move past the activation energy bump.

Q21.4.11

Write a nuclear reaction for each step in the formation of \(^{218}_{84}Po\) from \(^{238}_{98}U\) , 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

1α) \(^{238}_{92}U\rightarrow^{234}_{90}Th+^{4}_{2}He\)

2β) \(^{234}_{90}Th\rightarrow^{234}_{91}Pa+^{0}_{-1}e\)

3β) \(^{234}_{91}Pa\rightarrow^{234}_{92}U+^{0}_{-1}e\)

4α) \(^{234}_{92}U\rightarrow^{230}_{90}Th+^{4}_{2}He\)

5α) \(^{230}_{90}Th\rightarrow^{226}_{88}Ra+^{4}_{2}He\)

6α) \(^{226}_{88}Ra\rightarrow^{222}_{86}Rn+^{4}_{2}He\)

7α) \(^{222}_{86}Rn\rightarrow^{218}_{84}Po+^{4}_{2}He\)

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:

a. \(Pt^{2+}(aq)\) + \(Ag(s)\rightarrow\)

b. \(HCN(aq) + NaOH(aq)\rightarrow\)

c. \(Fe(NO_3)_3(aq)\) + \(NaOH(aq)\rightarrow\)

d. \(CH_4(g)\) + \(O_2(g)\rightarrow\)

S20.2.15

a. \(Pt^{2+}(aq)\) + \(Ag(s)\rightarrow\) \(Ag^{2+}(aq)\) + \(Pt(s)\) is a redox reaction because \(Pt^{2+}\) is reduced to Pt while Ag is oxidized to \(Ag^{2+}\).

b. \(HCN(aq) + NaOH(aq)\rightarrow\) \(H_2O(l) + NaCN(aq)\) is an acid base reaction because HCN is an acid (weak acid) while NaOH is a base (strong base).

c. \(Fe(NO_3)_3(aq)\) + \(3NaOH(aq)\rightarrow\) \(Fe(OH)_3(s) + 3NaNO_3(aq)\) is a precipitation reaction because the reactants form \(Fe(OH)_3(s)\), a solid .

d. \(CH_4(g)\) + \(2O_2(g)\rightarrow\) \(CO_2(g) + 2H_2O (g)\) is a redox reaction because oxygen is reduced to water while methane is oxidized to carbon dioxide.

Q20.5.10

Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.

S20.5.10

To find the solubility of a salt, we can use the Nernst equation, \(E^{o}_{cell}=\frac{RT}{nF}lnK\). We can calculated \(E^{o}_{cell}\) by subtracting the anode from the cathode. After that, can plug in constant values to solve for K (solubility).

The variables for the Nernst equation are as follows:

• \(E^{o}_{cell}\) = \(E^{o}_{Cathode}\) - \(E^{o}_{Anode}\)
• R (Gas Constant) = 8.3145 \(\frac{J}{mol K}\) or 0.08820574 \(\frac{L atm}{mol K}\) (Depending on units)
• T = Temperature in Kelvin
• n = Number of electrons transferred in the reaction
• F (Faraday's Constant)= 96,485 \(\frac{J}{V mol}\)
• K = Equilibrium (In this case the solubility)

Q24.6.6

Do strong-field ligands favor a tetrahedral or a square planar structure? Why?

S26.6.6

Strong-field ligands favor tetrahedral structures. Tetrahedral structures are divided into two rows while square planar structures are divided into four shorter rows. Having a strong-field ligand means that the structure will be low spin and each row must be filled with electron pairs before moving onto the next row. With a square planar structure, electrons can fill more rows and leave less unpaired electrons than in a tetrahedral structure.

Q14.7.11

A particular reaction was found to proceed via the following mechanism:

What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates.

- A + B -> C + D (slow)

- 2C -> E (fast)

- E + A -> 3B + F (fast)

Q14.7.11

The overall reaction is 2A + C -> D + F.

This reaction is catalytic because the addition of one (in this case B), speeds up the rate of a chemical reaction without being consumed by the reaction. It can also reappear as a product. Therefore, B is the catalyst in this reaction.

Intermediates are produced as products but are then consumed by the reaction, so they do not appear in the end as products. C and E are intermediates because C is a product from the first reaction but is consumed in the second reaction. Likewise, E is produced in the second reaction but is consumed by the third reaction. D, however, is not an intermediate because while it is produced as a product in the first reaction, it is not consumed by the other reactions and appears in the overall reaction.