Extra Credit 19

Q19.6A

Given the following voltaic cell diagram and using the standard gold electrode as the reduction half reaction, and metal M as an oxidizing half reaction, find the \(E^∘\) reduction half reaction:

\[Au^{3+} +3e^- →Au (s) \qquad E^∘=1.52V \\ M^{3+} +3e^- →M (s) \qquad E^∘=?\]

1. \(In^{3+}+3e^-→In \qquad E^∘=? \qquad E^∘_{cell}=1.86V\)

2. \(Al^{3+}+3e^-→Al \qquad E^∘=? \qquad E^∘_{cell}=3.20V\)

3. \(La^{3+}+3e^-→La \qquad E^∘=? \qquad E^∘_{cell}=3.90V\)

4. \(U^{3+}+3e^-→U \qquad E^∘=? \qquad E^∘_{cell}=3.18V\)

Tutorial

The \(E^∘\) value of a voltaic cell is given by the formula \(E^∘_{(cell)} = E^∘_{(cathode)} - E^∘_{(anode)}\)

To use this formula, we must identify which reaction will be the oxidation half reaction(anode) and which will be the reduction half reaction(cathode). The problem states that the Gold cell is the reduction half reaction and that the other metal present will undergo oxidation.

Because we know the reduction half reaction's \(E^∘\) value and we know the \(E^∘_{cell}\) value, we can use the formula \(E^∘_{(cell)} = E^∘_{(cathode)} - E^∘_{(anode)}\) to determine the \(E^∘\) value of the oxidation half reaction.

Solution

1. \(1.86V = 1.54V - E^∘_{anode} \\ E^∘_{anode} = 0.32V \)

2. \(3.20V = 1.54V - E^∘_{anode} \\ E^∘_{anode} =1.66V \)

3. \(3.90V = 1.54V - E^∘_{anode} \\ E^∘_{anode} = 2.36V \)

4. \(3.18V = 1.54V - E^∘_{anode} \\ E^∘_{anode} = 1.64V \)

Q19.59A

By referring back to figure 19-20, explain what would happen at each individual circumstance

1. zinc is wrapped around the head and tip of the iron nail
2. a hole is poked at the center of an iron nail
3. the nail is completely covered with copper

Tutorial

1. To determine which metal will be corrode (be oxidized), we must find the standard reduction potential (SRP) values for both of the metals. The metal with the smallest (most negative) SRP value will be preferentially oxidized over the other metal because the metal with a small SRP value will more willingly lose electrons.

2. Poking a hole in a piece of metal does not alter the chemical composition of the nail in any way, but it does increase the surface area of the nail. So, we must approach this problem looking for effects that increased surface area would have on a reaction, as we know that the reaction itself will not change.

3. By covering the iron nail completely in copper, the original metal (iron) is no longer exposed to the oxidizing agent. We must now consider the oxidation of copper with the oxidizing agent of air, as the iron is rendered unreactive by having no contact with air.

Solution

1. Zinc(-0.763V) has a lower SRP value than iron(-0.44V) does, so the oxidation of zinc is more favorable than the oxidation of iron. Because of this, the zinc wrapped around parts of the iron nail will be oxidized as the sacrificial anode, which would prevent the iron from oxidation.

2. Increased surface area of the nail would allow more oxidation of iron to occur because more of the metal is exposed to the oxidizing agent.

3. Because iron no longer comes into contact with air, it cannot be oxidized by the air. Copper may be oxidized instead, which would form a thin copper oxide layer around the nail and prevent further oxidation of the nail.

Q21.3D

Write the correct names for the following:

\(1. [Fe(OH)(H_2O)_3(NH_3)_2]^{2+} \\ 2. [Cu(ONO)_2(NH_3)_4] \\ 3. [Pt(H_2O)_2(NH_3)_2][PtCl_6] \\ 4. [Fe(ox)_2(H_2O)(NH_3)]^− \\ 5. Ag_2[HgCl_4]\)

Tutorial

The overall naming structure for complex ions is: (numerical prefix + ligand name) (metal ion name) (metal ion charge) where the prefix-ligand is repeated for all ligands in alphabetical order and metal ion names are altered to end in -ate if the complex ion has an overall negative charge.

Solution

1. Diamminetriaquahydroxoiron (III) ion

2. Tetraaminedinitritocopper (II)

3. Diamminediaquaplatinum (III) Hexachloroplatinate (III)

4. Ammineaquadioxalatoferrate (II) ion

5. Silver Tetrachloromercurate (II)

Q24.11A

The initial rate of the reaction \(A+B→C+D\) is determined for different initial conditions, with the results listed in the table.

1. What is the order of reaction with respect to A and to B?
2. What is the overall reaction order?
3. What is the value of the rate constant, k?

Tutorial

To determine the order of a reaction with respect to individual reactants from experimental data, the best method to evaluate the data is to find two experiments for which all reactant concentrations except one are held constant. Using these two experiments, determine by what factor the reactant and rate change. The mathematical relationship between these two values gives us the order of the reaction with respect to the reactant whose concentration we observed. This process can be repeated for all reactants in the reaction.

To determine the overall reaction order after finding the reaction orders for each reaction, all we must do is add together all of the individual reaction orders.

Once we have the generalized rate law \(rate = k [reactant\,1]^{order} [reactant\,2]^{order}\) we can plug in all measured values for any one experiment and determine the reaction's rate constant \(k\).

Solution

1. First, let's compare experiments 1 and 3 where the [B] is held constant so we can observe changes due to [A].

[A] is doubled from reaction 1 to reaction 3.

The rate almost exactly doubles from reaction 1 to reaction 3.

Because the change in [A] is reflected the same way in the reaction rate, we can conclude that the reaction is first order with respect to A.

Now let's examine experiments 1 and 2 where [A] is held constant so we can observe changes due to [B].

[B] is doubled from reaction 1 to reaction 2.

The rate quadruples from reaction 1 to reaction 2.

Because the difference of the rate (4x) is the square of the difference of [B] (2x) between these two reactions, we can conclude that the reaction is second order with respect to B

2. The overall reaction order is the sum of all individual reaction orders, therefore

\(Order = Order\,of\,A + Order\,of\,B \\ Order = 1+2 = 3\)

This reaction is third order.

3. To solve for the rate constant \(k\) we can plug in all measured values from experiment 1 to the generalized rate law for this reaction.

\(rate = k[A]^1[B]^2 \\ 3.35·10^{-4} = k[0.185]^1[0.144]^2 \\ k = .0873 M^{-2}s^{-1}\)

Q24.51A

The following observations of a reaction’s rate constant have been made: at \(T=325K\) \(k=3.2 · 10^{-6}M^{-1}s^{-1}\) ; at \(T=456K\) \(k=2.8 · 10^{-5}M^{−1}s^{−1}\). What is the activation energy of this reaction?

Tutorial

In this problem we are given reaction rates at multiple temperatures and are asked to find the activation energy of the reaction. The Arrhenius equation gives a relationship between all of these values.

\[ln(\frac{k_2}{k_1})=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})\]

Using this equation and the known constant R = 8.314J/mol*K, we can solve for the reaction's activation energy.

Solution

Let's assign \(T_1=325K\) \(k_1=3.2·10^{-6}M^{-1}s^{-1}\) and \(T_2=456K\) \(k_2=2.8·10^{-5}M^{−1}s^{−1}\)

Now we can plug these values directly into the Arrhenius equation

\[ln(\frac{2.8·10^{-5}M^{−1}s^{−1}}{3.2·10^{-6}M^{-1}s^{-1}})=-\frac{E_a}{8.314}(\frac{1}{456K}-\frac{1}{325K}) \\ E_a = 20401J = 20.4kJ\]

The activation energy for this reaction is 20.4kJ.

Q25.27A

As the museum authenticator, you have been given the task of determining which era a recently donated piece of pottery is from. If, through radiocarbon dating, you find that the disintegration rate is 13 dis min-1 g-1, what is the age of the pottery piece?

Tutorial

This problem provides the current rate of disintegration of Carbon-14, a radioactive material. We know that the half-life of Carbon-14 is 5730 years, which means that every 5730 years the activity (disintegrations per minute) decreases by 1/2.

For Carbon-14 dating problems we assume the initial activity of Carbon-14 to be 15 disintegrations per minute.

Because we are given the current activity and we know the half-life of the material in question, we can use the formula which relates the percent of radioactive material remaining with the rate constant and the time.

\[ln(\frac{Current Amount}{Original Amount}) = - k t \]

Because we know the fraction of radioactive material remaining, we only need to find the rate constant for the decay of Carbon-14 in order to be able to solve for the time. We can solve for the rate constant of this decay reaction using the half life of Carbon-14.

Solution

First, we can find the rate constant of the decay of Carbon-14

\(ln(\frac{1}{2})=-k(5730) \\ k = .000121 s^{-1}\)

Now we can use \(ln(\frac{Current Amount}{Original Amount}) = - k t \) to determine the time since the pottery was made

\(ln(\frac{13}{15})=-.000121t \\ t=1183 years\)

The pottery is around 1183 years old.

Q18.6

Which of the following ions will oxidize Br⁻ ion to Br₂?

1. Pb²⁺
2. H⁺
3. Au³⁺
4. MnO₄^-

Tutorial

For a material to be able to oxidize a compound (to steal an electron from the compound) , it must have a higher (more positive) standard reduction potential than the compound it is oxidizing.

If this condition is met, the reaction will be spontaneous because \(E^∘_{(cell)} = E^∘_{(cathode)} - E^∘_{(anode)}\) will give a positive \(E^∘_{(cell)}\) value, which is thermodynamically favorable.

A compound will oxidize Br^- to Br₂ if its half reaction has a higher SRP than the \(E^∘_{(red)}=1.09V\) for the Bromine half reaction.

Solution

1. \(E^∘_{Pb^{2+}}=-0.13 \\ E^∘_{(cell)} = -0.13V - 1.09V = -1.22V\)

This compound will not oxidize Br^- to Br₂.

2. \(E^∘_{H^{+}}=0 \\ E^∘_{(cell)} = 0V - 1.09V = -1.09V\)

This compound will not oxidize Br^- to Br₂.

3. \(E^∘_{Au^{3+}}=1.54 \\ E^∘_{(cell)} = 1.54V - 1.09V = 0.45V\)

This compound will oxidize Br^- to Br₂.

4. \(E^∘_{MnO_{4}^-}=1.54 \\ E^∘_{(cell)} = 2.87V - 1.09V = 1.78V\)

This compound will oxidize Br^- to Br₂.

Q21.4.3

Half-lives for the reaction \(A + B → C\) were calculated at three values of [A]₀, and [B] was the same in all cases. The data are listed in the following table:

Does this reaction follow first-order kinetics? On what do you base your answer?

Tutorial

To evaluate what order a reaction is, when given initial concentration and half life for several trials, we must examine the relationship between these two values.

Zero order reactions will show a directly proportional relationship between these two values. First order reactions will have the same half life for all initial concentrations. Second order reactions will show an inversely proportional relationship between these two values.

Solution

Because different initial concentrations of a reactant cause this reaction to have varying half lives, we know that this is not a first order reaction.

Instead, we can see that there is an inverse relationship between initial concentration and half life, leading us to believe that the reaction is second order.

Second order reactions express the relationship

\[t_{\frac{1}{2}}=\frac{1}{k·[A_0]}\]

Which is consistent with this data, therefore the reaction is second order.