Extra Credit 16

Q17.2.5

Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

1. \(Al(s)+Zr^{4+}(aq)\rightarrow Al^{3+}(aq)+Zr(s)\)
2. \(Ag^{+}(aq)+NO(g)\rightarrow Ag(s)+NO^{-3}(aq)\) (acidic solution)
3. \(SiO_{2}^{-3}(aq)+Mg(s)\rightarrow Si(s)+Mg(OH)_{2}(s)\) (basic solution)
4. \(ClO_{3}^{-}(aq)+MnO_{2}(s)\rightarrow Cl^{-}(aq)+MnO_{4}^{-}(aq)\) (basic solution)

S17.2.5

We can use the mnemonic "LEO goes GER" to remind us what oxidation is the loss of electrons and reduction is the gain of electrons. The oxidizing agent is the species that is the oxidant, or the one that gets reduced. The reducing agent is the reductant, which means it is the species that gets oxidized. Because electrons are negative particles, the loss of electrons results in an overall more positive product, while the gain of electron results in an overalll more negative product.

1. In this first equation, we can see that Al(s) is on one side of the equation, while Al³⁺ is on the other side. The aluminum metal goes from a neutral charge to a positive 3+ charge. This is an overall positive change, meaning that we had a loss of electrons. This tells us that Aluminum is the species that has been oxidized, as well as being the reducing agent. Across the reaction, we have Zr⁴⁺ in our reactants side and Zr(s) metal in our product. It goes from a positive 4+ charge to a neutral charge, meaning there was an overall negative charge. A negative charge means that the there was a gain of electrons, or reduction. Because Zr⁴⁺ is reduced, it is also the oxidation agent.

2. Ag⁺ goes from a positive 1 charge to a neutral charge as Ag(s) metal in the products. This is an overall negative charge, meaning there must have been a gain of electrons. Ag⁺ is therefore the reduced species. As the reduced species, Ag⁺ is also the oxidizing agent. As we look at the species NO, we need to keep in mind the different individual charges of each element. Oxygen in normal compounds carry a -2 charge, leaving N to be a +2 charge. In the product side, NO₃^- shows that the overall charge is -1. If the three Oxygens retained their -2 charge each, that leaves Nitrogen to have a +5 charge. There is an overall positive difference. This means there is a loss of electrons, meaning oxidation occurs and that NO₃^- is also the reducing agent.

3. SiO₃^2- has an overall charge of -2. Keeping our Oxygen 2- rule in mind, we find that Si has a charge of 4+. Silicon becomes a solid metal with a neutral charge in the products. That means there must have been a gain in electons, meaning that SiO₃^2- has been reduced and is also the oxidizing agent. Magnesium metal with a neutral charge becomes Mg(OH)₂. Hydroxide has a charge of -1 each, meaning that Mg has a charge of +2 to maintain the overall neutral charge in the compound. Becoming more positive is a loss of electrons, meaning that Mg has been oxidized and is the reducing agent.

4. The Chlorine in ClO₃^- has a charge of 5+ and has a charge of -1 in the products. Thus, it has gained electrons, making this the reduced species and also the oxidizing agent. The Manganese in Mn(O₂) has a 4+ charge and becomes MnO₄^- meaning that it now has a 7+ charge. This means that it has loss electrons, telling us that this is the oxidated species and also the reducing agent.

Q19.1.15

The standard reduction potential for the reaction \(\ce{[Co(H_{2}O)_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(H_{2}O)_{6}]^{+2}}(aq)\) is about 1.8 V. The reduction potential for the reaction \(\ce{[Co(NH_{3})_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(NH_{3})_{6}]^{+2}}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H₂O)₆]²⁺ and/or [Co(NH₃)₆]²⁺, can be oxidized to the corresponding cobalt(III) complex by oxygen.

S19.1.15

We first need to find the standard reduction potential of the oxidation of oxygen. The equation off the table is:

\(O _{2}(g)+ 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2} (aq)\) \(E^o= 0.7V\)

Because we are looking to use this for oxidation, we must flip this equation around since this is a standard reduction potential. When we flip equations, we must also flip the Eº value as a negative value.

\(H_{2}O_{2} (aq)\rightarrow O _{2}(g)+ 2H^{+}(aq) + 2e^{-}\) \(E^o= -0.7V\)

To calculate the overall Eº, we must add the two values together under the following equation. The cathode is where reduction occurs and the anode is where oxidation occurs.

\[E^{o}_{cell}= E^{o}_{cathode} + E^{o}_{anode}\]

1. For our first equation: \[\ce{[Co(H_{2}O)_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(H_{2}O)_{6}]^{+2}}(aq)\]\[E^{o}=1.8 V + (- 0.7V) = 1.1V\]

A positive Eº means that the process is spontaneous.

2. For our second equation: \[\ce{[Co(NH_{3})_{6}]^{+3}}(aq)+\ce{e}^{-}\ \rightarrow {[Co(NH_{3})_{6}]^{+2}}(aq)\] \[E^{o}=0.1V + (-0.7V) = -0.6V\]

A negative Eº means that the process is nonspontaneous.

Q19.3.6

How many unpaired electrons are present in each of the following?

1. \([CoF_{6}]^{3-}\) (high spin)
2. \([Mn(CN)_{6}]^{3-}\) (low spin)
3. \([Mn(CN)_{6}]^{4-}\) (low spin)
4. \([MnCl_{6}]^{4-}\) (high spin)
5. \([RhCl_{6}]^{3-}\) (low spin)

S19.3.6

For each question, we need to pay attention to the coordination number of the transition metal. This determines if the structure is an octahedral complex which will help us with determining the amount of unpaired electrons there are.

1. First we need to find the charge of the transition metal. Fluorine has a charge of -1 in compounds, meaning that Cobalt must have a charge of +3. Having a positive charge means that the metal must have lost some electrons and in this specific case, Cobalt lost 3 electrons. Transition metals lose electrons from the 4s orbital first because of the higher energy levels, before losing electrons in the 3d orbital. This means that Cobalt will have 6 total 3d electrons left. Using the crystal splitting theory, and keeping in mind that this is high spin, we find that there are 4 unpaired electrons.

2. We can continue our train of thought from the previous part. The charge of the Manganese metal is +3. That means that it has 4 total 3d electrons left. Being low spin, we fill up the lower level before going to the top. This means that there will be 2 unpaired electrons.

3. The charge of Manganese in this problem is +2. That means that it only loses electrons from the 4s orbital and keeps all its 3d electrons. Therefore, being low spin, there will be five 3d electrons and a total of one unpaired electron.

4. This problem is similar to the previous part. Mn also has a +2 charge in this question meaning that there will be five 3d electrons left. However, in this question, it is in a high spin state, meaning we will be filling all the spots before pairing up our electrons. This means that we have a total of 5 unpaired electrons.

5. Rh has a charge of +3 charge, meaning that it will only lose one of its 3d electrons after losing the two 4s electrons. It will have a total of six 3d electrons remaining. Because this complex is in a low spin state, there will be no unpaired electrons, as the entire lower level is filled.

Q12.4.6

What is the half-life for the first-order decay of phosphorus-32?

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

The rate constant for the decay is 4.85 × 10⁻² day⁻¹.

S12.4.6

This is a first order reaction, so we can use our half life equation below:

\[t_{1/2}=\frac{0.693}{k}\]

The rate constant is given to us in units per day. All we have to do, is to plug it into the equation.

\[t_{1/2}=\frac{0.693}{4.85*10^{-2}}\]

\[=14.3 days\]

Q21.2.1

Write the following isotopes in hyphenated form (e.g., “carbon-14”)

1. \(_{11}^{24}\textrm{Na}\)
2. \(_{13}^{29}\textrm{Al}\)
3. \(_{36}^{73}\textrm{Kr}\)
4. \(_{77}^{194}\textrm{Ir}\)

S21.2.1

Isotopes are named by the element followed by the molecular weight.

1. The element is Sodium and this isotope has a molecular weight of 24. Therefore, this isotope is named Sodium-24.

2. The element is Aluminum and this isotope has a molecular weight of 29. Therefore, this isotope is named Aluminum-29.

3. This element is Kryton and this isotope has a molecular weight of 73. Therefore, this isotope is named Kryton-73.

4. This element is Iridium and this isotope has a molecular weight of 194. Therefore, this isotope is named Iridium- 194.

Q21.5.4

Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.

S21.5.4

Nuclear chain reactions occur when one nuclear reaction causes an average of one or more subsequent nuclear reactions. To maintain a sustained controlled nuclear reaction, for every two or three neutrons released, only one must be allowed to strike another nucleus. Anymore, will lead to too much release, which can cause an explosion.

Q20.4.2

List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.

S20.4.2

Relationships between different factors are depicted in equations we use everyday. An equation that connects an electrochemical cell potential to another factor is the following equation:

\(\Delta G=-nFE\) which can be rearranged to \(E=-\frac{\Delta G}{nF}\)

This shows that there is a defined relationship between Gibbs Free Energy and the Cell potential.

In addition, there are factors that can affect Gibbs Free Energy. According to the Nernst Equation:

\[E=E^{o} - \frac{(RT)}{nF}lnQ\]

Two factors that will affect Gibbs Free Energy are temperature and concentrations/gas pressures. These are apparent in the Nernst Equation, as T stands for temperature and Q is the ratio of the concentrations and/or gas pressures of the products over reactants.

Q20.5.31

The silver–silver bromide electrode has a standard potential of 0.07133 V. What is K_sp of AgBr?

S20.5.31

First, we must find the standard reduction potentials of each side of the reaction. We are given that this reaction is done with a Silver to Silver Bromide electrode. We can find these following equations in a standard reduction potential chart:

\(Ag^{+}(aq) + e^{-} \rightarrow Ag(s)\) \(E^{o}=0.8 V\)

\(AgBr(s) + e^{-} \rightarrow Ag(s) + Br^{-}(aq)\) \(E^{o}= 0.07133 V\)

We also need to know how \(AgBr\) dissolves and divides into its ions:

\[AgBr(s) \rightarrow Ag^{+}(aq) + Br^{-}(aq)\]

\[K_{sp}= [Ag^{+}][Br^{-}]\]

We want to manipulate our two equations to get it in the form of our dissolving equation. First, we must flip our first equation so that we have the Ag⁺ ion on the correct side. As we flip our standard reduction equation, we must also flip our E° value.

\(Ag(s) \rightarrow Ag^{+}(aq) + e^{-}\) \(E^{o}=-0.8 V\)

To find our E°_cell value, we must add our two E° values according to the following equation.

\[E^{o}_{cell}= E^{o}_{cathode} + E^{o}_{anode}\]

\[E^{o}_{cell}= 0.01733V + (-0.8V)\]

\[E^{o}_{cell}= -0.7287 V\]
Now that we have our Eº_cell value, we can use the Nernst Equation to find our K value, which will resemble K_sp in this case.

\[E = E^{o}cell - \frac{0.0591}{n}logK\]

We can plug in our \(E^{o}\) value into the equation as well as our number of electrons. In this case it is one electron.

\[0 = -0.7287 - \frac{0.0591}{1}logK\]

We can add -0.7287 to the other side.
\[0.7287 = -0.0591logK\]

We divide 0.0591 onto the other side.
\[\frac{0.7287}{-0.0591} = log K\]

Solve the left side of the equation.
\[-12.33 = log K\]

Raise each side to the power of 10 to cancel out the log.
\[10^{-12.33} = K\]

Solve.

\[K_{sp}= 4.7 * 10^{-13}\]