Extra Credit 9

Q17.1.8

Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

a) \(SO_3^{2-}(aq) + 2Cu(OH)_2(s) \to SO_4^{2-}(aq) + 2Cu(OH)(s) + H_{2}O(l)\)

b) \(O_2(g) + 2Mn(OH)_2(s) \to 2MnO_2(s) + 2H_{2}O\)

c) \(2NO_3^{-}(aq) + 3H_2(g) \to 2NO(g) +2H_{2}O(l)+2OH^-(aq)\)

d) \(2Al(s) + 3CrO_2^{4-}(aq) + 12 H_{2}O(l) \to Al(OH)_3(s) + Cr(OH)^{4-}(aq) +6OH^-(aq)\)

Review

In order to identify the species that are oxidized or reduced, we need to understand their respective definitions:

\(Oxidation: loss\; of\; electrons\)

\(Reduction: gain \; of \; electrons\)

\(Reducing \; agents \; are \; the \; species \; that \; are \; oxidized\)

\(Oxidizing \; agents \; are \; the \; species \; that \; are \; reduced\)

Another way to memorize this is by using the acronym OIL RIG which stands for oxidation is losing electrons and reduction is gaining electrons

Now that we get these two concepts, we can delve into the topic of oxidation states which indicate the charge of the molecule. Positive oxidation states correlate to reduction and negative oxidation states correlate to oxidation.

Some important factors to know about oxidation states are that:

• Oxygen usually has an oxidation state of -2 (Exceptions: Peroxide)
• Hydrogen almost always has an oxidation state of +1 (sometimes -1)
• Elements that are by themselves have an oxidation state of 0 (Ex. \(H_{2}\), Zn, \(Cl_{2})\
• Hydroxide has an oxidation state of -1

Solutions

a) \(SO_3^{2-}(aq) + 2Cu(OH)_2(s) \to SO_4^{2-}(aq) + 2Cu(OH)(s) + H_{2}O(l)\)

• \(SO_3^{2-}(aq)\)
  • Set [Charge of] S + 3[Charge of] O = Overall Charge

S + 3(-2) = -2

S = 4

• \(Cu(OH)_2(s)\)
  • Set [Charge of] Cu + 2[Charge of] \(OH_2\)
  • Charge of OH = -1

Cu + 2(-1) = 0

Cu = +2

Products

• \(SO_4^{2-}(aq)\)
  • Set [Charge of] S + 4[Charge of] O = Overall Charge

S + 4(-2) = -2

S= +6

• \(2Cu(OH)(s)\)
  • Set [Charge of] Cu + [Charge of] OH = Overall Charge

Cu + (-1) = 0

Cu = +1

• \(H_{2}O(l)\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2

Reactants | Products

From the table above, we can see that that the charge of S goes from +4 to +6 indicating a loss in electrons (oxidation). We can also see that Cu goes from +2 to +1 from gaining an electron (reduction).

Since we only judge whether the reactants are oxidized or reduced, \(SO_3^{2-}(aq)\) was oxidized and is therefore the reducing agent and \(Cu(OH)_2(s)\) was reduced and therefore the oxidizing agent.

b) \(O_2(g) + 2Mn(OH)_2(s) \to 2MnO_2(s) + 2H_{2}O\)

Reactants

• \(O_2(g)\)
  • Oxidation state of \(O_2(g)\) is 0
• \(2Mn(OH)_2(s)\)
  • [Charge of] Mn + [Charge of] \(OH_2\) = Overall Charge

Mn + 2(-1) = 0

Mn = +2

Products

• \(2MnO_2(s)\)
  • [Charge of] Mn +2[Charge of] O = Overall Charge

Mn + 2(-2) = 0

Mn = +4

• \(2H_{2}O\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2

Reactants | Products

As seen from the table above, Mn goes from a +2 charge to a +4 charge. Therefore, \(2Mn(OH)_2(s)\) is being oxidized and is therefore the reducing agent.

Also, O can be seen going from 0 to -2. Therefore, \(O_2(g)\) is being reduced and is therefore the oxidizing agent.

c) \(2NO_3^{-}(aq) + 3H_2(g) \to 2NO(g) +2H_{2}O(l)+2OH^-(aq)\)

Reactants

• \(2NO_3^{-}(aq)\)
  • [Charge of] N + 3[Charge of] O = -1

N + 3(-2) = -1

N = +5

• \(3H_2(g)\)
  • Oxidation state of \(3H_2(g)\) = 0

Products

• \(2NO(g)\)
  • [Charge of] N + [Charge of] O = 0

N + (-2) = 0

N = +2

• \(2H_{2}O\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2
• \(2OH^-(aq)\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2

Reactants | Products

From the table above, we can see that the oxidation state of N goes from +5 to +1. Therefore, \(2NO_3^{-}(aq)\) is being reduced and is the oxidizing agent.

Also, the H can be seen from going from 0 to +1. Therefore, \(3H_2(g)\) being oxidized and therefore being a reducing agent.

d) \(2Al(s) + 3CrO_2^{4-}(aq) + 12 H_{2}O(l) \to Al(OH)_3(s) + Cr(OH)^{4-}(aq) +6OH^-(aq)\)

Reactants

• \(2Al(s)\)
  • Oxidation state of Al = 0
• \(3CrO_2^{4-}(aq)\)
  • 3[Charge of] Cr + 2[Charge of] O = Overall Charge

3Cr + 2(-2) = -4

Cr = 0

• \(12 H_{2}O(l)\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2

Products

• \(Cr(OH)^{4-}(aq)\)
  • [Charge of] Cr + [Charge of] OH = -4

Cr + (-1) = -4

Cr = -3

• \(6OH^-(aq)\)
  • Oxidation state of H = +1
  • Oxidation state of O = -2
• \(Al(OH)_3(s)\)
  • [Charge of] Al + 3[Charge of] OH = 0

Al + 3(-1) = 0

Al = 3

Reactants | Products

From the table above, Cr can be seen going from 0 to -3. Therefore \(3CrO_2^{4-}(aq)\) is being reduced and is the oxidizing agent.

Also, Al can be seen going from 0 to +3. Therefore, \(2Al(s)\) is being oxidized and is the reducing agent.

Q19.17

Which of the following elements is most likely to form an oxide with the formula MO₃: Zr, Nb, or Mo?

Solution

It is important to note that no negative or positive charge is given on MO₃ . Because of this you want the element you choose to balance out the charge from \(O_3\) . Mo is the most likely to form an oxide with \(O_3\). Since oxygen has an oxidation number of -2 and there are 3 of them, \(O_3\) provides a -6 charge. Looking at the periodic table of elements, Zr can be seen to have 2 4d orbital electrons, Nb with 3 4d orbital electrons, and Mo with 4 4d electrons. However, when forming an oxide, the 5s orbital electrons are also taken to account since they are on the same energy level (n=5). Including this, Mo has 6 electrons which is just enough to bond with \(O_3\).

Q19.2.9

Predict whether the carbonate ligand \(CO_{3}^2-\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.

Solution

\(CO_{3}^2-\) has a trigonal planar shape which explains its 120-degree angles it has between the Carbon atoms and the Oxygen atoms. Because of the 120-degree angles between the Oxygen atoms, the two Oxygens cannot bond to the same metal. Because of this, the carbonate ligand is most likely to be monodentate. \(CO_{3}^2-\) can be considered bidentate when it bonds to two different metals. It can "bite" two times but on two metals but since it cannot "bite" twice on the same metal it is monodentate

Q12.3.22

The following data have been determined for the reaction:

\(I^− + OCl^- \to IO^− + Cl^−\)

Determine the rate equation and the rate constant for this reaction.

Solution

Using the reactants, we can form the rate law of the reaction: $$ r=k[OCl^-]^n[I^-]^m $$

From there, we need to use the data to determine the order of both \([OCl^-]\) and \([I^-]\). In doing so, we need to compare \(r_1\) to \(r_2\) such that:

$$ \frac {r_1}{r_2} = \frac {(0.10^m)(0.050^n)}{(0.20^m)(0.050^n)} = \frac {3.05 \times 10^{-4}}{6.20 \times 10^{-4}} $$

$$ 0.5^m = 0.5 $$

$$ m = 1 $$

We can "cross out" the concentration of \([OCl^-]\) because it has the same concentration in both of the trials used.

Now that we know m (\([I^-]\)) has a first order of 1.

We cannot "cross out" \([I^-]\) to find \([OCl^-]\) because no two trials have the same concentration. In order to solve for n we will plug in 1 for m.

$$ \frac {r_1}{r_3} = \frac {(0.10^{1})(0.050^n)}{(0.30^{1})(0.010^n)} = \frac {3.05 \times 10^{-4}}{1.83 \times 10^{-4}} $$

$$ \frac {1}{3} (5^{n}) = 1.6666667 $$

$$ 5^{n} = 5 $$

$$ n = 1 $$

Since we know that orders of both n and m are equal to one, we can not substitute them into the rate law equation along with the respective concentrations (from either the first, second, or third reaction) and solve for the rate constant, k.

$$ r=k[OCl^-]^n[I^-]^m $$

$$ 3.05 * 10^{-4}= k[0.05]^1[0.10]^1 $$

$$ k = 6.1 * 10^{-2} \frac {L}{mol \times s} $$

Thus the overall rate law is: $$ r = (6.1 * 10^{-2} \frac {L}{mol \times s})[OCl^-][I^-] $$

The units for K depend on the overall order of the reaction. To find the overall order we add m and n together. By doing this we find an overall order of 2. This is why the units for K are $$ \frac {L}{mol \times s} $$

Q12.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

Solution

To solve this problem we first need to define homogeneous and heterogeneous catalysts. Homogeneous catalysts are in the same phase as the reactants while heterogeneous catalysts are in a different phase from the reactants.

• homogeneous catalysts:
  • catalyst interacts with the reactants and forms a uniform solution while also increasing the rate of a chemical reaction
  • take part in reaction and separate itself
• heterogeneous catalysts:
  • the atoms of the reactants will be together with the catalyst weakening their bonds before they form new bonds.

Q21.4.25

A \(\ce{^5_8}B\) atom (mass = 8.0246 amu) decays into a \(\ce{^4_8}B\) atom (mass = 8.0053 amu) by loss of a β⁺ particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

Solution

Using Einstein's equation, $$E=mc^2$$

we can alter it so that we can account for the change in mass indicated in the problem so that it looks like this: $$ΔE=Δmc^2$$

such that ΔE= change in energy, Δm= change in mass, c= speed of light ( \(~3.0 \times 10^8)\)

Since the problem states that this problem involves a loss of a β⁺ particle or by electron capture, we write the equation as follows: $$ \ce{^5_8}B \to \ce{^4_8}B + \ce{^0_1}β⁺ $$

Now we have to calculate for the Δm: $$Δm = m_{reactants} - m_{products} $$

To find the mass of the products we multiply by the masses of protons/neutrons by the number of each on the product side.

$$ (8.0246amu)- (8.0053amu + 0.00055) $$

$$Δm=0.01875 amu$$

Since in the equation, Δm is in kilograms. We convert the 0.01985 amu to kg.

In order to do so, we have to divide 0.01985 amu by Avogadro's number (\(6.022 \times 10^{23}\)) to convert into grams. Then divide by 1000 g to convert to kg.

$$\frac{0.01875}{6.022 \times 10^{23}} * \frac{1 kg}{1000 g} = 3.1136 \times 10^{-29} kg $$

Now we can plug in Δm and the speed of light, c into the equation yielding: $$ΔE= (3.1136 \times 10^{-29} kg)(3.00 /times 10^8 m/s)^2 $$

$$= 2.8022 \times 10^-12 J $$

To convert to volts, we multiply \(-2.8022 \times 10^-12\) back by Avogadro's number to the units to be J/mol

$$= 1.6875 \times 10^12 J/mol$$

To convert to volts, we need to divide by Faraday's constant: $$\frac {1.6875 \times 10^{12} J/mol}{96485 J/mol V} $$

which yields us the final answer$$= 17.5 MeV$$

Q21.3.12

Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.

1. \(Pb(s) | PbSO_4(s) | SO_4^{2-}(aq) || Cu^{2+}(aq) | Cu(s) \)
2. \(Hg(l) | Hg_2Cl_2(s) | Cl^-(aq) || Cd^{2+}(aq) | Cd(s) \)

Solution

To approach this problem, some important things to note are that:

• To the left of the cell diagram composes of the anode
• To the right of the cell diagram composes of the cathode
• the || portion of the cell diagram represents the salt bridge

Knowing this, we can use this information to write out both the half-reactions for the anode and cathode.

1. \(Pb(s) + SO_4^{2-}(aq) \to PbSO_4(s) + 2e^{-}\) anode

\(Cu^{2+}(aq) + 2e^{-} \to Cu(s)\) cathode

\(Cu^{2+}(aq) + Pb(s) + SO_4^{2-}(aq) \to PbSO_4(s) + Cu(s)\) overall reaction

2. \( 2Hg(l) + 2Cl^{-}(aq) \to Hg_{2}Cl_2(s) + 2e^{-}\) anode

\(Cd^{2+} + 2e^{-} \to Cd(s)\) cathode

\( 2Hg(l) + 2Cl^{-}(aq) + Cd^{2+}(aq) \to Hg_{2}Cl_{2}(s) + Cd(s) \) overall reaction

Q20.5.24

The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction \(NAD^{+}+H^{+}+2e^{−} \to NADH ; E°=−0.32V. \)

$$ acetate + CO_2+ 2H^{+} +2e^{-} \to pyruvate + H_2O ; E^° = −0.70 V$$

$$ pyruvate + 2H^{+} + 2e^{-} \to lactate ; E^° = −0.185 V $$

1. Would NADH be able to reduce acetate to pyruvate?
2. Would NADH be able to reduce pyruvate to lactate?
3. What potential is needed to convert acetate to lactate?

Solution

1. In order to start the problem, you need to flip the reaction for NADH. In doing so, the signs for E° switches and becomes positive indicating a spontaneous reaction.

$$NADH \to (NAD^{+}+H^{+}+2e^{−} ; E° = 0.32 V$$

The next step to take is to add the two E° (both the NADH and acetate) to check if they will create a spontaneous reaction.

$$ 0.32 V (NADH) + -0.70 V (acetate) = -0.38 V $$

Since the reaction's cell potential is negative, it indicates that the reaction is not spontaneous. Therefore, NADH will not be able to reduce acetate to pyruvate.

2. The same applies to the second part except this time, the E° of the NADH and pyruvate are added together.

$$ 0.32 V (NADH) + -0.185 V (acetate) = 0.135 V $$

Since this yields a positive value, that means the reaction is spontaneous. Therefore, NADH will be able to reduce pyruvate to lactate.

3. The third problem involves both half reactions in that of acetate and pyruvate. In order to convert acetate to lactate, we need to have a greater potential than the sum of both standard potentials in the problem.

$$ E° + ((-0.185 V) + (-0.70 V)) > 0 $$

$$ E° > 0.885 V $$

Therefore, a voltage of over 0.885 is needed to convert acetate to lactate.