Extra Credit 9

Question 17.1.8

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.

Answer17.1.8

First, to tackle this problem, we must break down the question into a series of tasks we are given. We are asked for the species that was a) oxidized, b) reduced, c) the oxidizing agent, and d) the reducing agent in each of the reactions of the previous problem.

Then, we must remember what it means for a species to be oxidized or reduced and for a species to be an oxidizing agent or reducing agent. Reduction-Oxidation (aka RedOx) Reactions are made of two parts that always occur together because one species cannot be oxidized without the other becoming reduced.

• A species is oxidized if its oxidation number increases because it loses an electron (i.e becomes more positive/ less negative).
  • Example: \(Zn^{+} \rightarrow Zn^{2+}\)
• A species is reduced if its oxidation number decreases because it gains an electron (i.e becomes more negative/ less positive).
  • Example: \(Zn^{2+} \rightarrow Zn^{+}\)
• The oxidizing agent is the species which accepts the electron (i.e the reduced species).
• The reducing agent is the species which gives away the electron (i.e the oxidized species).

(Additional information can be found on: Oxidation-Reduction Reactions, particularly Examples 1.3-1.5)

Next, we must recall the reactions from the previous problem (Q17.1.7):

1. \(SO_{3}^{2-}(aq)+Cu(OH)_{2}(s) \rightarrow SO_{4}^{2-}(aq)+Cu(OH)(s)\)
2. \(O_{2}(g)+Mn(OH)_{2}(s) \rightarrow MnO_{2}(s)\)
3. \(NO_{3}^{-}(aq)+H_{2}(g) \rightarrow NO(g)\)
4. \(Al(s)+CrO_{4}^{2-}(aq) \rightarrow Al(OH)_{3}(s)+Cr(OH)_{4}^{-}\)

Finally, to answer this question is to go through each reaction, separate them into half-reactions, take note of the changes in their oxidation states, and identify & label the different components.

1. \(SO_{3}^{2-}(aq)+Cu(OH)_{2}(s) \rightarrow SO_{4}^{2-}(aq)+Cu(OH)(s)\)
   • \(SO_{3}^{2-}(aq) \rightarrow SO_{4}^{2-}(aq)\)
     • Oxidation State: \((+2) \rightarrow (+4)\)
     • Since it becomes more positive, this species is oxidized
     • Since it gave up electrons to become more positive, it is a reducing agent
   • \(Cu(OH)_{2}(s) \rightarrow Cu(OH)(s)\)
     • Oxidation State: \((+1) \rightarrow (0)\)
     • Since it becomes less positive, this species is reduced
     • Since it gains electrons to become less positive, it is an oxidizing agent.
2. \(O_{2}(g)+Mn(OH)_{2}(s) \rightarrow MnO_{2}(s)\)
   • Oxidation States: \((O_{2}^{+0})+(Mn^{2+} O^{-2}H^{+1}) \rightarrow Mn^{4+}O^{4-}\)
     • O₂: \((0) \rightarrow (-4)\); reduced, oxidizing agent
     • Mn: \((+2) \rightarrow (+4)\); oxidized, reducing agent
     • Note: OH becomes water and it is not part of the reaction
3. \(NO_{3}^{-}(aq)+H_{2}(g) \rightarrow NO(g)\)
   • \(NO_{3}^{-}(aq) \rightarrow NO(g)\)
   • Oxidation States: \(N^{+5} O_{3}^{2-} \rightarrow N^{+2}O^{2-}\)
     • N: \((+5) \rightarrow (+2)\); reduced, oxidizing agent
     • O: \((-6) \rightarrow (-2)\); oxidized, reducing agent
4. \(Al(s)+CrO_{4}^{2-}(aq) \rightarrow Al(OH)_{3}(s)+Cr(OH)_{4}^{-}\)
   • \(Al(s) \rightarrow Al(OH)_{3}(s)\)
     • Oxidation State: \((0) \rightarrow (+3)\)
     • Oxidized therefore it is a reducing agent
   • \(CrO_{4}^{2-}(aq) \rightarrow Cr(OH)_{4}^{-}\)
     • Oxidation State: \((+6) \rightarrow (+3)\)
     • Reduced, therefore it is an oxidizing agent

Question 19.1.7

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

Answer 19.1.7

First, we must determine what oxide means.

• A simple definition of an oxide is that it is a compound with one or more oxygen atoms combined with another element.
• Additional information on Oxides can be found on Oxides.

Then, we must analyze what information we were given.

• Formula: \(MO_{3}\)
  • There is no element M which means we must find the atom closest in molar mass to O₃

Next, we must solve for what is wanted.

• O₃ = 3 oxygens = Multiply molar mass of oxygen by 3 = 48 g/mol

Finally, we must determine which amongst the choices (Zr, Nb, or Mo) is the right answer.

• Mo is closest in atomic weight (42 g/mol).

Question 19.2.9

Predict whether the carbonate ligand CO₃^2- will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.

Answer 19.2.9

First, we must analyze what information we were given.

• Carbonate ligand

Next, we must determine what information is wanted.

• What type of ligand the carbonate ligand will coordinate to a metal center as.

Then, we must define what a ligand is and the different types to be able to proceed.

• In a nutshell, a ligand can be either an ion or molecule that binds to a central metal atom to form a coordination complex.
• A monodentate ligand binds to the center through ONE atom.
• A bidentate ligand can bind to the center at two points.
• A tridentate ligand can bind to the center at three points.
• More information can be found on Ligands.

Finally, we would have to draw the coordination complexes possible with the ligand we were given. Here, we can denote the central atom as "M".

• It is possible for the carbonate ligand to bind to the central atom as a monodentate ligand.

• It is also possible for the carbonate ligand to bind to the central atom as bidentate ligand.

Question 12.3.22

The following data have been determined for the reaction:

\[I^{-}+OCl^{-} \rightarrow IO^{-}+Cl^{-}\]

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

Answer 12.3.22

To begin with, identify what is wanted.

• Rate equation
• Rate constant

Then, define what each of the terms means.

• Rate equation: \(rate = k[A]^{a}[B]^{b}\)
• Rate constant: \(k= \dfrac{rate}{[A]^{a}[B]^{b}}\)
• More information can be found on: Reaction Rate

Next, identify what is given.

• Reaction: \(I^{-}+OCl^{-} \rightarrow IO^{-}+Cl^{-}\)
• Initial Molarities of each reactant at each order
• Rates at each order

Finally, we solve this question by

1. writing out the rate equation:
   • \(rate=k[A]^{a}[B]^{b}\)
2. rearranging the rate equation to isolate k (rate constant)
   • \(k=\dfrac{rate}{[A]^{a}[B]^{b}}\)
3. determining the order of A
   • Here, \([A] = [I^{-}]\)
   • From the table above, we can see how from [A] increases by 2 and its Rate also increases by 2.
     • From this, we can derive the equation: 2^x=2.
       • What does x have to be for this equation to be true? 1.
       • Therefore, A's reaction order is 1. ([A]¹ )​​​
4. determining the order of B
   • Here, \(B = [OCl^{-}]\)
   • From the table above, we can see how [B] remains the same but its Rate increases by 2.
     • From this we can derive the equation, 1^x=2.
       • What does x have to be for this equation to be true? 0
       • Therefore, B's reaction order is 0. [B]⁰
5. inputting the information we have to the rate equation
   • \(3.05x10^{-4}=k[0.10]^{1}[0.50]^{0}\)
   • \(k= \dfrac{3.05x10^{-4}}{[0.10]^{1}[0.50]^{0}}\)
   • \(k= 3.05x10^{-3}\)
6. We also need the correct units!
   • Overall Rate of Reaction: (1) + (0) = 1
   • Units: Sec^-1
   • Therefore, k= 3.05x10^-3 Sec^-1

Question 12.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

Answer 12.7.2

First, what are catalysts?

• They are what increases the rate of a chemical reaction.

Next,

1. What are homogenous catalysts?
   • Catalysts that are the same (homo-) phase as the reactants
2. What are heterogenous catalysts?
   • Catalysts that are different (hetero-) phase as the reactants

*Note: phase is when you can see a boundary/difference between the components

Finally, compare their functions:

• With heterogenous catalysts, the reactions are adsorbed on to the surface of the catalyst and the products are consequently de-adsorbed after the reaction happens.
• With homogenous catalysts, it is less complicated as the catalyst reacts/ dissolves in a solvent of substrates and forms a uniform solution while also increasing the rate of a chemical reaction. There is no de-adsorption after the product is formed.

Question 21.4.25

A \(\ce{^{8}_{5}B}\) atom (mass = 8.0246 amu) decays into a \(\ce{^{8}_{4}B}\) atom (mass = 8.0053 amu) by loss of a \(\beta ^{+}\) particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

Answer 21.4.25

First, create the equation

• \(\ce{^{8}_{5}B} \rightarrow \ce{^{8}_{4}B}\ + \beta ^{+}\)

Then, subtract the masses of the reactants from the products.

• \((8.0246amu)- [(8.0053amu)+(0.00055amu)]\) \(= 0.01875 amu\)

Then, take the total mass and convert it to kg:

\(0.01875amu x 1.6605402x10^{27} kg = 3.113512874x10^{-29} kg\)

Then, use \(E=mc^{2}\) where m= mass in kg and c= 2.998x10⁸ m*sec^-1

• \(E=(3.113512874x10^{-29} kg)(2.998x10^{8} m*sec^{-1})^{2}\)

• \(E= 2.79796565 J\)

Next, convert Joules to eV (1 J = 6.25 * 10¹⁸ eV)

• \((2.79796565 J) * (6.25 * 10^{18} eV) = 17487285.28 eV\)

Finally, since the question wants it in millions of electron volts (MeV), convert eV to MeV.

• \(17487285.28 eV * \dfrac{1 MeV}{10^{6} eV} = 17.4873 MeV\)

Therefore, 17.5 MeV is produced by the reaction.

Question 20.3.12

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. \(\ce{Pb}(s)│\ce{PbSO_{4}}(s)│\ce{SO_{4}^{2-}}(aq)║\ce{Cu^{2+}}(aq)│\ce{Cu}(s)\)

1. \(\ce{Hg}(l)│\ce{Hg_{2}Cl_{2}}(s)│\ce{Cl^{-}}(aq)║\ce{Cd^{2+}}(aq)│\ce{Cd}(s)\)

Answer 20.3.12

First, write half-reaction for each cell diagram

1. \(\ce{Pb}(s)│\ce{PbSO_{4}}(s)│\ce{SO_{4}^{2-}}(aq)║\ce{Cu^{2+}}(aq)│\ce{Cu}(s)\)
   • \(Pb(s) + PbSO_{4} (s) + 2e^{-} \rightarrow SO_{4}^{2-}(aq)\)
   • \(Cu^{2+}(aq) \rightarrow Cu + 2e^{-}(s)\)
2. \(\ce{Hg}(l)│\ce{Hg_{2}Cl_{2}}(s)│\ce{Cl^{-}}(aq)║\ce{Cd^{2+}}(aq)│\ce{Cd}(s)\)
   • \(Hg(l) +Hg_{2}Cl_{2}(s)+e^{-} \rightarrow Cl^{-}(aq)\)
   • \(Cd^{2+}(aq) \rightarrow Cd+2e^{-}(s)\)

Second, write the overall reaction for each.

1. \(Pb(s) + PbSO_{4}(s) + Cu^{2+}(aq) \rightarrow Cu (s) + SO_{4}^{2-}(aq)\)
2. \(2Hg(l) +2Hg_{2}Cl_{2}(s) + Cd^{2+}(aq) \rightarrow Cd(s)+2Cl^{-}(aq)\)

Finally, state which half reactions occur at the anode and at the cathode.

*Remember not to PANIC (Positive is Anode and Negative Is Cathode) i.e, Oxidation is Anode and Reduction is Cathode

1. Anode
   • \(Cu^{2+}(aq) \rightarrow Cu + 2e^{-}(s)\)
   • \(Cd^{2+}(aq) \rightarrow Cd+2e^{-}(s)\)
2. Cathode
   • \(Pb(s) +PbSO_{4}(s) + 2e^{-} \rightarrow SO_{4}^{2-}(aq)\)
   • \(Hg(l) +Hg_{2}Cl_{2}(s)+e^{-} \rightarrow Cl^{-}(aq)\)

Question 20.5.24

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

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?

\(acetate + CO_{2} + 2H+ +2e^{-} \rightarrow pyruvate +H_{2}O\) \(E° = −0.70 V\)

\(pyruvate + 2H+ + 2e^{-} \rightarrow lactate\) \(E° = −0.185 V\)

Answer 20.5.24

First, we need to analyze what we were given. We were given the half reaction \(NAD^{+} + H^{+} + 2e^{-} → NADH\). The electron being in the reactant side shows that this is a reduction reaction.

1. Would NADH be able to reduce acetate to pyruvate?
   • Yes, because NADH will donate its electrons.
2. Would NADH be able to reduce pyruvate to lactate?
   • Yes, because NADH will donate its electrons.
3. What cell potential is needed to convert acetate to lactate?
   • Add the cell potentials together.
     • \((-0.70V) + (-0.185 V) = (-0.885V)\).