Extra Credit 9

Q17.1.8

The following in basic solution:

1. \(\ce{SO3^2-}(aq)+\ce{Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+\ce{Cu(OH)}(s)\)
2. \(\ce{O2}(g)+\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)\)
3. \(\ce{NO3-}(aq)+\ce{H2}(g)⟶\ce{NO}(g)\)
4. \(\ce{Al}(s)+\ce{CrO4^2-}(aq)⟶\ce{Al(OH)3}(s)+\ce{Cr(OH)4-}(aq)\)

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.

S17.1.8

1. The species that is oxidized is \(\ce{SO3^2-}\)(aq) because the Sulfur atom increases in oxidation state from +4 to +6 as it goes to \(\ce{SO4^2-}\)(aq). Since this is oxidized, and a reduction reaction occurs simultaneously, this also acts as a reducing agent for \(\ce{Cu(OH)2}\)(s). The copper atom decreases in oxidation state, from +2 to +1, so \(\ce{Cu(OH)2}\)(s) is reduced. This is also an oxidizing agent, as it oxidizes \(\ce{SO3^2-}\)(aq) by giving an oxygen atom.

If you write the half reactions

\(\ce{H2O}(l)+\ce{SO3^2-}(aq)⟶\ce{SO4^2-}(aq)+\ce{2H^1+}(aq)+\ce{2e^1-}\)

You can see it is an oxidation reaction because it loses electrons and oxidation is loss of electrons.

\(\ce{Cu(OH)2}(s)+\ce{H^1+}(aq)+\ce{e^1-}(aq)⟶\ce{Cu(OH)}(s)+\ce{H2O}(l)\)

You can see this is a reduction process because it gains an electron and reduction is the gain of electrons.

2. \(\ce{Mn(OH)2}\) is oxidized because Mn goes from an oxidation state of +2 to +4 when it goes to the product of \(\ce{MnO2}\). So it also acts as a reducing agent. \(\ce{O2}\) is reduced from where oxygen has an oxidation state of 0 to an oxidation state of -2. Since it is reduced, it also acts as an oxidizing agent for the \(\ce{MnO2}\) species.

If you write the half reactions

\(\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)+\ce{2H+}(aq)+\ce{2e-}(aq)\)

You can see that this is an oxidation reaction because it loses electrons.

3. H₂ is oxidized as it goes from an oxidation state of 0 to a state of +1 and it acts as a reducing agent. \(\ce{NO3-}\) is reduced because nitrogen goes from an oxidation state of +5 to +2. Therefore it is an oxidizing agent to H₂ as well.

\(\ce{NO3-}(aq)+\ce{4H+}(aq)+\ce{3e-}(aq)⟶\ce{NO}(g)+\ce{2H2O}(l)\)

You can see that since this half gains electrons, it is a reduction reaction half of the reaction.

4. \(\ce{Al}\)(s) is the oxidized species and thus the reducing agent because Aluminum goes from an oxidation state of 0 to an oxidation state of 3+. \(\ce{CrO4^2-}\)(aq) is the reduced species and the oxidizing agent because Cr does from an oxidation state of +6 to a state of +4.

If you write the half reactions

\(\ce{Al}(s)+\ce{3H2O}(l)⟶\ce{Al(OH)3}(s)+\ce{3H+}(aq)+\ce{3e-}(aq)\)

You can see this half loses electrons which means it is an oxidation reaction since oxidation is loss of electrons.

\(\ce{CrO4^2-}(aq)+\ce{4H+}(aq)+\ce{3e-}(aq)⟶\ce{Cr(OH)4-}(aq)\)

This half reaction gains electrons so it is a reduction reaction since reduction is the gain of electrons.

Q19.1.7

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

S19.1.7

Mo because Zr has an oxidation state of +4 and Nb has a oxidation state of +5 and those would not balance out the charge of 3 oxygens in the state of -2 which creates a charge of -6. Mo however has multiple oxidation states, the most common being +6 which balances out the -6 charge created by 3 oxygen ions. This is why its most likely to form an oxide with the formula MO₃ or \(\ce{MoO3}\).

Q19.2.9

Predict whether the carbonate ligand \(\ce{CO3^2−}\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.

S19.2.9

\(\ce{CO3^2−}\) can be either monodentate or bidentate, since two of its oxygen atoms have lone pairs as shown above and can form covalent bonds with a transition metal ion. In most cases carbonate is monodentate because of its trigonal planar geometry (there is 120 degrees between the oxygens so it's hard for both to bind to the same metal). However, in some cases it will bind to two different metals, making it bidentate.

Q12.3.22

The following data have been determined for the reaction:

I^-+OCl^-⟶IO^-+Cl^-

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

S12.3.22

First, the set of the equation depends on the concentration of [I^-] and [OCl^-] as they are tested. This would give the formula for the rate equation to equal some rate constant times these concentrations to some degree.

rate = k [I^-]^m[OCl^-]ⁿ where m and n are some number. To determine this number, we can look the experimental results. Between trial 1 and trial 2, the only experimental change is [I^-] which doubles. Looking at the rate and how that changes, it is seen that it changes by increasing by a factor of 2.033. If this is rounded down to 2, this would mean m=1, since if the rate increases by the same factor as [I^-], this means that the exponent it is raised to is 1.

In the third trial, the difference between the first trial is that [I^-] is increased by a factor of 3 and [OCl^-] decreases by a factor of 5. Since the exponent for [I^-] has already been determined as 1, if you divide the rate (1.83x10^-4) by 3 you can determine the exponent for [OCl^-]. 1.83x10^-4/3 is 6.1x10^-5. Then you take this number and divide the trial one rate. (3.05 × 10^-4)/(6.1x10^-5) is 5. Since trial 1 rate is 5 times greater than trial 3 (without factoring in I^- concentration), and the OCl^- concetration is 5 times greater in trial 1 as well, the exponent for this is also determined to be 1.

Therefore the rate law = rate = k [I^-][OCl^-]. From here, if you choose any trial, and plug in the values, you can determine k, the rate constant. Trial one gives 3.05 × 10^-4 = k (.1)(.05) so if you solve for k it equals .061 L/mol-s.

Q12.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

S12.7.2

https://chem.libretexts.org/LibreTex...9%3A_Catalysis

In homogeneous catalysts, the catalyst is in the same phase with the reactants of the reaction. The number of collision that occur between the catalysts and the reactants are at a maximum, because the catalyst can blend through the reaction mixture uniformly in the same state. Many can only be used at lower temperatures because they tend to decompose in solutions.

On the other hand, heterogeneous catalysts are in a different phase than the reacts. This means that one of the reacts has to interact in a process called adsorption so that the chemical bond in the reactant becomes weak enough to break. When a reactant adsorbs to the catalytic surface, the bonds can break and form new bonds that create the product. Then the products can desorb from the surface. It does not have a maximum collision interaction like in homogeneous catalysts because the phase is not the same.

Q21.4.25

A \[^{8}_{5}\text{B}\] atom (mass = 8.0246 amu) decays into a \[^{8}_{4}\text{B}\] 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.

S21.4.25

The equation can be written out as \[^{8}_{5}\text{B} →^{8}_{4}\text{B}+^{0}_{1}\text{e} \] and if you look at the mass on both sides, 8.0246 amu - 8.0053 amu - .00055 amu = .01875 amu unaccounted for so was released in energy.

https://chem.libretexts.org/LibreTex...uclear_Fission

Then we take the equation E=mc² which is used to solve for energy. The mass should be converted to kg, using the conversion factor 1 amu = 1.6605 X 10^-27 kg. You multiply this number by .01875 amu. This gives 3.1134 X 10^-29 kg. Then you take the c constant which equals 3 X 10⁸ m/s (also known as the speed of light). You plug these numbers in the equation to solve for the energy in joules because squaring the speed of light gives the units m²/s² and the mass units are kg so kg*m²/s² which are the units for joules.

(3.1134 x 10^-29 kg) (2.998 x 10⁸ m/s)²= 2.798 x 10^-12 J

Then the conversion factor from J to MeV (as is found in Worksheet 10) is 1.6022 x 10^-13J = 1 MeV. If you compute (2.798 x 10^-12 J )/ (1.6022 x 10^-13J) = 17.46 MeV. So this is approximately 17.5 millions of electron volts.

Q20.3.12

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)∣PbSO₄(s)∣SO₄²⁻(aq)∥Cu²⁺(aq)∣Cu(s)
2. Hg(l)∣Hg₂Cl₂(s)∣Cl⁻(aq) ∥ Cd²⁺(aq)∣Cd(s)

S20.3.12

These reactions are written in cell notation so based on that we can see that to the left of the double verticle lines, those are the species involved in the oxidation half-reaction, and to the right of the lines, those species are involved in the reduction reaction.

We can refer to a standard reduction table for the half reactions and number of electrons involved:

https://chem.libretexts.org/Core/Ana.../Voltaic_Cells

a. \(Pb(s) + SO4^{2-}(aq) \rightarrow PbSO_4(s)+ 2e^-(aq)\)

This is the spontaneous oxidation reaction that takes place a the anode, because oxidation reactions happen at the anode where electrons are released and move towards the cathode.

\(Cu^2(aq) + 2e^-(aq)\rightarrow Cu(s)\) This is the spontaneous reduction reaction which occurs at the cathode as it gains electrons that have moved from the anode.

The overall reaction would be

\(Pb(s)+ SO4^{2-}(aq)+Cu^2(aq) \rightarrow Cu(s)+PbSO_4(s)\)

The number of electrons on each side cancel out so there is no need to multiple by a factor.

b. \(2Hg(l) + 2Cl^-(aq) \rightarrow Hg_2Cl_2(s) + 2e^-(aq)\)

This is the spontaneous oxidation reaction because it is on the left of the double lines as written. You can see that it loses electrons like oxidation reactions do, and these electrons will flow to the cathode. This oxidation reaction is occurring at the anode.

\(Cd^2+(aq) + 2e^-(aq)\rightarrow Cd(s)\)

This is the spontaneous reduction reaction happening at the cathode. It gains electrons that have moved from the anode. Since it gains electrons, it is a reduction reaction.

The overall reaction is

\(2Hg(l) + 2Cl^-(aq) +Cd^2+(aq) \rightarrow Cd(s) +Hg_2Cl_2(s) \)

The number of electrons on each side cancel out so there is no need to multiple by a factor.

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

\[NAD^++H^++2e^-\rightarrow NADH (E^o=-0.32V)\]

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?

S17.2.2

In a standard reduction potential table, a species with lower E° means it's a stronger reducing agent (more spontaneous at oxidizing). And higher E^o means it's stronger oxidizing agent(more spontaneous at reducing).

https://chem.libretexts.org/LibreTex...ode_Potentials

a. No, NADH would not reduce acetate to pyruvate. The reaction for acetate to pyruvate has a lower E° of -0.70 V in comparison to the -0.32 V for the NADH half-reaction. This means that the acetate is a stronger reducing agent than NADH, and would be more likely to reduce that.

b. Yes NADH could reduce pyruvate to lactate because the E° of NADH is smaller and therefore is a stronger reducing agent than pyruvate so it is able to reduce it spontaneously.

c. In order to convert acetate to lactase, two reactions spontaneously occur and give the E = -0.70 V - .0185 V = -0.885 V. This means that in order for a cell to cause both of these reduction reactions to occur, the cell potention of -0.885 V or lower is needed for it to occur.