Extra Credit 9

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Problem 17.1.18

Identify the species that were oxidized, the species that were reduced, the oxidizing agent, and the reducing agent in each of the reactions below:

a) SO₃^2-(aq) + Cu(OH)₂(s) --> SO₄^2-(aq) + Cu(OH)(s)

b) O₂(g) + Mn(OH)₂(s) --> MnO₂(s)

c) NO₃^-(aq) + H₂(g) --> NO(g)

d) Al(s) + CrO₄^2-(aq) --> Al(OH)₃(s) + Cr(OH)₄^-(aq)

Answer:

a) SO₃^2-(aq) + Cu(OH)₂(s) --> SO₄^2-(aq) + Cu(OH)(s)

species oxidized: SO₃^2-

species reduced: Cu(OH)₂

oxidizing agent: Cu(OH)₂

reducing agent: SO₃^2-

b) O₂(g) + Mn(OH)₂(s) --> MnO₂(s)

species oxidized: Mn(OH)₂

species reduced: O₂

oxidizing agent: O₂

reducing agent: Mn(OH)₂

c) NO₃^-(aq) + H₂(g) --> NO(g)

species oxidized: H₂

species reduced: NO₃^-

oxidizing agent: NO₃^-

reducing agent: H₂

d) Al(s) + CrO₄^2-(aq) --> Al(OH)₃(s) + Cr(OH)₄^-(aq)

species oxidized: Al

species reduced: CrO₄^2-

oxidizing agent: CrO₄^2-

reducing agent: Al

Tutorial:

a) SO₃^2-(aq) + Cu(OH)₂(s) --> SO₄^2-(aq) + Cu(OH)(s)

Assign oxidation numbers to reactants
- O in SO₃^2-has an oxidation number of -2, and S in SO₃^2-has an oxidation number of +4.
- O in Cu(OH)₂has an oxidation number of -2, H in Cu(OH)₂ has an oxidation number of +1, and Cu in Cu(OH)₂ has an oxidation number of +2
Assign oxidation numbers to products
- O in SO₄^2-has an oxidation number of -2, and S in SO₄^2- has an oxidation number of +6.
- O in Cu(OH) has an oxidation number of -2, H in Cu(OH) has an oxidation number of +1, and Cu in Cu(OH) has an oxidation number of +1.
To find the species that is oxidized, determine which oxidation number increases. An increase in oxidation number indicates that an electron has been lost (increase in positive charge), and therefore indicates oxidation. In this example, S goes from an oxidation number of +4 (in the SO₃^2-compound) to an oxidation number of +6 (in the SO₄^2- compound) and is thus oxidized.
To find the species that is reduced, determine which oxidation number decreases. A decrease in oxidation number indicates that an electron has been gained by the species (increase in negative charge), and therefore indicates reduction. In this example, Cu (in the Cu(OH)₂compound) goes from an oxidation number of +2 to an oxidation number of +1 (in the Cu(OH) compound). It is therefore reduced.
The species that is oxidized is also known as the reducing agent. Because the oxidized species (in this example SO₃^2-) loses electrons, it "induces" reduction in another species, thus acting as the reducing agent.
The species that is reduced is also known as the oxidizing agent. Because the reduced species (in this example Cu(OH)₂) gains electrons, it "induces" oxidation, or the loss of electrons, in another species, thus acting as an oxidizing agent.

b) O₂(g) + Mn(OH)₂(s) --> MnO₂(s)

Assign oxidation numbers to reactants
- O₂ has an oxidation number of 0 because it is a pure element.
- O in Mn(OH)₂ has an oxidation number of -2, H in Mn(OH)₂ has an oxidation number of +1, and Mn in Mn(OH)₂ has an oxidation number of +2.
Assign oxidation numbers to products
- O in MnO₂has an oxidation number of -2, and Mn in MnO₂ has an oxidation number of +4.
To find the species that is oxidized, determine which oxidation number increases. An increase in oxidation number indicates that an electron has been lost (increase in positive charge), and therefore indicates oxidation. In this example, Mn goes from an oxidation number of +2 (in the Mn(OH)₂compound) to an oxidation number of +4 (in the MnO₂ compound) and is thus oxidized.
To find the species that is reduced, determine which oxidation number decreases. A decrease in oxidation number indicates that an electron has been gained by the species (increase in negative charge), and therefore indicates reduction. In this example, O (in the O₂compound) goes from an oxidation number of 0 to an oxidation number of -2 (in the MnO₂compound). O₂is therefore reduced.
The species that is oxidized is also known as the reducing agent. Because the oxidized species (in this example Mn(OH)₂) loses electrons, it "induces" reduction in another species, thus acting as the reducing agent.
The species that is reduced is also known as the oxidizing agent. Because the reduced species (in this example O₂) gains electrons, it "induces" oxidation, or the loss of electrons, in another species, thus acting as an oxidizing agent.

c) NO₃^-(aq) + H₂(g) --> NO(g)

Balance the equation.
- Separate the equation into balanced half reactions.
  - 5e^- + 6H⁺ + NO₃^- --> NO + 3H₂O
  - H₂ --> 2H⁺ + 2e^-
- Find the least common multiple of the coefficients of the electrons in the two equations, and multiply each equation by a coefficient so that the coefficient in front of the electron in both equations is the same.
  - (5e^- + 6H⁺ + NO₃^- --> NO + 3H₂O) x 2
  - (H₂ --> 2H⁺ + 2e^-) x 5
- Add the equations together and simplify.
  - 2H⁺ + 2NO₃^- + 5H₂--> 2NO + 6H₂O
- To balance in a basic condition, add 2OH^- to both sides of the equation. Note the OH^-reacts with the H⁺to produce two water molecules.
  - 2H₂O + 2NO₃^- + 5H₂--> 2NO + 6H₂O + 2OH^-
  - simplify: 2NO₃^- + 5H₂--> 2NO + 4H₂O + 2OH^-
For purposes of this tutorial, we will used the balanced equation in basic conditions: 2NO₃^- + 5H₂--> 2NO + 4H₂O + 2OH^-
Assign oxidation numbers to the reactants.
- O in NO₃^- has an oxidation number of -2, and N in NO₃^- has an oxidation number of +4.
- H in H₂has an oxidation number of 0 because it is a pure element.
Assign oxidation numbers to the products.
- O in NO has an oxidation number of -2, and N in NO has an oxidation number of +2.
- H in H₂O has an oxidation number of +1, and O in H₂O has an oxidation number of -2.
- H in OH^- has an oxidation number of +1, and O in OH^- has an oxidation number of -2.
To find the species that is oxidized, determine which oxidation number increases. An increase in oxidation number indicates that an electron has been lost (increase in positive charge), and therefore indicates oxidation. In this example, H goes from an oxidation number of 0 (in the H₂compound) to an oxidation number of +1 (in the water molecule and OH^-ion) and H₂is thus oxidized.
To find the species that is reduced, determine which oxidation number decreases. A decrease in oxidation number indicates that an electron has been gained by the species (increase in negative charge), and therefore indicates reduction. In this example, N (in the NO₃^- compound) goes from an oxidation number of +4 to an oxidation number of +2 (in the NOcompound). NO₃^-is therefore reduced.
The species that is oxidized is also known as the reducing agent. Because the oxidized species (in this example H₂) loses electrons, it "induces" reduction in another species, thus acting as the reducing agent.
The species that is reduced is also known as the oxidizing agent. Because the reduced species (in this example NO₃^-) gains electrons, it "induces" oxidation, or the loss of electrons, in another species, thus acting as an oxidizing agent.

d) Al(s) + CrO₄^2-(aq) --> Al(OH)₃(s) + Cr(OH)₄^-(aq)

Assign oxidation numbers to the reactants
- Al in Al(s) has an oxidation number of 0 because it is a pure element.
- O in CrO₄^2- has an oxidation number of -2, and Cr in CrO₄^2- has an oxidation number of +6.
Assign oxidation numbers to the products
- O in Al(OH)₃ has an oxidation number of -2, H in Al(OH)₃ has an oxidation number of +1, and Al in Al(OH)₃has an oxidation number of +3.
- O in Cr(OH)₄^-has an oxidation number of -2, H in Cr(OH)₄^- has an oxidation number of +1, and Cr in Cr(OH)₄^-has an oxidation number of +3.
To find the species that is oxidized, determine which oxidation number increases. An increase in oxidation number indicates that an electron has been lost (increase in positive charge), and therefore indicates oxidation. In this example, Al goes from an oxidation number of 0 (as Al(s)) to an oxidation number of +3 (in the Al(OH)₃ compound) and Al(s) is thus oxidized.
To find the species that is reduced, determine which oxidation number decreases. A decrease in oxidation number indicates that an electron has been gained by the species (increase in negative charge), and therefore indicates reduction. In this example, Cr (in the CrO₄^2-compound) goes from an oxidation number of +6 to an oxidation number of +3 (in the Cr(OH)₄^-compound). CrO₄^2-is therefore reduced.
The species that is oxidized is also known as the reducing agent. Because the oxidized species (in this example Al(s)) loses electrons, it "induces" reduction in another species, thus acting as the reducing agent.
The species that is reduced is also known as the oxidizing agent. Because the reduced species (in this example CrO₄^2-) gains electrons, it "induces" oxidation, or the loss of electrons, in another species, thus acting as an oxidizing agent.

Problem 19.1.7

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

Answer: Mo

Tutorial:

Mo has 4 electrons in its d-orbital (this can be determined simply by observing its specific placement in the d-block). It can form a +2 ion (by losing its two 5s electrons), +3 ions (by losing its two 5s electrons and losing one 4d electron), +4 ions (by losing its two 5s electrons and losing two 4d electrons), +5 ions (by losing its two 5s electrons and losing three 4d electrons), and +6 ions (by losing its two 5s electrons and losing four 4d electrons).
Zr only has 2 electrons in its d-orbital, and can therefore form only +2, +3, and +4 ions.
Nb has only 3 electrons in its d-orbital, and can therefore form only +2, +3, +4, and +5 ions.
The compound MO₃ contains 3 O^2- ions, producing a -6 charge. Therefore, the M ion in this molecule must be +6 in order for the MO₃ compound to be neutral (which it is).
Therefore, Mo is the only element that could possibly be M in MO₃ because it is the only element that can form a +6 ion.

Problem 19.2.9

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

Answer: Monodentate

Tutorial:

Monodentate ligands "bite" the metal center in only one place. Polydentate ligands "bite" the metal center in multiple locations. Polydentate ligands have multiple atoms that care able to donate a pair of electrons to the metal center, however, polydentate ligands must also possess the correct orientation and distance between the two binding atoms for binding in multiple locations on the metal center. In the carbonate ion, the distance between the two oxygen atoms is too short for chelation to occur.

Problem 12.3.22

The following data have been determined for a reaction:

I^-+ OCl^- --> IO^- + Cl^-

	1	2	3
[I^-]_initial (M)	0.10	0.20	0.30
[OCl^-]_initial(M)	0.050	0.050	0.010
Rate (mol/L/s)	3.05 x 10^-4	6.20 x 10^-4	1.83 x 10^-4

Determine the rate equation and rate constant for this reaction.

Answer:

rate = k[I^-][OCl^-]; k = .061 M^-1s^-1

Tutorial:

Use the data provided to determine n and m, the order of I^-and OCl^-respectively. This will be done by constructing rate laws based on the data in the table and dividing the two equations.
- Equation 1A: 3.05 x 10^-4 = k[.1]ⁿ[.05]^m
- Equation 2A: 6.20 x 10^-4 = k [.2]ⁿ[.05]^m
- Divide Equation 1A by Equation 2A.
  - .492 = [.5]ⁿ
  - Because .492 is approximately equal to .5, n=1, and the reaction is first order in I^-
- Equation 1B: 6.02 x 10^-4 = k [.2]¹[.05]^m
- Equation 2B: 1.83 x 10^-4 = k [.3]¹[.01]^m
- Divide Equation 1B by Equation 2B.
  - 3.39 = .667 [5]^m
  - 5.08 = [5]^m
  - Because 5.08 is approximately equal to 5.08, m = 1, and the reaction is first order in OCl^-
- rate = k[I^-][OCl^-]
Use the data provided to determine k. This is done by plugging in concentration values and the rate value and solving for the unknown k.
- Using data set 3:
  - 1.83 x 10^-4= k [.3][.01]
  - k = .061 M^-1s^-1
  - Units are determined by noting that the reaction is second order overall. Because both reactants have units in M, and because the desired units are M/s, k must be in M^-1s^-1to yield M/s overall.

Problem 12.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

Answer: Heterogeneous catalysts are in a different physical state from the reactants. Heterogeneous catalysts adsorb a reactant at its (referring to the heterogeneous catalyst) catalytic active site. The heterogeneous catalyst induces the bonds in the reactant to become weak and ultimately break. The reaction this occurs through a process of adsorption (not to be confused with absorption, which is when one substance is incorporated into another substance's physical structure). Homogeneous catalysts are in the same physical state as the reactants. Generally, the homogeneous catalysts can be found evenly dispersed in the reactant. Both homogeneous and heterogeneous catalysts increase the rate of a reaction by providing an alternate mechanism with a lower activation energy. This alternate mechanism may have more steps than the original mechanism, but the activation energy of the rate-determining step is lowered by the catalyst, homogenous or heterogeneous.

Problem 21.4.25

A boron-8 atom (mass = 8.0246 amu) decays into a beryllium-8 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?

Answer: 17.5 MeV

Tutorial:

Calculate the mass defect of the reaction. The mass defect, synonymous with the term "nuclear binding energy," is defined as the energy required to bind the nucleus or the mass released by the loss of particles in a particular reaction.
- Mass of reactant (boron-8): 8.0246 amu
- Mass of products (beryllium-8 + positron): 8.0053 amu + .00055 amu = 8.00585 amu
- Mass defect = mass_products - mass_reactants = 8.00585 amu - 8.0246 amu = -0.01875 amu
Convert the value of change in mass into MeV (electronvolts).
- (-0.01875 amu)(931 MeV/amu) = -17.5 MeV
- An answer of -17.5 MeV means that 17.5 MeV are produced by this reaction.

Problem 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.

a) Pb(s) ∣ PbSO₄(s) ∣ SO₄^2-(aq) ∥ Cu²⁺(aq) ∣ Cu(s)

b) Hg(l) ∣ Hg₂Cl₂(s) ∣ Cl^-(aq) ∥ Cd²⁺(aq) ∣ Cd(s)

Answer:

Equation a):

Anode half-reaction (oxidation): Pb(s) + SO₄^2-(aq) --> PbSO₄(s) + 2e^-
Cathode half-reaction (reduction): Cu²⁺(aq) + 2e^- --> Cu(s)
Overall reaction: Pb(s) + SO₄^2-(aq) + Cu²⁺(aq) --> PbSO₄(s) + Cu(s)

Equation b):

Anode half-reaction (oxidation): Cd(s) --> Cd²⁺(aq) + 2e^-
Cathode half-reaction (reduction): Hg₂Cl₂(s) + 2e^- --> 2Hg(l) + 2Cl^- (aq)
Overall reaction: Cd(s) + Hg₂Cl₂(s) --> Cd²⁺(aq) + 2Hg(l) + 2Cl^- (aq)

Tutorial:

Equation a):

Determine the standard cathode (reduction) half reactions in the proposed cell diagram. Using a standard reduction potential value chart, determine the standard reduction potential for each reduction half reaction.
- PbSO₄(s) + 2e^---> Pb(s) + SO₄^2-(aq) E° = -.356V
- Cu²⁺(aq) + 2e^- --> Cu(s) E° = .3419V
Compare the standard reduction potentials for the two reduction half-reactions. The reduction half-reaction with the smaller E° value will be the oxidation half-reaction (anode) and the reduction half-reaction with the larger E° value will be the reduction half-reaction (cathode).
- PbSO₄(s) + 2e^---> Pb(s) + SO₄^2-(aq) E° = -.356V (anode, oxidation)
- Cu²⁺(aq) + 2e^- --> Cu(s) E° = .3419V (cathode, reduction)
Reverse the reaction determined to be the oxidation half-reaction, and write the two half-reaction equations, one above the other.
- Pb(s) + SO₄^2-(aq) --> PbSO₄(s) + 2e^-
- Cu²⁺(aq) + 2e^- --> Cu(s)
If the number of electrons in the products and reactants does not cancel out, multiply by a coefficient so that the electrons cancels out. Then, add the equations together to obtain the overall equation.
- Pb(s) + SO₄^2-(aq) + Cu²⁺(aq) --> PbSO₄(s) + Cu(s)

Equation b):

Determine the standard cathode (reduction) half reactions in the proposed cell diagram. Using a standard reduction potential value chart, determine the standard reduction potential for each reduction half reaction.
- Cd²⁺(aq) + 2e^---> Cd(s) E° = -0.4030V
- Hg₂Cl₂(s) + 2e^- --> 2Hg(l) + 2Cl^- (aq) E° = .2682V
Compare the standard reduction potentials for the two reduction half-reactions. The reduction half-reaction with the smaller E° value will be the oxidation half-reaction (anode) and the reduction half-reaction with the larger E° value will be the reduction half-reaction (cathode).
- Cd²⁺(aq) + 2e^---> Cd(s) E° = -0.4030V (anode, oxidation)
- Hg₂Cl₂(s) + 2e^- --> 2Hg(l) + 2Cl^- (aq) E° = .2682V (cathode, reduction)
Reverse the reaction determined to be the oxidation half-reaction, and write the two half-reaction equations, one above the other.
- Cd(s) --> Cd²⁺(aq) + 2e^-
- Hg₂Cl₂(s) + 2e^- --> 2Hg(l) + 2Cl^- (aq)
If the number of electrons in the products and reactants does not cancel out, multiply by a coefficient so that the electrons cancels out. Then, add the equations together to obtain the overall equation.
- Cd(s) + Hg₂Cl₂(s) --> Cd²⁺(aq) + 2Hg(l) + 2Cl^- (aq)

Problem 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.

a) Would NADH be able to reduce acetate to pyruvate?

b) Would NADH be able to reduce pyruvate to lactate?

c) What potential is needed to convert acetate to lactate?

acetate + CO₂ + 2H⁺ +2e^- → pyruvate +H₂O E° = −0.70 V

pyruvate + 2H⁺ + 2e^- → lactate E° = −0.185 V

Answer:

a) no

b) yes

c) E° = -0.44 V

Tutorial:

For a):

If NADH is reducing acetate to pyruvate, NADH itself is oxidized. The equation involving NADH is this the oxidation half-reaction, and thinking of this situation as an electrochemical cell, this half-reaction would occur at the "anode."
Write the relevant reactions.
- NAD⁺ + H⁺ + 2e^- → NADH; E° = −0.32 V (anode, oxidation)
- acetate + CO₂ + 2H⁺ +2e^- → pyruvate +H₂O E° = −0.70 V (cathode, reduction)
Using the equation E°_cathode - E°_anode, we can find the overall E° value for the cell.
- E°_cathode - E°_anode = -.70 - (-.32) = -.70 + .32 = -.38V
Because the overall E° value for the cell is negative, we know that the reduction of acetate to pyruvate by NADH is not spontaneous.
- delta G° = -nFE°_cell
- delta G° > 0 if E°_cellis negative
- Because -.38 V (E°_cell) is negative, delta G° > 0 and the reaction is not spontaneous.

For b):

If NADH is reducing pyruvate to lactate, NADH itself is oxidized. The equation involving NADH is this the oxidation half-reaction, and thinking of this situation as an electrochemical cell, this half-reaction would occur at the "anode."
Write the relevant equations.
- NAD⁺ + H⁺ + 2e^- → NADH; E° = −0.32 V (anode, oxidation)
- pyruvate + 2H⁺ + 2e^- → lactate E° = −0.185 V (cathode, reduction)
Using the equation E°_cathode - E°_anode, we can find the overall E° value for the cell.
- E°_cathode - E°_anode = -0.185 - (-0.32) = -0.185 + 0.32 = .135 V
Because the overall E° value for the cell is positive, we know that the reduction of pyruvate to lactate by NADH is spontaneous.
- delta G° = -nFE°_cell
- delta G° < 0 if E°_cellis positive
- Because .135 V (E°_cell) is positive, delta G° < 0 and the reaction is spontaneous.

For c):

Finding the potential needed to convert acetate to lactate requires summing two half reactions. However, in order to do this, one cannot simply sum the reduction potentials of the half-reactions.
First, find the deltaG° of each half reaction using the equation deltaG° = -nFE°where n is the moles of electrons, F is Faraday's constant (96485 C/mol electrons) and E° is the potential for the half-reaction.
- acetate + CO₂ + 2H⁺ +2e^- → pyruvate +H₂O E° = −0.70 V
  - deltaG° = -nFE° = -(2 mol e^-)(96485 C/mol e^-)(-.70V) = 135 kJ
- pyruvate + 2H⁺ + 2e^- → lactate E° = −0.185 V
  - deltaG° = -nFE° = -(2 mol e^-)(96485 C/mol e^-)(-0.185) = 36 kJ
Next, add the deltaG° values.
- 135 kJ + 36 kJ = 171 kJ
Using the equation deltaG°_final = -nFE°_cellsolve for E°_cell.
- 171000 J = -nFE°_cell= -(4 mol e^-)(96485 C/mol e^-)(E°_cell)
  - Note that n = 4 because in the overall reaction, 4 moles of electrons are transferred (see below)
    - acetate + CO₂ + 2H⁺ +2e^- → pyruvate +H₂O
    - pyruvate + 2H⁺ + 2e^- → lactate
    - overall reaction: acetate + CO₂ + 4H⁺ +4e^---> lactate + H₂O
- E°_cell= -0.44 V
- A negative E°_cellvalue indicates a non-spontaneous reaction, so .44 V is needed to convert acetate to lactate.

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