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Extra Credit 8

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    Q17.1.7

    Balance the following in basic solution

    a. SO32-(aq)+Cu(OH)2(s)→SO42-(aq)+Cu(OH)(s)

    b. O2(g)+Mn(OH)2(s)→MnO2(s)

    c. NO3-(aq)+H2(g)→NO(g)

    d. Al(s)+CrO42-(aq)→Al(OH)3(s)+Cr(OH)4-(aq)

    Answer:

    a. SO32-(aq)+Cu(OH)2(s)→SO42-(aq)+Cu(OH)(s)

    1. First separate out into two half reaction based on the oxidized and reduced species by looking at the oxidation states of the elements

    oxidation: SO32-(aq)→SO42-(aq) reduction: Cu(OH)2(s)→Cu(OH)(s)

    2. Next balance the oxygen by adding water

    SO32-(aq) + H2O(l)→SO42-(aq) Cu(OH)2(s)→Cu(OH)(s) + H20(l)

    3. Add hydrogen ions to balance the hydrogens

    SO32-(aq) + H2O(l)→SO42-(aq) +2H+ Cu(OH)2(s) + H+→Cu(OH)(s) + H20(l)

    4. Add electrons to balance the charge on both sides

    SO32-(aq) + H2O(l)+ →SO42- +2H+ + 2e- Cu(OH)2(s)+ H++ e- →Cu(OH)(s) + H20(l)

    Note: make sure to scale up the reactions in order to get the same number of electrons in both reactions in order to cancel them out when you combine the reactions

    2Cu(OH)2(s) + 2H+ + 2e-→2Cu(OH)(s) + 2H20(l)

    5. Add the two half reactions and cancel out common terms on both sides

    SO32-(aq)+2Cu(OH)2(s)→SO42-(aq)+2Cu(OH)(s)+H2O(l)

    6. Add OH- to both sides of the reaction in order to create water with the H+ ions because this is in a basic solution. But in this problem, the H+ already cancelled out when the half reactions were added. Therefore, your answer is

    SO32-(aq)+2Cu(OH)2(s)→SO42-(aq)+2Cu(OH)(s)+H2O(l)

    b. O2(g)+Mn(OH)2(s)→MnO2(s) (Follow the same steps as above)

    1. Separate the half reactions (for the reduction half reaction, you have to put a water on right side because oxygen gas is reduced to water; you know water will be a product because that is the only way to balance the oxygens in the overall reaction)

    reduction: O2(g)→H2O(l) oxidation: Mn(OH)2(s)→MnO2(s)

    2. Add water to balance the oxygen

    O2(g)→2H2O(l) Mn(OH)2(s)→MnO2(s)

    3. Add hydrogen ions to balance the hydrogens

    O2(g)+4H+→2H2O(l) Mn(OH)2(s)→MnO2(s)+2H+

    4. Add electrons to balance the charge

    O2(g)+4H+→2H2O(l) +4e- Mn(OH)2(s)→MnO2(s)+2H++2e-

    Scale up the reactions

    O2(g)+4H+→2H2O(l) +4e- 2Mn(OH)2(s)→2MnO2(s)+4H++4e-

    5. Add the half reactions

    O2(g)+2Mn(OH)2(s)→2MnO2(s)+2H2O(l)

    6. Once again there is no need to add OH- because the H+ cancelled out when the half reactions were added.

    Therefore the answer is

    O2(g)+2Mn(OH)2(s)→2MnO2(s)+2H2O(l)

    c. NO3-(aq)+H2(g)→NO(g)

    1. Separate the half-reactions.

    oxidation: H2(g)→H+ reduction: NO3-(aq)→NO(g)

    2. Add water to balance oxygen

    H2(g)→H+ NO3-(aq)→NO(g)+2H2O(l)

    3. Add H+ to balance hydrogen

    H2(g)→2H+ NO3-(aq)+4H+→NO(g)+2H2O(l)

    4. Balance charge with electrons

    H2(g)→2H++2e- NO3-(aq)+4H++3e-→NO(g)+2H2O(l)

    Scale up the reactions

    3H2(g)→6H++6e- 2NO3-(aq)+8H++6e-→2NO(g)+4H2O(l)

    5. Add the half reactions

    3H2(g)+2NO3-(aq)+2H+→2NO+4H2O(l)

    6. Add OH- to make water with the H+

    3H2(g)+2NO3-(aq)+2H++2OH-→2NO+4H2O(l)+2OH-

    ----->3H2(g)+2NO3-(aq)+2H2O→2NO+4H2O(l)+2OH- (form the water)

    ----->3H2(g)+2NO3-(aq)→2NO+2H2O(l)+2OH- (cancel out the common water molecules on both sides)

    Therefore, the answer is

    3H2(g)+2NO3-(aq)→2NO+2H2O(l)+2OH-

    d. Al(s)+CrO42-(aq)→Al(OH)3(s)+Cr(OH)4-(aq)

    1. Separate the half reactions.

    oxidation: Al(s)→Al3+ reduction: CrO42-(aq)→Cr3+(aq)

    2. Add water to balance oxygen.

    Al(s)→Al3+(aq) CrO42-(aq)→Cr3+(aq)+4H2O(l)

    3. Add H+ to balance hydrogen.

    Al(s)→Al3+(aq) CrO42-(aq)+8H+→Cr3++4H2O(l)

    4. Balance charge with electrons

    Al(s)→Al3++3e- CrO42-(aq)+8H++3e-→Cr3+(aq)+4H2O(l)

    Scale up the reactions (not needed)

    Al(s)→Al3++3e- CrO42-(aq)+8H++3e-→Cr3+(aq)+4H2O(l)

    5. Add the half reactions

    Al(s)+CrO42-(aq)+8H+→Al3++Cr3++4H2O(l)(aq)

    6. Add OH- to form water with H+

    Al(s)+CrO42-(aq)+8H++8OH-→Al3+(aq)+Cr3+(aq)+4H2O(l)+8OH-

    -----> Al(s)+CrO42-(aq)+4H2O(l)→Al3+(aq)+Cr3+(aq)+8OH- (form the water and cancel common water molecules on both sides)

    Q19.1.6

    Which of the following is the strongest oxidizing agent: VO43-, CrO42-, or MnO4-?

    Answer: MnO4- is the strongest oxidizing agent because Mn has an oxidation state of +7 which means that it has been oxidized as much as it can. Now it can only be reduced by oxidizing other species. Mn has a highest oxidation state of +7 because oxygen has an O.S. of -2. Since there are 4 oxygens, 4(-2)=-8. In order to balance out the charge to get an overall charge of -1, the O.S. of Mn must be +7 because +7+(-8)=-1. The oxidation state of V is +5 and of Cr is +6 which can be determined using the same method. Since Mn has the highest oxidation state, it is the best oxidizing agent since it can accept the most electrons.

    Q19.2.8

    Specify whether the following complexes have isomers.

    1. tetrahedral [Ni(CO)2(Cl)2]
    2. trigonal bipyramidal [Mn(CO)4NO]
    3. [Pt(en)2Cl2]Cl2

    Answer:

    1. [Ni(CO)2(Cl)2]

    -ionization isomer:none because there is not an ion on the outside of the coordination sphere that could switch with one on the inside

    -coordination isomer: none because there is only one coordination complex so ligands can't switch between an anion and cation

    -linkage isomer: none because there is no ambidentate ligand present in order to bind differently

    -geometric isomer: none because it is tetrahedral so you can't get cis (90 degree) and trans (180 degree) angles between the two types of ligands.

    -optical isomer: none because there is a plane of symmetry since at least two of the ligands are the same

    2. trigonal bipyramidal [Mn(CO)4NO]

    -ionization isomer:none because there is not an ion on the outside of the coordination sphere that could switch with one on the inside

    -coordination isomer: none because there is only one coordination complex so ligands can't switch between an anion and cation

    -linkage isomer: none because there is no ambidentate ligand present in order to bind differently

    -geometric isomer: none because geometric isomers don't occur in trigonal bipyramidal geometries

    -optical isomer: none because there is a plane of symmetry

    3. [Pt(en)2Cl2]Cl2

    -ionization isomer:none because the anion ligand present on the inside is the same as the one on the outside

    -coordination isomer: none because there is only one coordination complex so ligands can't switch between an anion and cation

    -linkage isomer: none because there is no ambidentate ligand present in order to bind differently

    -geometric isomer: yes because the chloro ligands can be 900 degrees from each other (cis) or 180 degrees from each other (trans)

    -optical isomer: yes but only in the cis form of the complex because there is no plane of symmetry

    Q12.3.21

    The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.

    1. 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)
    2. 2NO(g)+O2(g)⟶2NO2(g)
    3. 3NO2(g)+H2O(l)⟶2HNO3(aq)+NO(g)

    The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 × 10−6 L2/mol2/s.

    Answer:

    1. Use the rate law in order to solve this problem

    rate=k[NO]2[O2] and (1/2)(delta[NO]2/delta t)=rate

    so

    (1/2)(delta[NO]2/delta t)=rate=k[NO]2[O2]

    (1/2)(delta[NO]2/delta t)=(5.8 x 10-6L2/mol2/s)(0.50)2(.75)=1.0875 x 10-6M/s

    (delta[NO]2/delta t)=2.175 x 10-6 M/s

    This is the rate of formation of NO2.

    Q12.7.1

    Account for the increase in reaction rate brought about by a catalyst.

    Answer:

    A catalyst lowers the Gibbs energy of activation which enhances the rate of forward and backward reactions. The catalyst forms an intermediate with the reactans in the initial step of the reaction and is released during product formation. Catalysts do not affect the enthalpies or the Gibbs energies of the reactants and products. They increase the rate of approach to equilibrium but do not change the equilibrium constant. The mechanism of the reaction changes which can result in multiple intermediates. The catalyst itself is not consumed by the reaction (meaning it is still present and unchanged even after the products form).

    Q21.4.24

    A 74Be atom (mass = 7.0169 amu) decays into a 73Li atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?

    Answer:

    74Be + 0-1ß→73Li (equation for electron capture reaction)

    You need to use the equation

    deltaE=(delta m)c2

    delta m= 7.0160-7.0169=-.0009 amu

    convert to kg: (-.0009)((1.6605 x 10-27kg)/amu))=-1.49 x 10-30 kg

    deltaE=(-1.49 x 10-30 kg)(2.998 x 108m/s)2=-1.34 x 10-13 J

    Convert to MeV/mol:

    (-1.34 x 10-13 J/nucleus)(1MeV/(1.6022 x 10-13 J)(6.022 x 1023 nuclei/mol)=-5.05x 10-23 MeV/mol

    Q20.3.11

    Sulfate is reduced to HS− in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?

    Answer:

    reduction: SO42−(aq)→ HS−(aq) oxidation: C6H12O6(aq) → HCO3−(g)

    1. Balance elements other than H and O (in this case, the C)

    SO42−(aq)→ HS−(aq) C6H12O6(aq) → 6HCO3−(g)

    2. Balance O by adding water

    SO42−(aq) → HS−(aq) + 4H2O(l) C6H12O6(aq) + 12H2O(l) → 6HCO3−(g)

    3. Balance H by adding H+

    SO42−(aq) + 9H+(aq) → HS−(aq) + 4H2O(l) C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq)

    4. Balance charge with electrons

    SO42−(aq) + 9H+(aq) + 8e− → HS−(aq) + 4H2O(l) C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq) + 24e−

    Scale up the reactions

    3SO42−(aq) + 27H+(aq) + 24e− → 3HS−(aq) + 12H2O(l) C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq) + 24e−

    5. Add the half reactions and cancel common terms

    C6H12O6(aq) + 3SO42−(aq) → 6HCO3−(g) + 3H+(aq) + 3HS−(aq)

    Q20.5.23

    For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?

    Answer:

    O2 (g) + 4H+ + 4e- -> 2H2O (l)

    E=Eo-(.0592V/n)logQ

    n=4 which is the number of electrons transferred

    Eo=1.23 V

    Q=[1]/([H+]4p{O2}) (assume partial pressure of O2 to be 1 bar)

    Q=[1]/[H+]4

    pH: 7 =-log10[H+] so [H+]=10-7

    E=1.23-(.0592V/4)log(1/[10-7]=1.1264 V

    Now for the .85M NaOH solution, get the ph by getting pOH first

    pOH=-log10[OH-]=-log10[.85]=.071

    pH=14-pOH=14-.071=13.93

    pH: 13.93 =-log10[H+] so [H+]=10-13.93

    E=1.23-(.0592V/4)log(1/[10-13.93]=1.024 V

    Extra Credit 8 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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