Extra Credit 8

S 17.1.7 Balance the following in basic solution

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

The first step is to break the reaction into its two separate half reactions, as followed:

SO3^2-(aq)→SO4^2(aq)- (oxidized) and Cu(OH)2(s)→Cu(OH)(s) (Reduced)

After you have broken them up you need to balance the oxygen by adding water:

SO3^2-(aq) + H2O(l)→SO4^2-(aq) and Cu(OH)2(s)→Cu(OH)(s) + H2O(l)

Next you are going to want to balance the hydrogen, you will add it to the side opposite to where you added water:

SO3^2-(aq) + H2O(l)→SO4^2-(aq) +2H+ (aq) and Cu(OH)2(s) + H+(aq)→Cu(OH)(s) + H2O(l)

After you have balanced the H+ ions then you can see the amount of electrons needed, if you lose electrons they will be on the product side and the solution is oxidized and if the solution gains electrons it will show up on the reactants side and will be reduced. The reactants for the oxidized reaction have a charge of 2- and the products with the two hydrogen have a zero charge so two electrons will be added to the product side in order to balance the electrons:

SO3^2-(aq) + H2O(l)+ →SO4^2-(aq) +2H+(aq) + 2e- and Cu(OH)2(s) + H+(aq) + e- →Cu(OH)(s) + H2O(l)

Before we can add the hydroxide in order to make these solutions basic we must first balance the electrons multiply Cu(OH)2 equations by two in order for the two reactions to have 2e-

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

Once the two reactions are balanced you can add the OH, add the OH to balance out the amount of hydrogen you have note you must add it to both sides of the reaction in order to balance it out:

SO3^2-(aq) + H2O(l)+ 2OH-(aq) → SO4^2-(aq) +2H+(aq) + 2OH(aq) + 2e- and 2Cu(OH)2(s) + 2H+(aq) +2OH(aq) + 2e-→2Cu(OH)(s) + 2H2O(l) +2OH-(aq)

The OH- and the H+ will react and form H2O after to can balance and cross out things that appear on both sides to get the final answer:

SO3^2-(aq) + H2O(l)+ + 2OH-(aq) → SO4^2(aq)- + 2H2O(l) + 2e- and 2Cu(OH)2(s) + 2H2O(l) + 2e- → 2Cu(OH)(s) + 2H2O(l) +2OH-(aq)

Finally, once you combine the two equations, you get:

SO3^2-(aq) +2Cu(OH)2(s) → SO4^2-(aq) + 2Cu(OH)(s) + H2O(l)

b.)Follow the same procedure as before to get:

O2(g) + Mn(OH)2(s) → MnO2(s)

Break them apart into reduced and oxidize, balance the oxygen by adding water and H+ lastly add the electrons lost or gained.

Oxidized: Mn(OH)2(s) → MnO2 (s) +2H^+(aq) +2e^-

Reduced: 4e^- +4H^+(aq) + O2(g) → 2H2O(l)

Now add the hydroxide in order to make it basic:

Oxidized: Mn(OH)2(s) + 2OH^- → MnO2 (s) +2H2O(l) +2e^-

Reduced: 4e^- + 4H2O(l) + O2(g) → 2H2O(l) +4OH^-

Balance the number of electrons so you can cancel them out and get your final equation.

Oxidized: 2Mn(OH)2(s) + 4OH^- → 2MnO2 (s) +4H2O(l) +4e^-

Reduced: 4e^- + 4H2O(l) + O2(g) → 2H2O(l) +4OH^-

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

c.) NO3^-(aq) + H2(g) → NO(g)

This equation the H2 is a spectator ion:

NO3-(aq) +H2(g) → NO(g)

NO3- → NO

NO3- → NO + 2H2O

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

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

Oxidized: 3H2O + Al + 3OH- → Al(OH)3 + 3H2O + 6e-

Reduced: 6e- +(2CrO4)2- + 8H2O → 2Cr(OH)4- + 8OH-

Final Formula: 8H2O(l) + Al(s) + 2CrO4 2-(aq) → Al(OH)3(s) + 2Cr(OH) 4 -(aq) +5OH-(aq)

In order to do this follow the same procedure as before.

Mistake on d.):

Al_(s)+CrO₄^2- _(aq) ⟶ Al(OH)_3(s)+Cr(OH)₄^-_(aq)

Oxidized: 6H₂O + 2Al → 2Al(OH)₃ + 6H⁺ + 6e-

Reduced: 6e- +3CrO₄^2- +12H⁺ → 3Cr(OH)₄^-

Final Formula: 12H₂O_(l) + 2Al_(s) + 3CrO₄^2-_(aq) → 2Al(OH)_3(s) + 3Cr(OH)₄^- _(aq) +6OH^-_(aq)

S 19.1.6 Which of the following is the strongest oxidizing agent: VO4^3−, CrO4^2−, or MnO4^−?

The answer to this would be MnO4^− because electronegativity increases as you go across the periodic table from left to right and bottom to top the electronegativity. Electronegativity is the measurement of an atoms willingness to attract a bonding pair of electrons. That means that Mn which is the furthest left is the most willing to attract an electron. You may be confused due to the fact that you see oxidizing in it, which usually looses an electron, but it says oxidizing agent, which means it oxidizes another element or compound and itself will be reduced, gaining an electron.

Alternatively, I believe \(\MnO_(4)^(-)\) is the correct answer because it is shown to be a strong oxidizing agent in the activity series. While it is true that the more electronegative an atom is the more it will want to attract more electrons, that is not the entire reason for \(\MnO_(4)^(-)\) being correct.

We define oxidation as a loss/donation of electrons to another compound in need. An oxidizing agent is a compound that undergoes reduction (a gain of electrons). It is the compound that pushes the oxidizing compound to oxidize.

Going back to the activity series table, we see that compared to \(\MnO_(4)^(-)\) the other choices are placed lower in the table, thus indicating that \(\MnO_(4)^(-)\) is the strongest out of the three to be an oxidizing agent.

S 19.2.8

Specify whether the following complexes have isomers.

1. Tetrahedral [Ni(CO)₂(Cl)₂]: Tetrahedral cannot have an isomer due to the structure, it does not make the correct angles in order to have cis/trans fac/mer.
   1. In addition, tetrahedrals have bond angles of about 109.5 all around which does not work with cis/fac (\(90^{\circ}\)) or trans/mer (\(90^{\circ}\), \(120^{\circ}\), \(90^{\circ}\)).
2. Trigonal bipyramidal [Mn(CO)₄NO]: Though trigonal bipyramidal can have isomers in this case it does not due to the fact that there is only one NO.
   1. In addition, [Mn(CO)₄NO] does not have the ligand format of (MA₄B₂) needed to get an isomer of a octehedral or (MA₂B₂) to get an isomer in square planar.
3. [Pt(en)₂Cl₂]Cl₂: Yes, there are four bonds with the en and two with the Cl this even number of bonds means that they can move around the central atom creating cis/trans isomers. Also if it is cis it has an optical isomer.

S 12.3.21

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

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Solution:

1. This is a rate law problem.

Thus: rate=k [A]^x • [B]^y

2. x and y are the order of the reaction, in this problem they give you that tell you that NO is a second and O2 is a first.

Thus: rate=k[NO]²[O₂]

3. If it is a second order reaction you raise it to the power of two; if it is a first order then you raise it to a power of one. A is the concentration of NO which is given in the problem as 0.75 and B is the concentration of O2 which is given as 0.5. They have given you k in the problem it is the rate constant(stays the some for this reaction) The equation would be:

rate = [0.75]^2·[0.5]^1[5.8x10^-6]=1.6x10^-6

S12.7.1 Account for the increase in reaction rate brought about by a catalyst.

The catalyst reduces the size of the activation energy (Gibbs free energy). By increasing the forward and reverse reactions, and by increasing the amount of collisions which in turn would increase the reaction rate.

It should be noted that the catalysts will not affect the energy levels where the reactants and products are. They may speed up a reaction, but cannot affect how much of each compound is reacted. Also, a catalyst is no consumed by a reaction and will sometimes be left out of the balances equation.

S12.4.24

A 7 Be atom (mass = 7.0169 amu) decays into a 7 Li atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?

Use the equation E=mc^2. To find the amount of energy created during this reaction you need to take the difference of the products and the reactants

7.0160 - 7.0169 = -0.0009

Then you plug everything into the above equation

E=(-0.0009amu)(2.998x10^8m/s)^2(1.6605x10^-27 kg)(6.022x10^23)=-2697.177 J/mol= -1.68x10^15 MeV

I disagree with the answer:

E=(-0.0009amu)(2.998x10⁸m/s)^2(931.5MeVs²m^-2/amu) = -7.53x10¹⁶ MeV

S20.3.11

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?

First you need to write the two half reactions:

Reduction: SO4^2-→HS^-

Oxidation: C₆H₁₂O₆→6HCO₃^-

Next you want to add water and hydrogen to balance the reactions.

Reduction: SO₄^2- +9H⁺ →HS^- + 4H2O

Oxidation: C₆H₁₂O₆ + 12H₂O→6HCO₃^- + 30H⁺

Next you want to add the electrons:

Reduction: SO₄^2-+9H⁺ +8e- →HS^- + 4H2O

Oxidation: C₆H₁₂O₆ + 12H₂O→6HCO₃^- + 30H⁺ +24e-

Now you want to balance the electrons and cross out things on both sides of the reactions to get:

Final Equation: 3SO₄^2-_(aq) + C₆H₁₂O_6(aq) →3HS^-_(aq) +6HCO₃^-_(aq)+3H⁺_(aq)

S 20.5.23

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?

The reactions we are using is

O2+4H⁺ → 2H₂O

For the first reactions we have a pH of 7.00 when the pH is seven then we know that the amount of hydrogen and the amount of hydroxide is 1.00x10^-7 using the equations with Q because these are initial conditions E_cell = E knot- 0.0592/n(lnQ).

E_cell= 1.23- 0.0592/4(ln(1/(10^7)^4))=

Next we are finding the E of cell at a different pH in this case the amount of H+ we have will be different using

OH- +H+→ H2O

[0.85]+ H⁺ → [10^-14]= 1.17x10^-14

Now plug these new numbers in the equations from above and you get:

E_cell= -0.6687