Extra Credit 8

Q17.1.7

Balance the following in basic solution:

1. SO₃^2-_(aq) + Cu(OH)_2(s) → SO₄^2-_(aq) + Cu(OH)_(s)

2. O_2(g) + Mn(OH)_(s) → MnO_2(s)

3. NO_3(aq)+H_2(g) → NO_(g)

4. Al_(s)+CrO₂^2-(aq) → Al(OH)_3(s)+Cr(OH)₄^-_(aq)

Q17.1.7

QUESTION 1:

SO₃^2-_(aq) + Cu(OH)_2(s) → SO₄^2-_(aq) + Cu(OH)_(s)

Step 1: Assign oxidation states for the elements within each of the compounds.

Reaction side:

⇒ SO₃^2-: S has a charge of +4, O has a charge of -2

⇒Cu(OH)₂: Cu has a charge of +2, (OH) has an overall charge of -1

Product side:

⇒SO₄^2-: S has a charge of +6, O has a charge of -2

⇒Cu(OH): Cu has a charge of +1, (OH) has a charge of -1

Step 2: After assigning oxidation states, identify which elements have lost or gained electrons. Those that gain electrons would be reduced and those that lose electrons would be oxidized. A good way to remember this is by the acronym OIL RIG. Oxidation Is Losing (electrons) and Reduction Is Gaining (electrons). In this case, the element being reduced is Cu in Cu(OH)_2(s) (making it the oxidizing agent) and the element being oxidized is S in SO₃^2-_(aq) (making it the reducing agent).

Step 3: Make half reactions that isolate the oxidation and reduction reactions.

OXIDATION: SO₃^2-_(aq) → SO₄^2-_(aq)

REDUCTION: Cu(OH)_2(s) → Cu(OH)_(S)

Step 4: Balance elements in the reactions other than O and H, we'll do this balancing later. ***For this particular reaction the elements (except O&H) are balanced.

Step 5: Balance the oxygen by adding water molecules to the side with less oxygen.

OXIDATION: SO₃^2-_(aq) + H₂O → SO₄^2-_(aq)

REDUCTION: Cu(OH)_2(s) → Cu(OH)_(s) + H₂O

Step 6: Balance the hydrogen atoms including the ones added during step 5.

OXIDATION: SO₃^2-_(aq) + H₂O → SO₄^2-_(aq) + 2H⁺

REDUCTION: Cu(OH)_2(s) + H⁺→ Cu(OH)_(s) + H₂O

Step 7: The charges of each side need to be equal to each other. This includes the charges of your original compounds as well as the water molecules and hydrogen atoms added later. To achieve equal charges on both sides, you need to add electrons. Looking at the oxidation reaction, the left side has an overall charge of 2- while the right side has an overall charge of 0. This means we will add two electrons to the right side.

Looking at the reduction reaction, the left side has an overall charge of +1 and the right side has an overall charge of 0. This means we'll add one electron to the left side.

OXIDATION: SO₃^2-_(aq) + H₂O → SO₄^2-_(aq) + 2H⁺ + 2e^-

REDUCTION: Cu(OH)_2(s) + H⁺ + e^-→ Cu(OH)_(s) + H₂O

Step 8: The electrons need to be equal to each other from both the oxidative and reduction half reactions. In order to make them equal, we need to multiply the entire reduction half reaction by a coefficient of 2.

REDUCTION: 2(Cu(OH)_2(s) + H⁺ + e^-→ Cu(OH)_(s) + H₂O) ⇒ 2Cu(OH)_2(s) + 2H⁺ + 2e^-→ 2Cu(OH)_(s) + 2H₂O

Now we have the half reactions:

OXIDATION: SO₃^2-_(aq) + H₂O → SO₄^2-_(aq) + 2H⁺ + 2e^-

REDUCTION: 2Cu(OH)_2(s) + 2H⁺ + 2e^-→ 2Cu(OH)_(s) + 2H₂O

Step 9: Since this is in a basic solution we are going to add an hydroxide(OH^-) molecule for every hydrogen atom on both sides of the half reactions. When the hydrogen and hydroxide combine, it creates a water molecule. See below:

OXIDATION: SO₃^2-_(aq) + H₂O + 2OH^- → SO₄^2-_(aq) + 2H⁺ + 2OH^- + 2e^- ⇒ SO₃^2-_(aq) + H₂O + 2OH^- → SO₄^2-_(aq) + 2H₂O + 2e^-

REDUCTION: 2Cu(OH)_2(s) + 2H⁺ + 2OH^-+ 2e^-→ 2Cu(OH)_(s) + 2H₂O +2OH^- ⇒ 2Cu(OH)_2(s) + 2H₂O+ 2e^-→ 2Cu(OH)_(s) + 2H₂O +2OH^-

Step 10: Combine the half reactions.

Overall: SO₃^2-_(aq) + 2Cu(OH)_2(s) + 3H₂O+ 2e^- + 2OH^-→ SO₄^2-_(aq) + 2Cu(OH)_(s) + 4H₂O + 2e^- + 2OH^-

Step 11: Cancel any common terms on either side of the equation, leaving:

Overall net equation(balanced): SO₃^2-_(aq) + 2Cu(OH)_2(s) → SO₄^2-_(aq) + 2Cu(OH)_(s) + H₂O

QUESTION 2: (Same rules as question 1 applies)

reaction: O_2(g) +Mn(OH)_2(g) → MnO_2(s)

Half reactions:

OXIDATION: Mn(OH)_2(g) → MnO_2(s)

REDUCTION: O_2(g) → MnO_2(s)

Balanced Half Reactions:

OXIDATION: Mn(OH)_2(g) → MnO_2(s) + 2H⁺ + 2e^-

REDUCTION: O_2(s) + 2e^- → MnO_2(s)

Include OH^- :

OXIDATION: Mn(OH)_2(g) +2OH^-→ MnO_2(s) + 2H⁺ + 2OH^- + 2e^-

REDUCTION: O_2(s) + 2e^- → MnO2(s)

Combine and cancel for overall reaction:

Overall Net Equation: Mn(OH)_2(g) + O_2(s) + 2OH^- → MnO_2(s) + 2H₂O

QUESTION 3:

Reaction: NO_3(aq)+H_2(g) → NO_(g) + H⁺

Half Reactions:

OXIDATION: H_2(g) → + H⁺

REDUCTION: NO_3(aq) → NO_(g)

Balance oxygen and hydrogen:

OXIDATION: 2H_2(g) → + 4H⁺

REDUCTION: NO₃(aq) +2H⁺ → NO_(g) + H₂O

Balance charges with electrons:

OXIDATION: 2H_2(g) → + 4H⁺ + 4e^-

REDUCTION: NO₃(aq) +2H⁺ + 2e^-→ NO_(g) + H₂O

Balance electrons:

OXIDATION: 2H_2(g) → + 4H⁺ + 4e^-

REDUCTION: 2(NO₃(aq) +2H⁺ + 2e^-→ NO_(g) + H₂O)

NEW REDUCTION RXN: 2NO₃(aq) + 4H⁺ + 4e^-→ 2NO_(g) + 2H₂O

Add OH^-:

OXIDATION: 2H₂(g) + 4OH^-→ + 4H⁺ + 4OH^- + 4e^-

REDUCTION: 2NO₃(aq) + 4H⁺ + 4OH^-+ 4e^-→ 2NO_(g) + 2H₂O +4OH^-

Add H⁺/OH^-:

OXIDATION: 2H₂(g) + 4OH^-→ 4H₂O + 4e^-

REDUCTION: 2NO₃(aq) + 4H₂O + 4e^-→ 2NO_(g) + 2H₂O +4OH^-

Combine half reactions:

Overall: 2NO₃(aq) + 2H₂(g) + 4H₂O + 4OH^- + 4e^-→ 2NO_(g) + 2H₂O +4OH^- +4H₂O + 4e^-

Cancel common terms for overall net equation:

Overall Net Equation: 2NO₃(aq) + 2H₂(g) → 2NO_(g) + 2H₂O

QUESTION 4:

Reaction: Al_(s)+CrO4^2-(aq) → Al(OH)_3(s)+Cr(OH)₄^-(aq)

Half Reactions:

OXIDATION: Al_(s) → Al(OH)_3(s)

REDUCTION: CrO4^2-(aq) → Cr(OH)₄^-(aq)

Add oxygen and hydrogen: *other elements already balanced.

OXIDATION: Al_(s) + 3H₂O→ Al(OH)_3(s) + 3H⁺

REDUCTION: CrO4^2-(aq) + 4H⁺→ Cr(OH)₄^-(aq) *the oxygen is already balanced.

Balance charges with electrons:

OXIDATION: Al_(s) + 3H₂O→ Al(OH)_3(s) + 3H⁺ + 3e^-

REDUCTION: CrO4^2-(aq) + 4H⁺ + 3e^- → Cr(OH)₄^-(aq)

Add OH^-:

OXIDATION: Al_(s) + 3H₂O→ Al(OH)_3(s) + 3H⁺ + 3OH^- + 3e^-

REDUCTION: CrO4^2-(aq) + 4H⁺ + 4OH^- + 3e^- → Cr(OH)₄^-(aq)

Add OH^-:

OXIDATION: Al_(s) + 3H₂O→ Al(OH)_3(s) + 3H₂O + 3e^-

REDUCTION: CrO4^2-(aq) + 4H₂O + 3e^- → Cr(OH)₄^-(aq)

Combine half reactions:

Overall: Al_(s) + CrO4^2-(aq) + 3H₂O + 4H₂O + 3e^- → Al(OH)_3(s) +Cr(OH)₄^-(aq) + 3H₂O + 3e^-

Cancel common terms for overall net equation:

Overall Net Equation: Al_(s) + CrO4^2-(aq) + 4H₂O → Al(OH)_3(s) +Cr(OH)₄^-(aq)

Q19.1.6

Which of the following is the strongest oxidizing agent: VO₄^3-, CrO₄^2-, or MnO₄^-

Q19.1.6

MnO₄^- is the strongest oxidizing agent.

Why? - First, what is an oxidizing agent? An oxidizing agent is a chemical species that has a strong tendency for acquiring electrons and being reduced, which brings about oxidation. An oxidizing agent causes another species to be oxidized while it itself gets reduced. The strength of how much a species "wants" electrons is measured by it's reduction potential. These values are determined experimentally but arranged in an activity series table. This table shows that the strongest oxidizing agents with the largest reduction potentials are at the top of the table and the strongest reducing agents (worst oxidizing) with the lowest reduction potentials are at the bottom. MnO₄^- is at the top of the table compared with the others and therefore is the best oxidizing agent.

Q19.2.8/Q19.2.8

Specify whether the following complexes have isomers.

a. tetrahedral [Ni(CO)₂(Cl)₂]- No isomers. Tetrahedral complexes don't have isomers in this case because all the ligands are the same distance away from each other.

b. trigonal bipyramidal [Mn(CO)₄NO]-No isomers. This complex would have cis and trans isomers if it had 4 of the same ligands and 2 of a different ligand (MA₄B₂) but this complex does not have any isomers.

c. [Pt(en)₂Cl₂]Cl₂ - This is a geometric isomer for planar complexes. This means it can form a cis and trans structure. Trans means 'opposite', so ligands (which are elements structured around the central metal ion) are opposite from each other in orientation(180^o). Cis infers the ligands are on the "same side" or right next to each other in orientation (90^o).

Q12.3.21/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.

a. 4NH_3(g) + 5O_2(g) → 4NO_(g) + 6H₂O_(g) (fast)

b. 2NO_(g) + O_2(g) → 2NO_{2(g) *second order NO(.75M *first order O2(.50M)}

c. 3NO_2(g) + H₂O_(l) → 2HNO_3(aq) + NO_(g) (fast)

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.

Rate constant: 5.8 x 10^-6 L/mol/s.

first order: ln[A]_t =ln[A]_o - kt

Second Order: [A]_t=[A]_o - kt

Because equation B is the rate limiting step, the rate law equation is:

Rate= k[NO]²[O₂]

The answer to the question is now a simple plug and chug;

rate= 5.8 × 10−6[0.75]²[0.5]= 0.00000163 M^-2s^-1

^{(original author for phase 1 did not complete this problem. I (Sanjoyita Mallick) completed it.}

Q21.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

Q21.7.2

Homogeneous catalyst: This catalyst is the same phase as the reactants. This is helpful because it allows the catalyst to be soluble in regards to the reaction.

Heterogeneous catalyst: This catalyst is in a different phase from the reactants.( gas, solid, or liquid). This allows for easy separation of reaction mixtures.

Q21.4.24

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

Q21.4.24

(This conversion is important to be wary of) = Units: 1 amu : 1u : 931.5 MeV

The reaction described above in the question is as follows: ⁷₄ B + ⁰_-1e → ⁷₃ Li + x rays ( ⁰_-1e <-- the symbol for electron capture)

(7.0169amu) - (7.0160amu)=0.0009amu (This is the mass defect)

We will now multiply the mass defect with 931.5 MeV/ 1 amu = .83835 MeV (This is how much energy is produced by the reaction)

Q20.3.13

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?

Q20.3.13

The steps to balance redox reactions are as follows:

1. Balance the following redox reaction in acidic conditions.
2. Step 1: Separate the half-reactions. ...
3. Step 2: Balance elements other than O and H. ...
4. Step 3: Add H₂O to balance oxygen. ...
5. Step 4: Balance hydrogen by adding protons (H⁺). ...
6. Step 5: Balance the charge of each equation with electrons.
7. Step 6: Get both equations to have the same amount of electrons so it can be canceled
8. Step 7: Cancel common species in both equations

After following these steps we get our individual half reactions to be:

OXIDATION: C₆H₁₂O_6(aq) +12H₂O_(l) → 6HCO₃^-_(g) + 30H⁺ + 24e^-

REDUCTION: SO₄^2-_(aq) + 9H⁺ + 8e^- → HS^-_(aq) + 4H₂O_(l)

_{After cancelling out the species that are common to both equations, the overall equation is:}

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

Q20.5.23

For the reduction of oxygen to water, E° = 1.23 V. (Part 1) What is the potential for this half-reaction at pH 7.00? (Part 2)What is the potential in a 0.85 M solution of NaOH?

Q20.5.23

Reduction of water: O₂(g) + 4 H⁺(aq) + 4e→ 2 H₂O(l) [Reduction is the gain of electrons. Oxygen gains electron to become water.]

We're going to use the Nerst equation: E_cell=E^o_cell - (.0592V/n) lnQ

(Part 1)

To find Q: 1/pO₂ [H⁺]⁴

To find [H⁺]: pH=-log[H⁺]----> 7=-log[H⁺]---->[H⁺]=1.94

E_cell=1.23V - (.0592V/2) Log(1/pO₂[1.94]⁴)----> 1.23V - (.0296V) Log(.0706)----> 1.23V - (.0296V)(-1.151)----> 1.26V

(Part 2)

pOH = - log [OH^-]= -log[0.85]= 0.07

14- pOH= 13.93 (this is the pH)

We will now convert the pH into a concentration;

10^-13.93=1.1748976e-14

E_cell=1.23V - (.0592V/2) Log(1/pO₂[1.1748976e-14]⁴)---->2.879V

^{(original author for phase 1 did not complete this problem. I (Sanjoyita Mallick) completed it.}