Extra Credit 6

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Phase II: most will be edited in red.

Q17.1.5B

Balance the following in acidic solution:

a. \(\ce{H2O2 + Sn^2+\rightarrow H2O + Sn^4+}\)

1. Separate into two equations where one is oxidized and reduced (determine using oxidation number)

Reduction: \(\ce{H2O2\rightarrow H2O}\)

Oxidation: \(\ce{Sn^2+\rightarrow Sn^4+}\)

2. Balance the number of elements to make sure that the coefficients are balanced for both equations (use H⁺ and H₂O when needed because it's acidic)

\(\ce{H2O2 + 2H^+\rightarrow 2H2O}\)

\(\ce{Sn^2+\rightarrow Sn^4+}\)

3. Balance the charge using electrons

For the first equation, hydrogen has a charge of 2+ so you will need to add 2 electrons to balance it. For the second equation, since Sn on the left side has a charge of 2+ and the Sn on the right side has a charge of 4+, the overall charge of the equation is 2+ so we will need to add 2 electrons to balance that equation.

\(\ce{H2O2 + 2H^+ +2e^{-}\rightarrow 2H2O}\)

\(\ce{Sn^2+\rightarrow Sn^4+ + 2e^-}\)

4. Because the electron is already balanced, last step is to combine the equations by crossing out common terms.

For this equation, we will cancel out the 2 electrons on both equations and since there are no other common terms we can combine the rest into one equation.

\(\ce{H2O2 + 2H^+ + Sn^2+\rightarrow 2H2O + Sn^4+}\)

b. \(\ce{ PbO2 + Hg\rightarrow Hg2^2+ + Pb^2+}\)

1. Separate into two equations where one is oxidized and reduced

Reduction: \(\ce{PbO2\rightarrow Pb^2+}\)

Oxidation: \(\ce{Hg\rightarrow Hg2^2+}\)

2. Balance the number of elements and use H⁺ and H₂O when needed because it's only acidic

\(\ce{PbO2 + 4H^+\rightarrow Pb^2+ + 2H2O}\)

\(\ce{2Hg\rightarrow Hg2^2+}\)

3. Balance the charge using electrons

For equation one, on the left side, there is a total charge of 4+ from the hydrogen and on the right side, there is a total charge of 2+ from Pb. Therefore, the overall charge of equation one is 2+ so we will need to add 2 electrons. For equation 2, there is no charge on the left side and a charge of 2+ on the right side so the overall charge of equation 2 is 2+ and so we will add 2 electrons too.

\(\ce{PbO2 + 4H^+ + 2e^{-}\rightarrow Pb^2+ + 2H2O}\)

\(\ce{2Hg\rightarrow Hg2^2+ + 2e^-}\)

4. Because the electron is already balanced, last step is to combine the equations

All we need to do is cancel out the 2 electrons on both sides and combine all the other terms together.

\(\ce{PbO2 + 2Hg + 4H^+\rightarrow Hg2^2+ + Pb^2+ + 2H2O}\)

c. \(\ce{Al + Cr2O7^2-\rightarrow Al^3+ + Cr^3+}\)

1. Separate into two equations where one is oxidized and reduced

Oxidation: \(\ce{Al\rightarrow Al^3+}\)

Reduction: \(\ce{Cr2O7^2-\rightarrow Cr^3+}\)

2. Balance the number of elements (use H⁺ and H₂O when needed because it's acidic)

\(\ce{Al\rightarrow Al^3+}\)

\(\ce{Cr2O7^2- + 14H^{+}\rightarrow 2Cr^3+ + 7H2O}\)

3. Balance the charge using electrons

For equation one, since there is no charge on the left side and a 3+ charge on the right side, we will need to all 3 electrons. For equation 2, on the left side, there is a 12+ charge and on the right, there is a 6+. Therefore, the overall charge is 6+ so we will need to add 6 electrons.

\(\ce{Al\rightarrow Al^3+ + 3e^-}\)

\(\ce{Cr2O7^2- + 14H^+ + 6e^{-}\rightarrow 2Cr^3+ + 7H2O}\)

4. Balance the amount of electrons from each equation so they can cross out each other

Since there are 3 electrons in the first equation and 6 electrons in the second equation, we will need to multiply equation one by 2 so in the last step, we are able to cancel out the 6 electrons.

\(\ce{2Al\rightarrow 2Al^3+ + 6e^-}\)

\(\ce{Cr2O7^2- + 14H^+ + 6e^{-}\rightarrow 2Cr^3+ + 7H2O}\)

5. Combine both equations

We can now cancel out the 6 electrons on both side. Since there is no other common terms, we can just combine both equations together.

\(\ce{2Al + Cr2O7^2- + 14H^{+}\rightarrow 2Al^3+ + 2Cr^3+ + 7H2O}\)

Q19.1.4

Why are the lanthanoid elements not found in nature in their elemental forms?

Lanthanoid are the elements belonging to the f-block (inner transition metals) of the periodic table. In lanthanides, the general valence shell electron configuration is of form \(\ce{4f^{1-12} 5d^{0-1} 6s^2}\). The lanthanoid series is from lanthanum (La) to lutetium (Lu).
The element exists in its elemental form when it is very less reactive in nature or when it does not combine easily with other elements. Lanthanoids are highly electropositive in nature and readily gives electrons to form compounds with other electronegative elements. As a result of this, they exists in the combined form rather than in the elemental state.

Q19.2.6

Name each of the compounds or ions, including the oxidation state of the metal.

\(\ce{[Co(CO3)3]^3-}\) (note that \(\ce{CO3^2-}\) is bidentate in this complex)
- CO₃ ligand has a charge of -2 (there are 3 of them so the total charge is -6)
- The total charge of the complex ion is -3 so to even it out from the CO₃ligands, the Co metal oxidation state must be +3
- Name: tricarbonatocobaltate(iii)
  - Explanation:
  - CO₃^2- is called a carbonate, therefore, as a ligand its name becomes carbonato
  - There are three carbonato ligands so we use latin number to indicate that which is tri
  - Cobalt metal has oxidation state of +3 so it is indicated with roman numeral in brackets after the name
  - We add prefix -ate because the whole complex ion is an anion
\(\ce{[Cu(NH3)4]^2+}\)
- NH₃ ligand has a total charge of 0
- The total charge of the complex ion is +2 so the Cu metal oxidation state must be +2
- Name: tetraamminecopper(ii)
  - NH₃ as a ligand is called ammine
  - There are four ammine ligands so we use latin number to indicate that which is tetra
  - Copper metal has oxidation state of +2 so it is indicated with roman numeral in brackets after the name
\(\ce{[Co(NH3)4Br2]2(SO4)3}\)
- Each SO₄ (sulfate) ion has a -2 charge and there are 3 of them (total charge: -6), that means to balance out this charge, each complex cobalt ion needs to have the charge of +3 (because there are 2 complex ions in each compound)
- Br ligands have a charge of -1 (in the complex ion, total charge of Br: -2) and NH₃ ligand has a total charge of 0
- To reach the total charge of +3 with the negative charged ligands, cobalt metal has the oxidation state of +5
- Name: tetraaminedibromocobalt(v) sulfate
  - Refer to nomenclature of coordination complex for ligand names
  - Ammine is written first because the ligands are arranged in alphabetical order
\(\ce{[Pt(NH3)4][PtCl4]}\)
- NH₃ is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH₃first, followed by the one with Cl (the same order as the formula)
- Pt will have an oxidation number of +2 in each complex
- Name: tetraammineplatinum(ii) tetrachloroplatinate(ii).
\(\ce{[Cr(en)3](NO3)3}\)
- We take the same approach, except (en) is a polydentate ligand with a prefix in its name (ethylenediamine), so "bis" is used instead of "bi", and parentheses are added
- NO₃ion has a charge of -1 and (en) ligand has 0 charge, therefore chromium metal in this compound has the oxidation state of +3
- Name: tris(en)chromium(iii) nitrate
\(\ce{[Pd(NH3)2Br2]}\) (square planar)
- Each Br ligands have a charge of -1 and the overall ion has a charge of 0, therefore it Pd has oxidation state of +2
- Name: diamminedibromopalladium(ii)
\(\ce{K3[Cu(Cl)5]}\)
- Name: potassium pentachlorocuprate(ii)
- The 3 K⁺ ions have a total charge of +3 and Cl^-ligands -5, therefore, copper in this compound has a charge of +2
\(\ce{[Zn(NH3)2Cl2]}\)
- Name: diamminedichlorozinc(ii)
- Same approach as number 6 and 7, therefore, zinc will have an oxidation state of +2

Q12.3.19

For the reaction \(\ce{Q\rightarrow W + X}\), the following data were obtained at 30 °C:

[Q]_initial (M)	0.170	0.212	0.357
Rate (mol/L/s)	6.68 × 10⁻³	1.04 × 10⁻²	2.94 × 10⁻²

What is the order of the reaction with respect to [Q], and what is the rate equation?
- Order reaction: 2 because when you use the ratio trial 3:2, it will look like this:
- (\(\dfrac{2.94*10^{-2}}{1.04*10^{-2}}\)) = (\(\dfrac{0.357^{x}}{0.212^{x}}\))
- 2.82 = 1.7^x
- x = 2 so the order of reaction is 2
- Rate reaction equation: Rate=k[Q]²
What is the rate constant?
- To find the rate constant (k) simply plug and calculate one of the trials into the rate equation
- 1.04 x 10^-2=k[0.212]²
- k=0.2314

Q12.6.1

Why are elementary reactions involving three or more reactants very uncommon?

Reactions involving three reactants or more requires the collision of three (or more) particles at the same place and time
This type of reaction known as termolecular is very uncommon because all the reactants must simultaneously collide with each other, with sufficient energy and correct orientation, to produce a reaction

Q21.4.22

A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of \(\ce{^{238}_{92}U}\) and 2.52 mg of \(\ce{^{206}_{82}Pb}\). Calculate the age of the ore. The half-life of \(\ce{^{238}_{92}U}\) is 4.5 × 10⁹ yr.

Formula we're going to use: \(\ce{N_{Now}=N_{Orig}e^{-λt}}\)
- N_Nowis the number of uranium atoms (or mol) present
  - 5.37 mg of \(\ce{^{238}_{92}U}\) = 0.00537g / 238 g/mol = 2.26 x 10^-5 mol
- N_Orig is the number of uranium atoms (or mol) originally (number of uranium atoms + lead atoms currently)
  - 2.52 mg of \(\ce{^{206}_{82}Pb}\) = 0.00252g / 206 g/mol = 1.22 x 10^-5mol
  - Sum of Pb and U = 2.26 x 10^-5 mol + 1.22 x 10^-5mol = 3.48 x 10^-5 mol
- t is the time (wanted)
- λ is the decay rate of uranium (\(\ce{λ=ln(2)/t_{1/2}}\)) (t_1/2= 4.5 x 10⁹ yr - given)
  - λ = ln(2)/4.5 x 10⁹=1.54 x 10^-10 year^-1
Plug and calculate these numbers into the equation and solve for t
- ln(2.26 x 10^-5 mol/3.48 x 10^-5 mol) = -1.54 x 10^-10 year^-1(t)
- t = 2.8 x 10⁹ years

Q20.3.10

Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)₆]^3-, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.

The gel contains phenolphthalein, therefore it can turn pink
Iron is easily oxidized under basic condition
The iron in [Fe(CN)6]^3- (Fe³⁺) is easily reduced into [Fe(CN)6]^4- (Fe²⁺)
Reduction:
- \(\ce{[Fe(CN)6]^3- + e^{-}\rightarrow [Fe(CN)6]^4-}\)
Oxidation:
- \(\ce{Fe(s) + 2OH^{-}\rightarrow Fe(OH)2 + 2e^-}\)
Overall reaction:
- \(\ce{2[Fe(CN)6]^3- + Fe(s) + 2OH^{-}\rightarrow Fe(OH)2 + 2[Fe(CN)6]^4-}\)

Q20.5.21

The reduction of Mn(VII) to Mn(s) by H₂(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:

\(\ce{MnO4^- (aq) + e^{-}\rightarrow MnO4^2- (aq)}\); E° = +0.56 V (purple → dark green)
\(\ce{MnO4^2- (aq) + 2e^- + 4H^+ (aq)\rightarrow MnO2(s)}\); E° = +2.26 V (dark green → dark brown solid)
\(\ce{MnO2(s) + e^- + 4H^+ (aq)\rightarrow Mn^3+ (aq)}\); E° = +0.95 V (dark brown solid → red-violet)
\(\ce{Mn^3+ (aq) + e^{-}\rightarrow Mn^2+ (aq)}\); E° = +1.51 V (red-violet → pale pink)
\(\ce{Mn^2+ (aq) + 2e^{-}\rightarrow Mn(s)}\); E° = −1.18 V (pale pink → colorless)

Is the reduction of MnO₄⁻ to Mn³⁺(aq) by H₂(g) spontaneous under standard conditions? What is E°cell?
- Yes, it is spontaneous
  - \(\ce{MnO4^- (aq) + e^{-}\rightarrow MnO4^2- (aq)}\); ΔG° = -F(0.56)
  - \(\ce{MnO4^2- (aq) + 2e^- + 4H^+ (aq)\rightarrow MnO2(s)}\); ΔG° = -2F(2.26)
  - \(\ce{MnO2(s) + e^- + 4H^+ (aq)\rightarrow Mn^3+ (aq)}\); ΔG° = -F(0.95)
  - Total: \(\ce{MnO4^- (aq) + 4e^- + 8H^+ (aq)\rightarrow Mn^3+ (aq)}\); ΔG° = -F(0.56+2(2.26)+0.95)
  - ΔG° = -nFE°, plug and solve for E°
    - E°cell = (0.56+2(2.26)+0.95)/4 = 1.51 V
Is the reduction of Mn³⁺(aq) to Mn(s) by H₂(g) spontaneous under standard conditions? What is E°cell?
- No, it is non spontaneous
- ΔG° = -nFE°
  - n = number of electrons involved in electrode reaction
  - F = 1 faraday = 96500 C
  - E° = standard reduction potential
    - \(\ce{Mn^2+ + 2e^{-}\rightarrow Mn}\); E° = -1.029 V
      - ΔG° = -2 x 96500 x -1.029 = 198597 J
    - \(\ce{Mn^3+ + e^{-}\rightarrow Mn^2+}\); E° = 1.51 V
      - ΔG° = -1 x 96500 x 1.51 = -145715 J
  - Combine both equations:
    - \(\ce{Mn^3+ + 3e^{-}\rightarrow Mn}\); ΔG° = -145715 + 198597 = 52882 J
  - Because ΔG° = -nFE° → 52882 = -3 x 96500 x E°
  - E° = -52882/289500 = -0.183 V

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Phase II: most will be edited in red.

Q17.1.5B

a. \(\ce{H2O2 + Sn^2+\rightarrow H2O + Sn^4+}\)

b. \(\ce{ PbO2 + Hg\rightarrow Hg2^2+ + Pb^2+}\)

c. \(\ce{Al + Cr2O7^2-\rightarrow Al^3+ + Cr^3+}\)

Q19.1.4

Q19.2.6

Q12.3.19

Q12.6.1

Q21.4.22

Q20.3.10

Q20.5.21