Extra Credit 6

Q17.1.5B

Balance the following in acidic solution:

a) \(H_2O_2 + Sn^{2+} \rightarrow H_2O + Sn^{4+}\)

b) \(PbO_2 + Hg \rightarrow Hg_2^{2+} + Pb^{2+}\)

c) \(Al + Cr_2O_7^{2-} \rightarrow Al^{3+} + Cr^{3+}\)

^Answers:

a) i) Write the half reactions

\(Sn^{2+} \rightarrow Sn^{4+}\) \(H_2O_2 \rightarrow H_2O\)

ii) Balance oxygen with H₂O

\(Sn^{2+} \rightarrow Sn^{4+}\) \(H_2O_2 \rightarrow H_2O + H_2O\)

iii) Balance hydrogen with H⁺

\(Sn^{2+} \rightarrow Sn^{4+}\) \(H_2O_2 + {2H^+} \rightarrow H_2O + H_2O\)

iv) Balance charge with e^-

\(Sn^{2+} \rightarrow Sn^{4+} + {2e^-}\) \(H_2O_2 + 2H^+ +{2e^-} \rightarrow H_2O + H_2O\)

v) Combine half reactions

\(Sn^{2+} + H_2O_2 + 2H^+ +2e^- \rightarrow Sn^{4+} + 2e^- + H_2O + H_2O\)

vi) Simplify the reaction

\(Sn^{2+} + H_2O_2 + 2H^+ \rightarrow Sn^{4+} + 2H_2O\)

vii) Check if number of atoms and charges balance.

The charges on both sides equal 4+, and the number of atoms are equal on both sides. 

b) i) Write the half reactions

\(Hg \rightarrow Hg_2^{2+}\) \(PbO_2 \rightarrow Pb^{2+}\)

ii) Balance elements other the O and H

\(2Hg \rightarrow Hg_2^{2+}\) \(PbO_2 \rightarrow Pb^{2+}\)

iii) Balance oxygen with H₂O

\(2Hg \rightarrow Hg_2^{2+}\) \(PbO_2 \rightarrow Pb^{2+}+2H_2O\)

iv) Balance hydrogen with H⁺

\(2Hg \rightarrow Hg_2^{2+}\) \(PbO_2+4H^+ \rightarrow Pb^{2+}+2H_2O\)

v) Balance charge with e^-

\(2Hg \rightarrow Hg_2^{2+}+2e^-\) \(2e^-+PbO_2+4H^+ \rightarrow Pb^{2+}+2H_2O\)

vi) Combine half reactions

\(2Hg + 2e^- + PbO_2 +4H^+\rightarrow Hg_2^{2+} +2e^- + Pb^{2+}+2H_2O\)

vii) Simplify the reaction

\(2Hg + PbO_2 +4H^+\rightarrow Hg_2^{2+} + Pb^{2+}+2H_2O\)

viii) Check if number of atoms and charges balance.

The charges on both sides equal 4+, and the number of atoms are equal on both sides. 

c) i) Write the half reactions

\(Al \rightarrow Al^{3+}\) \(Cr_2O_7^{2-} \rightarrow Cr^{3+}\)

ii) Balance elements other the O and H

\(Al \rightarrow Al^{3+}\) \(Cr_2O_7^{2-} \rightarrow 2Cr^{3+}\)

iii) Balance oxygen with H₂O

\(Al \rightarrow Al^{3+}\) \(Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O\)

iv) Balance hydrogen with H⁺

\(Al \rightarrow Al^{3+}\) \(14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O\)

v) Balance charge with e^-

\(Al \rightarrow Al^{3+}+3e^-\) \(6e^-+14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O\)

vi) Balance e^-

\(2Al \rightarrow 2Al^{3+}+6e^-\) \(6e^-+14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O\)

vii) Combine half reactions

\(6e^-+14H^++Cr_2O_7^{2-}+2Al \rightarrow 2Cr^{3+}+ 7H_2O+2Al^{3+}+ 6e^-\)

viii) Reduce

\(14H^++Cr_2O_7^{2-}+2Al \rightarrow 2Cr^{3+}+ 7H_2O+2Al^{3+}\)

ix) Check if number of atoms and charges balance.

The charges on both sides equal 12+, and the number of atoms are equal on both sides. 

Q19.1.4

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

^Answer:

^{Lanthanoids (in group 6 and 7) are not often found in their elemental form in nature because of where they are found in the earth and their characteristics. They are often found in the crust of the earth and as a result, they are hard to remove. Also, the elements have a tendency to combine with other lanthanoids, so they often are not found by themselves. This is the reason for them to be called "Rare Earth Metals."}

Q19.2.6

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

1. \([Co(CO_3)_3]^{-3}\)
2. \([Cu(NH_3)_4]^{+2}\)
3. \([Co(NH_3)_4Br_2]_2(SO_4)\)
4. \([Pt(NH_3)_4][PtCl_4]\)
5. \([Cr(en)_3](NO_3)_3\)
6. \([Pd(NH_3)_2Br_2]\)
7. \(K_3[Cu(Cl)_5]\)
8. \([Zn(NH_3)_2Cl_2]\)

^Answers:

1. \([Co(CO_3)_3]^{3-}\)

i) Identify your ligands

3 carbonate ligands

ii) Write out prefixes and name of metal. If the complex is an anion, the metal has the suffix "-ate".

tricarbonatocobaltate

iii) Determine oxidation number by counting charge of ligands and ion charge

\([Co(CO_3)_3]^{3-}\) \(Co = x\) \( 3CO_3^- = 3(-2)\)

\(3(-2) + x = -3\) \(x= +3\) So the oxidation number of Co is +3

iv) Include the oxidation state of the metal in roman numerals

tricarbonatocobaltate(III) ion

2. \([Cu(NH_3)_4]^{2+}\)

i) Identify your ligands Prefixes for Ligands

4 ammonia ligands

ii) Write out prefixes and name of metal. If the complex is an anion, the metal has the suffix "-ate".

tetraamminecopper

iii) Determine oxidation number by counting charge of ligands and ion charge

\([Cu(NH_3)_4]^{2+}\) \(Cu=x\) \(4NH_3=4(0)\)

\(4(0) + x = 2\) \(x= +2\) So the oxidation number of Cu is +2

iv) Include the oxidation state of the metal in roman numerals

tetraamminecopper(II) ion

3. \([Co(NH_3)_4Br_2]_2(SO_4)\)

i) Identify your ligands

4 ammonia ligands 2 bromine ligands

ii) Write out prefixes and name of metal. When you have more than one ligand, name them in alphabetical order (based on the ligand name, not the prefix).

tetraamminedibromocobalt

iii) Determine oxidation number by counting charge of ligands and anion and setting equal to 0

\([Co(NH_3)_4Br_2]_2(SO_4)\) \(Co=x\) \(4NH_3=4(0)\) \(Br_2=2(-1)\) \(SO_4=-2\)

\(2[ 4(0) + 2(-1) + x] +-2= 0\) \(-6 + 2x = 0\) \(x = +3\) So the oxidation number of Co is +3

iv) Include the oxidation state of the metal in roman numerals. Add anion

tetraamminedibromocobalt(III) sulfate

4. \([Pt(NH_3)_4][PtCl_4]\)

i) Identify your ligands for anion and cation

Cation: 4 ammonia ligands Anion: 4 chlorine ligands

ii) Write out prefixes and name of metal from each. If the complex is an anion, the metal has the suffix "-ate".

Cation: tetraammineplatinum Anion: tetrachloroplatinate

iii) Determine oxidation number by counting charge of ligands and set equal to 0

Cation: \([Pt(NH_3)_4]\) \(Pt= x\) \(4NH_3=4(0)\)

\(4(0)+ x =\)

Anion: \([PtCl_4]\) \(Pt=y\) \(4Cl=4(-1)\)

\(4(-1)+ y =\)

4(0)+ x + 4(-1)+ y= 0 \(x +y+ -4 = 0\) \(x+y = 4\)

Pt has common oxidation states of +2 and +4. Using this we can see that the Pt in the anode and cathode must be equal to +2

iv) Write the cation and anion. Include the oxidation state of the metal in roman numerals

tetraammineplatinum(II) tetrachloroplatinate(II)

5. \([Cr(en)_3](NO_3)_3\)

i) Identify your ligands

3 ethylenediamine ligands

ii) Write out prefixes for ligands with prefixes in the name already (bis, tris, ect.) and name of metal

tris(ethylenediamine)chromium

iii) Determine oxidation number by counting charge of ligands and setting equal to anion

\([Cr(en)_3](NO_3)_3\) \(Cr=x\) \(3(en)=3(0)\) \(3NO_3=3(-1)\)

\(3(0)+ x -3= 0\) \(x=+3\) So the oxidation number of Cr is +3

iv) Write the cation and anion. Include the oxidation state of the metal in roman numerals

tris(ethylenediamine)chromium(III) nitrate

6. \([Pd(NH_3)_2Br_2]\)

i) Identify your ligands

2 ammonia ligands 2 bromine ligands

ii) Write out prefixes and name of metal. When you have more than one ligand, name them in alphabetical order (based on the ligand name, not the prefix).

diamminedibromopalladium

iii) Determine oxidation number by counting charge of ligands

\([Pd(NH_3)_2Br_2]\) \(Pd=x\) \(2NH_3=2(0) \(2Br=2(-1)\)

\(2(0) + 2(-1) + x = 0\) \(x=+2\) So the oxidation number of Pd is +2

iv) Include the oxidation state of the metal in roman numerals

diamminedibromopalladium(II)

7. \(K_3[Cu(Cl)_5]\)

i) Identify your ligands

5 chlorine ligands

ii) Write out prefixes and name of metal. If the complex is an anion, the metal has the suffix "-ate".

pentachlorocuprate

iii) Determine oxidation number by counting charge of ligands and cation

\(K_3[Cu(Cl)_5]\) \(3K=3(+1)\) \(Cu=x\) \(5Cl=5(-1)\)

\(5(-1)+ x + 3(1) = 0\) \(x=+2\) So the oxidation number of Cu is +2

iv) Write the cation and anion. Include the oxidation state of the metal in roman numerals

potassium pentachlorocuprate(II)

8. \([Zn(NH_3)_2Cl_2]\)

i) Identify your ligands

2 ammonia ligands 2 chlorine ligands

ii) Write out prefixes and name of metal. When you have more than one ligand, name them in alphabetical order (based on the ligand name, not the prefix).

diamminedichlorozinc

iii) Determine oxidation number by counting charge of ligands

\([Zn(NH_3)_2Cl_2]\) \(Zn=x\) \(3NH=3(0)\) \(2Cl=2(-1)\)

\(2(0) + 2(-1) + x = 0\) \(x=+2\) So the oxidation number is +2

iv) Include the oxidation state of the metal in roman numerals

diamminedichlorozinc(II)

Q12.3.19

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

1. What is the order of the reaction with respect to [Q], and what is the rate equation?
2. What is the rate constant?

Answers:

1. We will use a "guess and check" method by using the rate law formula (\(rate=k[Q]^n\)) and try to find the constant k, letting us know that we found the correct order for the reaction.

i) Take ratio of rate divided by concentration \(\frac{rate}{[Q]}=k\)

\(\frac{6.68x10^{-3}}{.17}= .0393\) \(\frac{1.04 × 10^{-2}}{.212} = .0491\) \(\frac{2.94 × 10^{-2}}{.357} = .0824\)

Since the ratios are not the same, the reaction is not first order

Try again with 2nd order (square the Molarity and divide) \(\frac{rate}{[Q]^2}=k\)

\(\frac{6.68x10^{-3}}{.17^2} = .231\) \(\frac{1.04 × 10^{-2}}{.212^2} = .231\) \(\frac{2.94 × 10^{-2}}{.357^2} = .231\)

Since these values are the same then your reaction is 2nd order

\(rate=k[Q]^2\)

2. The ratio value that we got is the rate constant: .231 M^-1s^-1

Q12.6.10

Experiments were conducted to study the rate of the reaction represented by this equation.

\(2NO+2H_2 \rightarrow N_2+ 2H_2O\)

Initial concentrations and rates of reaction are given here.

Consider the following questions:

1. Determine the order for each of the reactants, NO and H₂, from the data given and show your reasoning.
2. Write the overall rate law for the reaction.
3. Calculate the value of the rate constant, k, for the reaction. Include units.
4. For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H₂ had been consumed.
5. The following sequence of elementary steps is a proposed mechanism for the reaction.

Step 1: \(NO + NO \rightleftharpoons N_2O_2\)

Step 2: \(N_2O_2 + H_2 \rightleftharpoons N_2O + N_2O\)

Step 3: \(N_2O + H_2 \rightleftharpoons N_2 + H_2O\)

Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

Answers

1. i) Find the order for [NO] by using experiment 3 and 4 where [H₂] is constant

Notice that [NO] doubles from experiment 3 to 4 and the rate quadruples. So the order for [NO] is 2

ii) Find the order for [H₂] by using experiment 1 and 2 where [NO] is constant

Notice that [H₂] doubles from experiment 1 to 2 and the rate doubles as well. So the order for [H₂] is 1

2. Put in the order for each product as the exponents for the corresponding reactant.

\(rate = k [NO]^2 [H_2]\)

3. Put in the concentrations and the rate from one of the experiments into the rate law and solve for k. (Here, experiement 1 is used but any of them will work)

\(rate = k [NO]^2 [H_2]\) \(.00018 = k [.006]^2 [.001]\) \(k = 5000 M^{-2}s^{-1}\)

4. Plug in values for experiment 2 into the rate law equation and solve for the concentration of NO

\(.00036=5000[NO]^2[.001]\) \([NO]^2= 7.2 x 10^{-5}\) \([NO] = .0085 M\)

5. Write the rate laws for each step and then see which matches the rate law we found in question 2. The rate determining step (the slow step) is the one that gives the rate for the overall reaction. Because of this, only those concentrations will influence the overall reaction, contrary to what we would believe if we just looked at the overall reaction.

Step 1: \(NO + NO \rightleftharpoons N_2O_2\)

\(rate =k_1[NO]^2\) This rate law is not the same as the one we calculate in question 2 so this can not be the rate determining step

Step 2: \(N_2O_2+H_2 \rightleftharpoons N_2O + N_2O\)

\(rate = k_2[N_2O_2][H_2]\)

Since \(N_2O_2\) is an intermediate you must replace it in the rate law equation. Intermediates can not be in the rate law because they do not appear in the overall reaction. Here you can take the reverse of equation 1 (k_-1) and substitute the other side (the reactants of equation 1) for the intermediate in the rate law equation.

\[rate_1 = rate_{-1}\]

\[k_1[NO]^2 = k_{-1}[N_2O_2]\]

\[[N_2O_2] = \frac{k_1[NO]^2}{k_{-1}}\]

\(rate= \frac{k_2k_{1}[NO]^2[H_2]}{k_{-1}}\)

Overall: \(rate={k[NO]^2[H_2]}\) This is the same so it is the rate determining step.

Q21.4.22

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

Answer:

Since we have the mass of both of the metals we are able to find the percentage of uranium left in the sample by dividing the uranium left by the total mass in the sample.

i) Determine the percentage of U in the sample.

\(\frac{5.37}{5.37+2.52} = 68.06% U\)

Now, since we have the percentage of uranium in the sample we have all the information neccesary to plug into our half life formula. We will use 1 as the initial (A_o) mass (100% U), .6806 as the current (A) mass (68.06% left), 4.5x10⁹ years as the given half life (h) and solve for t.

ii) Use formula for half life: \(A = A_0 (.5^{\frac{t}{h^2}})\)

\(.6808 = .5^{\frac{t}{(4.5x10^9)^2}}\)

To solve for t take the natural log of both sides and drop down the exponent the solve normally

\(ln(.6808) = ln(.5^{\frac{t}{(4.5x10^9)^2}})\)

\(-.38449 = \frac{t}{(4.5x10^9)^2}ln(.5)\)

\(-.38449 = \frac{t}{(4.5x10^9)^2}(-.69315)\)

\(.554701 = \frac{t}{(4.5x10^9)^2}\)

t = 2,498,044,246 years

t = 2,500,000,000 (2 sig figs)

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:

1. First you must determine the half reactions and balance them properly so you can write the balanced overall reaction

i) Write out the Reduction half reaction with just Sulfate and HS^-

\(S{O_4^{-2}}(aq) \rightarrow HS^-(aq)\)

ii) Balance oxygen with H₂O

\(S{O_4^{-2}}(aq) \rightarrow HS^-(aq) + 4H_2O(l)\)

iii) Balance hydrogen with H+

\(S{O_4^{-2}}(aq)+ 9H^+(aq) \rightarrow HS^-(aq) + 4H_2O(l)\)

iv) Balance charge with e-

\(S{O_4^{-2}}(aq)+ 9H^+(aq) + 8e^- \rightarrow HS^-(aq) + 4H_2O(l)\)

Notice that this reaction has its electrons on the reactants side. This is because this half reaction is the reduction (gain of electrons). You can use this as a method to check that you are placing them on the correct side.

2. i) Write out the Oxidation half reaction

\(C_6H_{12}O_6(aq) \rightarrow 6HCO_3^{-1}(aq)\)

ii) Balance elements other then H and O

\(C_6H_{12}O_6(aq) \rightarrow 6HCO_3^{-1}(aq)\)

iii) Balance oxygen with H₂O

\(C_6H_{12}O_6(aq)+ 12H_2O(l) \rightarrow 6HCO_3^{-1}(aq)\)

iv) Balance hydrogen with H+

\(C_6H_{12}O_6(aq)+ 12H_2O(l) \rightarrow 6HCO_3^{-1}(aq)+ 30H^+(aq)\)

v) Balance charge with e-

\(C_6H_{12}O_6(aq)+ 12H_2O(l) \rightarrow 6HCO_3^{-1}(aq)+ 30H^+(aq) + 24e^-\)

Notice that this reaction has its electrons on the products side. This is because this half reaction is the oxidation (loss of electrons). You can use this as a method to check that you are placing them on the correct side. Also, because the electrons are on opposite sides of each other in the two half reactions you are able to cancel them out.

3. i) Balance out electrons of each half reaction by multiplying the reduction half reaction by 3 to make the electrons on each 24.

\(3S{O_4^{-2}}(aq)+ 27H^+(aq) + 24e^- \rightarrow 3HS^-(aq) + 12H_2O(l)\) \(C_6H_{12}O_6(aq)+ 12H_2O(l) \rightarrow 6HCO_3^{-1}(aq)+ 30H^+(aq) + 24e^-\)

ii) Add two half reactions together to get Overall Reaction

\(C_6H_{12}O_6(aq) + 3SO_4^{2−}(aq) \rightarrow 6HCO_3^{-1}(g) + 3H^+(aq) + 3HS^-(aq)\)

4. Check if number of atoms and charges balance.

The charges on both sides equal -6, and the number of atoms are equal on both sides. ✓

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:

1. \(MnO_4^-(aq) + e^- \rightarrow MnO_4^{2-}(aq)\) ; E° = +0.56 V (purple → dark green)
2. \(MnO_4^{2-}(aq) + 2e^- + 4H^+(aq) \rightarrow MnO_2(s)\) ; E° = +2.26 V (dark green → dark brown solid)
3. \(MnO_2(s) + e^- + 4H^+(aq) \rightarrow Mn^{+3}(aq)\) ; E° = +0.95 V (dark brown solid → red-violet)
4. \(Mn^{3+}(aq) + e^- \rightarrow Mn^{2+}(aq)\) ; E° = +1.51 V (red-violet → pale pink)
5. \(Mn^{2+}(aq) + 2e^- \rightarrow Mn(s)\) ; E° = −1.18 V (pale pink → colorless)

1. Is the reduction of MnO₄^-2 to Mn⁺³(aq) by H₂(g) spontaneous under standard conditions? What is E°cell?
2. Is the reduction of Mn⁺³(aq) to Mn(s) by H₂(g) spontaneous under standard conditions? What is E°cell?

Answer:

1. In order for a reaction to be spontaneous, ΔG<0. \[ΔG= -nFE^ocell\] So E^o must be a positive value.

Take the E^o values for equation 2 and 3 and take the Cathode (Reduction) and subtract the Anode (oxidaton) E^o values

\(E^ocell=(2.26)-(.95)=1.31V\)

Since this value is positive, we know that the reaction is spontaneous

2. Take the E^o values for equation 4 and 5 and take the Cathode (Reduction) and subtract the Anode (oxidaton) E^o values

\(E^ocell=(1.51)-(-1.81)=3.32V\)

Since this value is positive, we know that the reaction is spontaneous