Extra Credit 7

Q17.1.6

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

previous problems:

1. \(\ce{H2O2 + Sn^2+ ⟶ H2O + Sn^4+}\)
2. \(\ce{PbO2 + Hg ⟶ Hg2^2+ + Pb^2+}\)
3. \(\ce{Al + Cr2O7^2- ⟶ Al^3+ + Cr^3+}\)

S17.1.6

a. In part (a), after separating and balancing, we get the following half reactions.

1) \(\ce{Sn^2+ ⟶ 2e^- + Sn^4+}\) (oxidation)

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

In the first half reaction, we see that Sn²⁺ loses electrons and turns into Sn⁴⁺. Since Sn²⁺ lost electrons it means it was oxidized (oxidation reaction) and in turn gave its electrons to H₂O₂. H₂O₂ gained electrons which means it was reduced (reduction reaction) making it the oxidizing agent and Sn²⁺ the reducing agent. Oxidizing agents gain electrons, and reduce their oxidation states. Reducing agents lose electrons and increase their oxidation states (i.e., become more oxidized).

Summary:

oxidized: Sn²⁺

reduced: H₂O₂

oxidizing agent: H₂O₂

reducing agent: Sn²⁺

b. In part (b), after separating and balancing, we get the following half reactions.

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

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

In the second half reaction, we see that Hg loses electrons and turns into Hg_2²⁺. Since Hg lost electrons it means it was oxidized (oxidation reaction) and in turn gave its electrons to PbO₂. PbO₂ gained electrons which means it was reduced (reduction reaction) making it the oxidizing agent and Hg the reducing agent. Oxidizing agents gain electrons, and reduce their oxidation states. Reducing agents lose electrons and increase their oxidation states (i.e., become more oxidized).

Summary:

oxidized: Hg

reduced: PbO₂

oxidizing agent: PbO₂

reducing agent: Hg

c. In part (c), after separating and balancing, we get the following half reactions.

1) [ \(\ce{Al ⟶ Al^3+ + 3e^-}\) ] x2 to cancel electrons from both half-reactions (oxidation)

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

In the first half reaction, we see that Al loses electrons and turns into Al³⁺. Since Al lost electrons it means it was oxidized (oxidation reaction) and in turn gave its electrons to Cr₂O_7^2-. Cr₂O_7^2- gained electrons which means it was reduced (reduction reaction) making it the oxidizing agent and Al the reducing agent. Oxidizing agents gain electrons, and reduce their oxidation states. Reducing agents lose electrons and increase their oxidation states (i.e., become more oxidized).

Summary:

oxidized: Al

reduced: Cr₂O_7^2-

oxidizing agent: Cr₂O_7^2-

reducing agent: Al

Q19.1.5

Which of the following elements is most likely to be used to prepare La by the reduction of La₂O₃: Al, C, or Fe? Why?

S19.1.5

Based on the reduction half-reaction chart we see that Al is the strongest reducing agent with a E°(V) value of -1.66, Fe has a E°(V) value of -0.036 (assuming Fe³⁺+ 3e^- → Fe (s) half reaction) and C has a E°(V) value of +0.207. Out of Al, C, and Fe we see that Al has the largest negative standard reduction potential, making it the strongest reducing agent, most capable of displacing La from La₂O₃

Q19.2.7

Name each of the compounds or ions given in Exercise Q19.2.5.

Compounds/ions from Q19.2.5:

1. [Co(en)₂(NO₂)Cl]⁺
2. [Co(en)₂Cl₂]⁺
3. [Pt(NH₃)₂Cl₄]
4. [Cr(en)₃]³⁺
5. [Pt(NH₃)₂Cl₂]

S19.2.7

1. Bis(ethylenediamine)chloronitrocobalt(III) ion

When determining oxidation states we know that ethylenediamine has a charge of 0, Chlorine has a charge of -1 and NO₂ has a charge of -1. Using these values we can set up an equation to determine the oxidation state of Cobalt.

Co + (-1) + (-1) + 2(0) = 1

Co + (-2) = 1

Co = +3 which gives us the III roman numeral

2. Dichlorobis(ethylenediamine)cobalt(III) ion

When determining oxidation states we know that ethylenediamine has a charge of 0 and Chlorine has a charge of -1. Using these values we can set up an equation to determine the oxidation state of Cobalt.

Co + 2(-1) + 2(0) = 1

Co + (-2) = 1

Co = +3 which gives us the III roman numeral

3. Diammineplatinum(IV) tetrachloride

When determining oxidation states we know that NH₃ has a charge of 0 and Chlorine has a charge of -1. Using these values we can set up an equation to determine the oxidation state of Platinum.

Pt + 4(-1) + 2(0) = 0

Pt + (-4) = 0

Pt = +4 which gives us the IV roman numeral

4. Tris(ethylenediamine)chromium(III)

When determining oxidation states we know that ethylenediamine has a charge of 0. Using these values we can set up an equation to determine the oxidation state of Chromium.

Cr + 3(0) = 3

Cr = +3 which gives us the III roman numeral

5. Diamminedichloroplatinum(II)

When determining oxidation states we know that NH₃ has a charge of 0 and Chlorine has a charge of -1. Using these values we can set up an equation to determine the oxidation state of Platinum.

Pt + 2(-1) + 2(0) = 0

Pt + (-2) = 0

Pt = +2 which gives us the II roman numeral

Q12.3.20

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N₂O₅, dissolved in chloroform, CHCl₃, is 6.2 × 10⁻⁴ min⁻¹.

\(\ce{2N2O5 ⟶ 4NO2 + O2}\)

What is the rate of the reaction when [N₂O₅] = 0.40 M?

S12.3.20

Rate of a reaction can be solved using the following equation:

rate = k [reactants]¹ Since the overall reaction is first-order N₂O₅ must be to the first power. k is the rate constant given in the problem (6.2 x 10^-4 min^-1)

rate = (6.2 x 10^-4 min^-1) (0.40 M)¹

rate = 0.000248 M/min

or 2.5 M/min

or 2.5 mol/L/min

Q12.6.11

The reaction of CO with Cl₂ gives phosgene (COCl₂), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:

• \(\ce{Cl2(g) ⇌ 2Cl(g)}\) (fast, k₁ represents the forward rate constant, k₋₁ the reverse rate constant)
• \(\ce{CO(g) + Cl(g) ⟶ COCl(g)}\) (slow, k₂ the rate constant)
• \(\ce{COCl(g) + Cl(g) ⟶ COCl2(g)}\) (fast, k₃ the rate constant)

1. Write the overall reaction.
2. Identify all intermediates.
3. Write the rate law for each elementary reaction.
4. Write the overall rate law expression.

S12.6.11

1. The reaction of CO and Cl₂ gives us overall reaction for the formation of phosgene (COCl₂):

\(\ce{CO(g) + Cl2(g) ⟶ COCl2(g)}\)

2. Intermediates:

Reaction intermediates are formed in one step and later consumed in a future step. They are not present in the overall reaction. From the three elementary steps given in the problem we can see that COCl(g) and Cl(g) are the intermediates for this reaction since they are formed in the equilibrium step and consumed in the second and third step. They are not present in the overall reaction.

3. Rate law for each elementary reaction:

Determine the rate law for each reaction by taking in consideration the k (rate constant) for each reaction and include the reactants.

rate law for step 1:

rate = k₁ [Cl₂] = k_-1 [Cl]²

rate law for step 2:

rate = k₂ [CO][Cl]

rate law for step 3:

rate = k₃ [COCl][Cl]

4. Overall rate law expression

Taking account all the rate laws we determined for each elementary reaction we put them all together to make the overall rate law expression:

k₁ [Cl₂] = k_-1 [Cl]²

= k₁ [Cl₂] = k_-1 [Cl]²

= k₁ [Cl₂]/k_-1 = [Cl]²

= ( k₁ [Cl₂]/k_-1 )^1/2 = [Cl] <--substitute into second step rate expression

k₂ [CO][Cl]

= k₂ ( k₁ [Cl₂]/k_-1 )^1/2 [CO]

= k [Cl₂]^1/2 [CO]

k₃ [COCl][Cl]

= k₃ [COCl] ( k₁ [Cl₂]/k_-1 )^1/2

= (k₃(k₁/k_-1)^1/2) [COCl][Cl₂]^1/2

= (k₃(k₁/k_-1)^1/2) [COCl][Cl] (COCl and Cl are intermediates)

Overall rate law expression: = (k₃(k₁/k_-1)^1/2\

Q21.4.23

Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this ²³⁹Pu was the capture of neutrons by ²³⁸U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 10⁹ years ago?

S21.4.23

²³⁹Pu is an isotope of Plutonium and has a half-life of 24,110 years. The fission of an atom of uranium-235 produces 2-3 neutrons which can be absorbed by uranium-238 to produce plutonium-239. This shows that no Pu-239 could remain since the formation of the earth. Because of this, present plutonium could not have been formed with the ²³⁸Uranium.

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?

S20.3.11

First, sulfate (SO₄2−) is reduced to HS^- in the presence of glucose (C₆H₁₂O₆).

reduction: \(\ce{SO4^2-(aq) + 9H^+(aq) + 8e^- → HS^-(aq) + 4H2O(l)}\)

Second, glucose is oxidized to bicarbonate.

oxidation: \(\ce{C6H12O6(aq) + 12H2O(l) → 6HCO3^-(g) + 30H^+(aq) + 24e^-}\)

Finally, when combining the two half-reactions we get the following overall reaction:

After multiplying the reduction reaction by 3 to cancel the electrons from both half-reactions, subtracting the H⁺'s from both sides, and canceling out the H₂O we get the following half-reactions:

3SO₄^2-(aq) → 3HS^-(aq) + 12H₂O(l)

C₆H₁₂O₆(aq) + 12H₂O(l) → 6HCO₃^-(g) + 3H⁺(aq)

Finally, when combining the two half-reactions we get the following overall reaction:

overall: \(\ce{C6H12O6(aq) + 3SO4^2-(aq) → 6HCO3^-(g) + 3H^+(aq) + 3HS^-(aq)}\)

Q20.5.22

Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:

\(\ce{Mn^3+(aq) + e^- → Mn^2+(aq)}\) E°=1.51 V

\(\ce{Mn^3+(aq) + 2H2O(l) → MnO2(s) + 4H^+(aq) + e^-}\) E°=0.95 V

1. What is E° for the disproportionation reaction?
2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
3. How could you prevent the disproportionation reaction from occurring?

S20.5.22

1. E° for the disproportionation reaction:

E°cell = E°cathode - E°anode

The cathode is where reduction occurs and the anode is where oxidation occurs. In this case the reduction reaction is \(\ce{Mn^3+(aq) + e^- → Mn^2+(aq)}\) (E°=1.51 V) and the oxidation reaction is \(\ce{Mn^3+(aq) + 2H2O(l) → MnO2(s) + 4H^+(aq) + e^-}\) (E°=0.95 V).

E°cell = 1.51 V - 0.95 V

E°cell = 0.56V (spontaneous)

2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0?

Disproportionation is more thermodynamically favored at a low pH than at pH 7.0 because as the pH goes up the rate for the forward reaction will slow because the radicals will become more persistent.

Disproportionation is less thermodynamically favored at low pH because as pH increases by 1.0, [H⁺] decreases by a factor of 10 because pH = -log[H⁺]. Thus, at pH 7.0, the concentration of [H⁺] is lower, which according to Le Chatelier's principle, makes the forward reaction more favorable since H⁺ is a product in the overall reaction. In other words, increasing pH decreases the value of Q, which according to the Nernst equation, increases the overall cell potential for a reaction.

3. Preventing the disproportionation reaction from occurring:

Disproportionation reactions occur because one state is unstable when it comes to both its reduced and oxidized form which is what makes it disproportionate. The disproportionate reaction occurs to gain more stability, you can prevent a disproportionation reaction from occurring by removing the water. The key to disproportionation reactions are the oxidation states of oxygen which both reduce and oxidize the Mn(III).