Extra Credit 6

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Q17.1.5B

Balance the following in acidic solution:

\(\displaystyle H_2O_2+Sn^{2+}\longrightarrow H_2O+Sn^{4+}\)
\(\displaystyle PbO_2+Hg\longrightarrow Hg_2^{2+}+Pb^{2+}\)
\(\displaystyle Al+Cr_2O_7^{2-}\longrightarrow Al^{3+}+Cr^{3+}\)

Q17.1.5B Answer

1. \(\displaystyle H_2O_2+Sn^{2+}\longrightarrow H_2O+Sn^{4+}\)

Step 1: assign oxidation numbers

H₂O₂ The oxidation number of O in compounds is usually -2, but it is -1 in peroxides. This makes H +1.
H₂O Not a peroxide so O is -2, and H is +1.
Sn²⁺Because of the charge it is 2+
Sn⁴⁺ Because of the charge it is 4+

Step 2: split into half reactions \(\displaystyle H_2O_2\longrightarrow H_2O\) and \(\displaystyle Sn^{2+} \longrightarrow Sn^{4+}\)

Step 3: determine which half reaction is reducing (gaining electrons) and which is oxidizing (losing electrons), and balance the electrons.

\(\displaystyle H_2O_2\longrightarrow H_2O\) : Oxygen goes from -1 to -2 so it is gaining 1 electron. The new reaction is \(\displaystyle H_2O_2+e^{-1}\longrightarrow H_2O\)

\(\displaystyle Sn^{2+} \longrightarrow Sn^{4+}\) : Sn goes from +2 to +4 so it loses 2 electrons. The new reaction is \(\displaystyle Sn^{2+} \longrightarrow Sn^{4+}+2e^{-1}\)

Step 4: balance the reactions with regards to species

\(\displaystyle H_2O_2+e^{-1}\longrightarrow H_2O\) : there are two oxygens on the reactant side and only one on the product side. To balance oxygens add H₂O. The new reaction is \(\displaystyle H_2O_2+e^{-1}\longrightarrow 2H_2O\) .

To balance the Hydrogens add H⁺ to the side that is lacking. The new balanced reaction is \(\displaystyle H_2O_2+e^{-1}+2H^{+}\longrightarrow 2H_2O\).

\(\displaystyle Sn^{2+} \longrightarrow Sn^{4+}+2e^{-1}\): already balanced

Step 5: Balance the two half reactions together by matching the coefficient before the electrons.

\(\displaystyle H_2O_2+e^{-1}+2H^{+}\longrightarrow 2H_2O\) : 1 electron

\(\displaystyle Sn^{2+} \longrightarrow Sn^{4+}+2e^{-1}\) : 2 electrons

\(\displaystyle 2(H_2O_2+e^{-1}+2H^{+}\longrightarrow 2H_2O)=2H_2O_2+2e^{-1}+4H^{+}\longrightarrow 4H_2O\) : 2 electrons!

Step 6: Conjoin the two new balanced half reactions, and because the electrons balance eachother out, they can be removed from the final reaction.

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

2. \(\displaystyle PbO_2+Hg\longrightarrow Hg_2^{2+}+Pb^{2+}\)

Step 1: assign oxidation numbers

PbO₂ The oxidation number of O in compounds is usually -2 which makes Pb +4
Hg No charge
Hg₂²⁺the charge it is +2, but theres two molecules so \(\displaystyle \frac{2+}{2}\) = +1
Pb²⁺ Because of the charge it is +2

Step 2: split into half reactions \(\displaystyle PbO_2\longrightarrow Pb^{2+}\) and \(\displaystyle Hg\longrightarrow Hg_2^{2+}\)

Step 3: determine which half reaction is reducing (gaining electrons) and which is oxidizing (losing electrons), and balance the electrons.

\(\displaystyle PbO_2\longrightarrow Pb^{2+}\): lead goes from +4 to +2 so it gains 2 electron. The new reaction is \(\displaystyle PbO_2+2e^{-1}\longrightarrow Pb^{2+}\)

\(\displaystyle Hg\longrightarrow Hg_2^{2+}\): Hg goes from 0 to +2 so it loses 2 electrons. The new reaction is \(\displaystyle Hg\longrightarrow Hg_2^{2+}+2e^{-1}\)

Step 4: balance the reactions with regards to species

\(\displaystyle PbO_2+2e^{-1}\longrightarrow Pb^{2+}\): there are two oxygens on the reactant side and none on the product side. To balance oxygens add H₂O to the product side until the oxygens are equal. The new reaction is \(\displaystyle PbO_2+2e^{-1}\longrightarrow Pb^{2+}+2H_2O\)

To balance the Hydrogens add H⁺ to the side that is lacking. The new balanced reaction is \(\displaystyle PbO_2+2e^{-1}+2H^{+}\longrightarrow Pb^{2+}+2H_2O\).

\(\displaystyle Hg\longrightarrow Hg_2^{2+}+2e^{-1}\): there is one Hg on the reactant side and two on the product so and an Hg to the reactant side. The new balanced equation is \(\displaystyle 2Hg\longrightarrow Hg_2^{2+}+2e^{-1}\)

Step 5: Balance the two half reactions together by matching the coefficient before the electrons.

\(\displaystyle PbO_2+2e^{-1}+2H^{+}\longrightarrow Pb^{2+}+2H_2O\). : 2 electrons

\(\displaystyle 2Hg\longrightarrow Hg_2^{2+}+2e^{-1}\) : 2 electrons

Step 6: Conjoin the two new balanced half reactions, and because the electrons balance each other out they can be removed from the final reaction.

\(\displaystyle PbO_2+2H^{+}+2Hg\longrightarrow Pb^{2+}+2H_2O+Hg_2^{2+}\).

3. \(\displaystyle Al+Cr_2O_7^{2-}\longrightarrow Al^{3+}+Cr^{3+}\)

Step 1: assign oxidation numbers

Al Has no charge.
Al³⁺ Because of the charge it is +3
Cr₂O₇^2-Oxygen is -2. Times -2 by 7, because there are 7 molecules, so it's -14. Subtract -2 from -14, because that's the charge of the molecule, which means Cr₂ has a combined charge of -12. Divide -12 by two because there are two Cr molecules, so each Cr molecule has a charge of -6.
Cr³⁺ Because of the charge it is +3

Step 2: split into half reactions \(\displaystyle Al\longrightarrow Al^{3+}\) and \(\displaystyle Cr_2O_7^{2-}\longrightarrow Cr^{3+}\)

Step 3: determine which half reaction is reducing (gaining electrons) and which is oxidizing (losing electrons), and balance the electrons.

\(\displaystyle Al\longrightarrow Al^{3+}\): Al goes from 0 to +3 so it is losing 3 electrons. The new reaction is \(\displaystyle Al\longrightarrow Al^{3+}+3e^{1-}\)

\(\displaystyle Cr_2O_7^{2-}\longrightarrow Cr^{3+}\): Cr goes from -6 to +3 so it gains 9 electrons. The new reaction is \(\displaystyle Cr_2O_7^{2-}+9e^{9-}\longrightarrow Cr^{3+}\)

Step 4: balance the reactions with regards to species

\(\displaystyle Al\longrightarrow Al^{3+}+3e^{1-}\): already balanced

\(\displaystyle Cr_2O_7^{2-}+9e^{9-}\longrightarrow Cr^{3+}\): there are 7 oxygens on the reactant side and none on the product side. To balance oxygens add H₂O. The new reaction is \(\displaystyle Cr_2O_7^{2-}+9e^{9-}\longrightarrow Cr^{3+}+7H_2O\).

To balance the Hydrogens add H⁺ to the side that is lacking. The new balanced reaction is \(\displaystyle Cr_2O_7^{2-}+9e^{9-}+14H^{+}\longrightarrow Cr^{3+}+7H_2O\)

Step 5: Balance the two half reactions together by matching the coefficient before the electrons.

\(\displaystyle Al\longrightarrow Al^{3+}+3e^{1-}\) : 3 electrons

\(\displaystyle Cr_2O_7^{2-}+9e^{9-}+14H^{+}\longrightarrow Cr^{3+}+7H_2O\) : 9 electrons

\(\displaystyle 3(Al\longrightarrow Al^{3+}+3e^{1-})=3Al\longrightarrow 3Al^{3+}+9e^{1-}\): 9 electrons!!!

Step 6: Conjoin the two new balanced half reactions, and because the electrons balance eachother out, they can be removed from the final reaction.

\(\displaystyle Cr_2O_7^{2-}+9e^{9-}+14H^{+}+3Al\longrightarrow Cr^{3+}+7H_2O+3Al^{3+}\)

Q19.1.4

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

Q19.1.4 Answer

Because Lanthanoid elements have too many protons. Anything above Uranium in the periodic table is too unstable to be found in nature.

Q19.2.6

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

[Co(CO₃)₃]³⁻ (note that CO₃²⁻ is bidentate in this complex)
[Cu(NH₃)₄]²⁺
[Co(NH₃)₄Br₂]₂(SO₄)₃
[Pt(NH₃)₄][PtCl₄]
[Cr(en)₃](NO₃)₃
[Pd(NH₃)₂Br₂] (square planar)
K₃[Cu(Cl)₅]
[Zn(NH₃)₂Cl₂]

Q19.2.6 Answer

To name coordination compounds name the cations and then anions. In the brackets the metal is always first but when writing the name the metal is always written last. Add prefixes to the different ligands if there is more than one of them. Refer to table 2.4.1 for the proper prefixes. List the anionic ligands first, the neutral ligands, second, and the cationic ions third. If there is more than one anionic, neutral, or cationic ligand list them in alphabetic order. Put the charge of the metal in parentheses outside of the name in roman numerals. If things are arranged in a certain geometric way add cis and trans to the front if they apply. If their is a bridging ligand write tris. If there are not two species that neutralize each other add ion to then end of the name.

tricarbonatocobaltate(III) ion;
tetraaminecopper(II) ion;
tetraaminedibromocobalt(III) sulfate;
tetraamineplatinum(II) tetrachloroplatinate(II);
tris-(ethylenediamine)chromium(III) nitrate;
diaminedibromopalladium(II);
potassium pentachlorocuprate(II);
diaminedichlorozinc(II)

Q12.3.19

For the reaction Q⟶W+XQ⟶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?
What is the rate constant?

Q12.3.19 Answer

1. use the formula \(\displaystyle \frac{rate_2}{rate_1}=k\frac{Q_2}{Q_1}^{x}\)

where \(\displaystyle Q_1\) is [Q] intial and \(\displaystyle rate_1\) is the coinciding rate. In this problem \(\displaystyle Q_1\) is .170 and is \(\displaystyle 6.68*10^{-3}\)

\(\displaystyle Q_2\) is [Q] is a different molarity and \(\displaystyle rate_2\) is its coinciding rate. In this problem I'm going to use .357 as \(\displaystyle Q_2\) and its coinciding \(\displaystyle rate_2\) is \(\displaystyle 2.94*10^{-2}\)

\(\displaystyle rate_2\) should be larger a larger number than \(\displaystyle Q_1\)

now insert these into the equation \(\displaystyle \frac{.170}{.357}=k\frac{6.68*10^{-3}}{2.94*10^{-2}}^{x}\)

To find the order of the reaction we need to figure out what x is. To find x ignore k (the rate constant) this gives you \(\displaystyle 2.1=4.41^{x}\)

This means that x is 2 and the the order of the reaction is 2.

2. write the rate formula for the original reaction using the now known order of reaction: \(\displaystyle rate=k[Q]^{2}\)

insert one column of the table to find k. It does not matter which column k will be the same for all of them. This gives you \(\displaystyle 6.68*10^{-3}=k[.17]^{2}\) now solve for k.

k equals 2.31

Q12.6.10

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

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

Initial concentrations and rates of reaction are given here.

Experiment	Initial Concentration [NO] (mol/L)	Initial Concentration, [H₂] (mol/L)	Initial Rate of Formation of N₂ (mol/L min)
1	0.0060	0.0010	1.8 × 10⁻⁴
2	0.0060	0.0020	3.6 × 10⁻⁴
3	0.0010	0.0060	0.30 × 10⁻⁴
4	0.0020	0.0060	1.2 × 10⁻⁴

Consider the following questions:

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

Step 1: \(\displaystyle 2NO \leftrightharpoons N_2O_2\)

Step 2: \(\displaystyle N_2O_2+H_2 \leftrightharpoons H_2O+N_2O\)

Step 3: \(\displaystyle N_2O+H_2 \leftrightharpoons N_2+H_2O\)

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.

Q12.6.10 Answer

1. use the formula \(\displaystyle \frac{rate_2}{rate_1}=k\frac{Q_2}{Q_1}^{x}\) to find the order of each reactant.

to find the order of NO find the two initial concentrations where the concentration of H₂ remained the same and input this data into the equation with the correct corresponding rates. Write the bigger concentration as \(\displaystyle rate_2\). This gives you the equation \(\displaystyle \frac{1.2*10^{-4}}{.3*10^{-4}}=k\frac{.002}{.001}^{x}\) ignore the k constant and simplify, this gives you \(\displaystyle 4=2^{x}\) which means that x = 2 and that NO is second order.
to find the order of H₂find the two initial concentrations where the concentration of NO remained the same and input this data into the equation with the correct corresponding rates. Write the bigger concentration as \(\displaystyle rate_2\). This gives the equation \(\displaystyle \frac{3.6*10^{-4}}{1.8*10^{-4}}=k\frac{.002}{.001}^{x}\) ignore the k constant and simplify, this gives you \(\displaystyle 2=2^{x}\) which means that x = 1 and that H₂is first order.

2. the correct overall rate law for the reaction is \(\displaystyle rate=k[NO]^{2}[H_2]\)

3. to calculate the value of the k constant input one row of data from the table in the rate law. It does not matter which row you choose k will be the same for all rows. \(\displaystyle 1.8*10^{-4}=k[.006]^{2}[.001]\) simplified this gives you \(\displaystyle 1.8*10^{-4}=k*(3.6*10^{-8})\) this means that k is \(\displaystyle 5000 M^{-2}L^{-2}min^{-1}\)

4. Let y equal the volume of the reaction vessel in liters. In doing so, we can calculate the moles of each reactant.

NO concentration: \(\mathrm{(0.0060 mol/liters) x (y liters) = 0.0060 mol}\)

H₂ concentration: \(\mathrm{(0.0010 mol/liters) x (y liters) = 0.0010 mol}\)

Now that both reactants are both in moles. We can subtract the moles of H₂ from the moles of NO.

\(\mathrm{0.0060 mol - 0.0010 mol}\)

Now that we got 0.0050 mol, we can now divide by the volume of the container (y). The volume for both reactants is y so we can let y = 1 for the final answer.

\(\displaystyle .005 \frac{mol}{L}\)

5. If step I gives N₂O₂ in adequate amounts, steps 1 and 2 combine to give \(\displaystyle 2NO+H_2 \rightarrow H_2O+N_2O\). This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry. This means that step II is the rate determining step.

Q21.4.22

A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of \(\displaystyle U_{92}^{238}\) and 2.52 mg of \(\displaystyle Pb_{82}^{206}\). Calculate the age of the ore. The half-life of \(\displaystyle U_{92}^{238}\) is \(\displaystyle 4.5*10^{9}\) yr.

Q21.4.22 Answer

First Determine the percentage of U in the sample. You can use the following equation because of the law of the conservation of mass and the mass that was uranium has turned into lead.

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

2. Then use the formula for half life \(\displaystyle A=A_0(.5\frac{t}{h^{2}})\) For your initial mass use 1 (100% U) and .6806 as your current mass (68.06% left) and solve for t

\(\displaystyle .6808=.5^{\frac{t}{(4.5*10^{9})^{2}}}\) t = 2,498,044,246 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.

440px-Corrosão_de_ferro_por_sacrifício_em_contacto_com_cobre.jpg

Q20.3.10 Answer

Identify half Reactions:

oxidation: \(\displaystyle {Fe(s) + 2OH^- ⟶ Fe(OH)2 + 2e^-}\)

reduction: \(\displaystyle {[Fe(CN)6]^3- + e^- ⟶ [Fe(CN)6]^4-}\)

the overall reaction is: \(\displaystyle {2[Fe(CN)6]^3- + Fe(s) + 2OH^- ⟶ Fe(OH)2 + 2[Fe(CN)6]^4-}\)

The gel contains Phenolphthalein so it has the ability to turn pink. Iron is easily oxidized under basic conditions.

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:

\(\displaystyle MnO_4^{−}(aq) + e^{-} \rightarrow MnO_4^{-2}(aq)\); E° = +0.56 V (purple → dark green)
\(\displaystyle MnO_4^{-2}(aq) + 2e^{-}+ 4H^{+}(aq) \rightarrow MnO_2(s)\); E° = +2.26 V (dark green → dark brown solid)
\(\displaystyle MnO_2(s) + e^{-} + 4H^{+}(aq) → Mn^{+3}(aq)\); E° = +0.95 V (dark brown solid → red-violet)
\(\displaystyle Mn^{3+}(aq) + e^{-} \rightarrow Mn^{2+}(aq)\); E° = +1.51 V (red-violet → pale pink)
\(\displaystyle Mn^{2+}(aq) + 2e^{-} \rightarrow Mn(s)\); 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?
Is the reduction of Mn³⁺(aq) to Mn(s) by H₂(g) spontaneous under standard conditions? What is E°cell?

Q20.5.21 Answer

1. Take the \(\displaystyle E^{o}\)values for equation 2 and 3 and subtract anode \(\displaystyle E^{o}\) values from the cathode \(\displaystyle E^{o}\) values

\(\displaystyle E^{o}cell=(2.26)−(.95)=1.31V\)

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

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

\(\displaystyle E^{o}cell=(1.51)−(−1.81)=3.32V\)

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

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Q17.1.5B

Q17.1.5B Answer

1. \(\displaystyle H_2O_2+Sn^{2+}\longrightarrow H_2O+Sn^{4+}\)

2. \(\displaystyle PbO_2+Hg\longrightarrow Hg_2^{2+}+Pb^{2+}\)

3. \(\displaystyle Al+Cr_2O_7^{2-}\longrightarrow Al^{3+}+Cr^{3+}\)

Q19.1.4

Q19.1.4 Answer

Q19.2.6

Q19.2.6 Answer

Q12.3.19

Q12.3.19 Answer

Q12.6.10

Q12.6.10 Answer

Q21.4.22

Q21.4.22 Answer

Q20.3.10

Q20.3.10 Answer

Q20.5.21

Q20.5.21 Answer