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.

S17.1.6

Given:

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

\(PbO_2+Hg\rightarrow Hg^{2+}_2+Pb^{2+}\)

\(Al+Cr_2O^{2-}_7\rightarrow Al^{3+}+Cr^{3+}\)

Wanted:

• Species that undergoes oxidation
• Species that undergoes reduction
• Oxidizing Agent
• Reducing Agent

A species that undergoes oxidation gives up an electron, thus becoming more positive in charge.

A species that undergoes reduction gains an electron, thus becoming more negative in charge.

An oxidizing agent (oxidant) gains electrons, thus it undergoes reduction.

An reducing agent (reductant) loses electrons, thus it undergoes oxidation.

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

Step 1: Make sure half-equations are balanced

Step 2: Separate equation into two half-reactions

\[H_2O_2\rightarrow H_2O\]

\[Sn^{2+} \rightarrow Sn^{4+}\]

Step 3: Make sure oxygen is balanced by adding H₂O

\[H_2O_2\rightarrow 2H_2O\]

\[Sn^{2+} \rightarrow Sn^{4+}\]

Step 4: Add H⁺ to balance hydrogen

\[H_2O_2+2H^+\rightarrow 2H_2O\]

\[Sn^{2+} \rightarrow Sn^{4+}\]

Step 5: Balance charges of the equations with e^-

\[H_2O_2+2H^++2e^-\rightarrow 2H_2O\]

\[Sn^{2+} \rightarrow Sn^{4+}+2e^-\]

Step 6: Half-equations with electron on reactant side are being reduced, and half-equations with electron on product side are being oxidized.

Oxidized: \(Sn^{2+}\)

Reduced: \(H_2O_2\)

Reducing Agent: \(Sn^{2+}\)

Oxidation Agent: \(H_2O_2\)

b) \(PbO_2+Hg\rightarrow Hg^{2+}_2+Pb^{2+}\)

Step 1: Make sure half-equations are balanced

Step 2: Separate equation into two half-reactions

\[PbO_2\rightarrow Pb^{2+}\]

\[Hg\rightarrow Hg^{2+}_2\]

Step 3: Make sure oxygen is balanced by adding H₂O and balance the Mercury

\[PbO_2\rightarrow Pb^{2+}+2H_2O\]

\[2Hg\rightarrow Hg^{2+}_2\]

Step 4: Add H⁺ to balance hydrogen

\[PbO_2+4H^+\rightarrow Pb^{2+}+2H_2O\]

\[2Hg\rightarrow Hg^{2+}_2\]

Step 5: Balance charges of the equations with e^-

\[PbO_2+4H^++4e^-\rightarrow Pb^{2+}+2H_2O\]

\[2Hg\rightarrow Hg^{2+}_2+4e^-\]

Step 6: Half-equations with electron on reactant side are being reduced, and half-equations with electron on product side are being oxidized.

Oxidized: \(Hg\)

Reduced: \(PbO_2\)

Reducing Agent: \(Hg\)

Oxidation Agent: \(PbO_2\)

c) \(Al+Cr_2O^{2-}_7\rightarrow Al^{3+}+Cr^{3+}\)

Step 1: Make sure half-equations are balanced

Step 2: Separate equation into two half-reactions

\[Al\rightarrow Al^{3+}\]

\[Cr_2O^{2-}_7\rightarrow Cr^{3+}\]

Step 3: Make sure oxygen is balanced by adding H₂O and balance the Cromium

\[Al\rightarrow Al^{3+}\]

\[Cr_2O^{2-}_7\rightarrow 2Cr^{3+}+7H_2O\]

Step 4: Add H⁺ to balance hydrogen

\[Al\rightarrow Al^{3+}\]

\[Cr_2O^{2-}_7+14H^+\rightarrow 2Cr^{3+}+7H_2O\]

Step 5: Balance charges of the equations with e^-

\[Al\rightarrow Al^{3+}+3e^-\]

\[Cr_2O^{2-}_7+14H^++14e^-\rightarrow 2Cr^{3+}+7H_2O\]

Step 6: Half-equations with electron on reactant side are being reduced, and half-equations with electron on product side are being oxidized.

Oxidized: \(Al\)

Reduced: \(Cr_2O^{2-}_7\)

Reducing Agent: \(Al\)

Oxidation Agent: \(Cr_2O^{2-}_7\)

Chang's Review 17.1.6

Overall the tutorial was easy to follow and well organized. Improvements I would add is instructions on how to determine the oxidation states of the elements in order to figure out which species is going through oxidation or reduction to create the half reactions. I would also highlight the changes you made when it comes to balancing your equations such as highlighting the number to indicate what you balanced. Also in part C there was a mistake on the # of electrons added to the reduction side. Since Cr₂O₇^2- and 14H⁺ gave a overall charge of +12 and the 2Cr³⁺ has a overall charge of +6 since water has no charge, you only need to add 6 e^- rather than 14 e^- to the reduction reaction.

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

An activity series is a list of elements in decreasing order of their reactivity. Elements on the top of the list are good reducing agents because they easily give up an electron, and elements on the bottom of the series are good oxidizing agents because they are highly electronegative would really want to accept an electron.

Step 1: Compare Aluminum, Carbon, and Iron on an activity series. Many activity series include carbon and hydrogen as references. An activity series can be found here

The activity series goes in the order (from top to bottom): Aluminum, Carbon, and Iron.

Step 2: Identify which element is the best reducing agent.

Elements on the top of the list are the best reducing agents, because they give up electrons the best.

Aluminum is the best reducing agent.

Solution: Aluminum will be the best reducing agent to prepare La by the reduction of La₂O₃ because it is the most reactive in the series amongst the three elements.

Chang's Review 19.1.5

I agree with the answer because aluminum is the best reducing agent because it is at the top of the activity series. A change I would make is providing a screenshot of the activity series with a red box around aluminimum and iron so the reader can locate them more easily. The link was helpful but I think including an image would be even better. I provided a small example below.

Q19.2.7

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

S19.2.7

Given:

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

Wanted:

Names of the above compounds.

1. [Co(en)₂(NO₂)Cl]⁺

Step 1: Attain the names of the ligands and metal cation. Names can be found here.

Co: Cobalt

en: Ethylenediamine

NO₂: Nitro

Cl: Chloro

Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.

(en)₂: bis(Ethylenediamine)

Step 3: Find the charges of the ligands. Charges can be found here.

en: 0

NO₂: -1

Cl: -1

Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.

Co + 2(en) + (NO₂) + Cl = 1

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

Co = 3

Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.

Chlorobis(ethylenediamine)nitrocobalt(III)

2. [Co(en)₂Cl₂]⁺

Step 1: Attain the names of the ligands and metal cation. Names can be found here.

Co: Cobalt

en: Ethylenediamine

Cl: Chloro

Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.

(en)₂: bis(Ethylenediamine)

Cl₂: dichloro

Step 3: Find the charges of the ligands. Charges can be found here.

en: 0

Cl: -1

Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.

Co + 2(en) +2(Cl) = 1

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

Co = 3

Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.

Dichlorobis(Ethylenediamine)cobalt(III)

3. [Pt(NH₃)₂Cl₄]

Step 1: Attain the names of the ligands and metal cation. Names can be found here.

Pt: Platinum

NH₃: Ammine

Cl: Chloro

Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.

(NH₃)₂: diammine

Cl₄: tetrachloro

Step 3: Find the charges of the ligands. Charges can be found here.

NH₃: 0

Cl: -1

Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.

Pt + 2(NH₃) + 4(Cl) = 0

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

Pt = 4

Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.

Diamminetetrachloroplatinum(IV)

4. [Cr(en)₃]³⁺

Step 1: Attain the names of the ligands and metal cation. Names can be found here.

Cr: Cromium

en: ethylenediamine

Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.

(en)₃: tris(ethylenediamine)

Step 3: Find the charges of the ligands. Charges can be found here.

en: 0

Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.

Cr + 3(en) = 3

Cr + 3(0) = 3

Cr = 3

Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.

Tris(ethylenediamine)cromium(III)

5. [Pt(NH₃)₂Cl₂]

Step 1: Attain the names of the ligands and metal cation. Names can be found here.

NH₃: Ammine

Cl: Chloro

Pt: Platinum

Step 2: Add the appropriate pre-fixes to each ligand depending on the number.

(NH₃)₂: diammine

Cl₂: dichloro

Step 3: Find the charges of the ligands. Charges can be found here.

NH₃: 0

Cl: -1

Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.

Pt + 2(NH₃) + 2(Cl) = 0

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

Pt = 2

Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.

Diamminedichloroplatinum(II)

Chang's Review

I really like the step wise solution and the hyperlinks that explains how to figure out the prefixes and the naming of the ligands. Personally I like figuring out the charge of the metal ion first before proceeding onward. I liked how the tutorial showed how to derive the charge of the metal ion step by step and how each ligand was named so that it is easier to piece together the nomenclature. Bolding the answer would make it stand out more.

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.

\[2N_2O_5\rightarrow 4NO_2+O_2\]

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

S12.3.20

Given:

Rate constant (k): 6.2 × 10⁻⁴ min⁻¹

Order of N₂O₅: first order

Concentration of \(N_2O_5\): 0.40 M

Wanted:

Rate of the reaction

\(rate= k[A]^m[B]^n\) shows that the rate of the reactants depend on the concentration of the reactants. The values of m and n are reaction order with respect to A and B respectively. Order does not depend on stoichiometric values, but rather trials. The rate constant (k) does not depend on concentrations of the reactants (A and B), but is affected by surface area and temperature.

Step 1: Write down the overall equation with the given values.

\(rate= 6.2*10^{-4}[N_2O_5]\)

Step 2: Plug in given molarity value and solve.

\(rate= 6.2*10^{-4}[0.40]\)

\(rate=\frac{0.00025M}{min}\)