Extra Credit 13

Q17.2.2

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions.

Mg(s)│Mg²⁺(aq)║Cu²⁺(aq)│Cu(s)

Ni(s)│Ni²⁺(aq)║Ag⁺(aq)│Ag(s)

Answer:

Mg (s)= oxidized ; Cu²⁺(aq) = reduced ; Mg(s)= reducing agent ; Cu²⁺(aq)= oxidizing agent

Ni(s)= oxidized ; Ag⁺(aq) = reduced ; Ni(s) = reducing agent ; Ag⁺(aq) = oxidizing agent

You know Mg is oxidized because oxidation is a loss of electrons, which means a gain in positive charge. Mg loses 2 electrons to become Mg²⁺.The species that is oxidized is a reducing agent because it donates electrons in the redox reaction. Reduction is a gain of electrons Cu²⁺ gains 2 electrons to become Cu. Cu is the oxidizing agent because it acts as an electron acceptor which allows it to be reduced.

You know Ni is oxidized because oxidation is a loss of electrons and, which means a change to a more positive charge. Ni becomes Ni²⁺. Ni is a reducing agent because it allows reduction to occur by donating an electron. Ag⁺ is reduced because it gains an electron and therefore gains a negative charge to become Ag. Ag⁺ is the oxidizing agent because it accepts the electron that is donated by Ni.

Q19.1.11

Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na₂Cr₂O₇ is required in the titration. What percentage of the ore sample was iron?

Answer:

In order to find the percentage of the ore sample that was iron we must find the weight in grams of iron used, which we have the ability to find since we titrated it against a known concentration of 0.0100 M Na₂Cr₂O_7.

First you must write a balanced reaction based on the information given in the problem.

6Fe²⁺_(aq) + Cr₂O₇^2-_(aq) + 14H₃O_(aq) ⟶ 6Fe³⁺_(aq) + 2Cr³⁺_(aq) +21H₂O_(l)

We are working with a 2.5 g sample of iron and 19.7 ml of Na₂Cr₂O₇ that is 0.01 M.

Since we are titrating the ore against the 0.01 M Cr₂O₇^2- we can use the molarity and volume to find moles of Cr₂O₇^2- used. By multiplying 0.01 M and 19.17 ml, which is 0.01917 L, we find that 1.97x10^-4 moles of Cr₂O₇^2- were used.

Based on the stochiometry of the problem we know for every one mole of Cr₂O₇^2- used 6 moles of iron are needed. We can multiply 1.97x10^-4 moles by 6 to find the number of moles of iron used, which is 1.1502x10^-3 moles.

We know a very basic conversion to go from moles to grams. We multiply the moles of iron by the molar mass of iron, 1.1502x10^-3 moles x 55.487g = .06424 g of iron.

Finally to find the percentage of iron used we simply divide the mass used, 0.06424 by the total mass 2.5 to get .02569 g. We multiply this by 100 to get a percentage of 2.57%.

Q19.3.1

Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH₃)₆]Cl₃.

Answer:

Oxidation state: Co³⁺

We can determine the oxidation state of the metal by first looking at the charge of the ligand (NH₃) which in this case is neutral. We then look at the oxidation state of the balancing anion Cl^-. Our overall complex is not charged therefore our metal needs to balance out the -3 charge coming from the 3 Cl^-.

Number of d electrons: 6

We determine the number of electrons by looking at the charge on Co³⁺ we know the first 2 electrons are lost by the S block because of their lower bond energy and the last one is lost in the d block resulting in 6 electrons.

Number of unpaired electrons: 0

Since this is an octahedral transition metal complex we have to take into account the fact that there are two spin options for this complex; high or low. We know this complex is low spin because NH₃ is a strong field ligand which creates a high delta O and allows for electron pairing on the T₂G level since the energy pairing here is lower than the energy required to move the electron to the Eg block. The oxidation state 3+ also affects the spin of the complex. The higher the oxidation state the stronger the ligand field.

Q12.4.3

Use the data provided in a graphical method to determine the order and rate constant of the following reaction:

2P⟶Q+W

Answer:

Because the question asked for a graphical analysis I used an excel spread sheet to analyze the data. I found the slope of each line and determined that the reaction is second order.

Q12.7.6

For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:

(a)

energy required- initial energy= E^a

35-10 = 25 kJ

(b)

energy required- initial energy= E^a

20-10=10 kJ

Q21.5.1

Write the balanced nuclear equation for the production of the following transuranium elements:

1. berkelium-244, made by the reaction of Am-241 and He-4
2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons
3. lawrencium-257, made by the reaction of Cf-250 and B-11
4. dubnium-260, made by the reaction of Cf-249 and N-15

Answer:

1. ²⁴¹₉₅Am + ⁴₂He ⟶ ²⁴⁴₉₇Bk + ¹₀n
2. ²³⁹₉₄Pu + 15 ¹₀n ⟶ ²⁵⁴₁₀₀Fm + 6⁰_-1e
3. ²⁵⁰₉₈Cf + ¹¹₅B ⟶ ²⁵⁷₁₀₃Lr + 4¹₀n
4. ²⁴⁹₉₈Cf + ¹⁵₇N ⟶ ²⁶⁰₁₀₅Db + 4¹₀n

Edit: When balancing the equation one must ensure the sum of mass numbers and atomic numbers (respectively) on either side of the reaction must be equal to each other.

Q20.3.15

Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.

1. 2Cl⁻_(aq) + Br²_(l) → Cl²_(g) + 2Br⁻_(aq)
2. 2NO2(g) + 2OH−(aq) → NO2−(aq) + NO3−(aq) + H2O(l)
3. 2H2O(l) + 2Cl-(aq) → H2(g) + Cl2(g) + 2OH−(aq)
4. C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)

Answer:

1. oxidation/ anode : 2Cl^-_(aq) → Cl_2(g) + 2e^- 1.35872

reduction/ cathode: 2e^- + Br_2(l)→ 2Br^-_(aq) 1.0873

E^o_cell= cathode- anode

E^o_cell= 1.0873-1.35827

E^o_cell=-.27097

A negative E^o_cell indicates a non-spontaneous reaction. The spontaneous reaction would be the reverse. The spontaneous cell diagram would be written as

Pt_(s)| Br^-_(aq)|Br_2(l)||Cl_2(g)|Cl^-_(aq)|Pt_(s)

2. oxidation/ anode: 2OH^-_(aq) + No_2(g)→ NO₃^-_(aq) + H₂O_(l) + e^- -.8

reduction/ cathode: e^- + NO_2(g)→ NO_2(aq) 1.07

E^o_cell= cathode- anode

E^o_cell= 1.07 - (-.8)= 1.87

A positive E^o_cell indicates a spontaneous reaction.

Pt_(s)| NO_2(g)| NO^-_2(aq)|| NO_2(g)|NO^-_3(aq)|Pt_(s)

3. oxidation/ anode: 2Cl^-_(aq) → Cl_2(g) + 2e^- -1.358

reduction/ cathode: 2e^-+ 2H₂O_(l)→H_2(g) + 2OH^-_(aq) -.83

E^o_cell= cathode- anode

Edit: E^o_cell= E^o_{cathode -} E^o_anode

E^o_cell= -.83 - 1.38= -2.188

A negative E^o_cell indicates a nonspontaneous reaction. The spontaneous reaction would be the reverse. The spontaneous cell diagram would be written as

Pt_(s)| H₂O_(l) |H_2(g) || Cl^-_(aq) | Cl_(g) | Pt_(s)

4. oxidation/ anode: 6H₂O_(l) + C₃H_8(g) +5O_2(g) → 3CO_2(g) + 20H⁺_(aq)+ 20e^-

reduction/ cathode: 20e^-+ 20H⁺_(aq)+5O_2(g)→ 10H₂O_(l)

This combustion reaction for propane is not very commonly defined in terms of an oxidation reduction reaction, however it is obviously occurring here. I could not find standard reduction potentials for these reactions so I found the Gibbs energy of formations for each element and added them up to determine ΔG^o.

ΔG^o=∑ΔG of products−∑ΔG of reactants

[4(-228.61) + 3(-394.39)] - [ (23.4)+ (0)] = -2361.01

This reaction has a very negative ΔG^o so it is spontaneous.

Pt_(s)|C₃H_8(g) , CO_2(g)|| 5O_2(g), 4H₂O_(g)| Pt_(s)

Q20.5.28

Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CN^- as a complexing agent for Mn³⁺/Mn²⁺?

Mn³⁺_(aq) + e^- → Mn²⁺_(aq) E° = 1.51 V

Mn(CN)₆^3- _(aq) + e− → Mn(CN)₆^4-_(aq) E° = −0.24 V

Answer

The reaction of Mn³⁺ to Mn²⁺ is spontaneous, which we see from the positive E_cell value. CN^- was used to stabilize the species so that it would become non-spontaneous, which we see through the negative E_cell value. To find the net thermodynamic stablilization energy we need to find the energy difference between each reaction and then relate this to ΔG_o.

1.51 - (-0.24) = 1.75 V

ΔG^o=−nFE^o

ΔG^o= -1(96,485)(1.75)

ΔG^o= -168849 kJ