Extra Credit 27

Q17.3.6

Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br₂(l) is the same as for Br₂(aq).

S17.3.6

Pt(s)│H₂(g)│H⁺(aq)║Br₂(aq)│Br^-(aq)│Pt(s)

1. Split the cell diagram into two half reactions :

Br₂(l) + 2e^- ⇌ 2Br^-(aq) E^o_Cell = +1.09V

2H⁺(aq) + 2e^- ⇌H₂(g) E^o_Cell = 0.00V

2. Determine which species is the anode and which is the cathode :

H₂ is the anode because the anode corresponds to the oxidation reaction, and since hydrogen loses an electron and goes from a 0 to 1+ charge it is the anode.

Br₂ is the cathode because the cathode corresponds to the reduction reaction, and since bromine gains an electron and goes from a 0 to 1- charge it is the cathode.

3. Determine cell potential

E^o_Cell= E^o_Red,Cathode+E^o_Red,Anode

E^o_Cell= 1.09V - 0.00V

E^o_Cell= +1.09

Since E^o_Cell > 0, this means ΔG < 0 meaning this is a spontaneous reaction.

Q19.1.25

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M₃O₄ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M₂O₃, to permit estimation of the metal’s two oxidation states.)

1. Sc₂O₃
2. TiO₂
3. V₂O₅
4. CrO₃
5. MnO₂
6. Fe₃O₄
7. Co₃O₄
8. NiO
9. Cu₂O

S19.1.25

General explanation on how to solve these problems:

a) Sc₂O₃

(1) Determine the oxidation number of Oxygen

O₂ has a general and common oxidation number of -2.

(2) Set up the chemical formula like an equation, solving for Sc_2. Put any subscripts as coefficients in the equation. For example, O₃ would thus be 3(O) and Sc₂ would thus be 2 (Sc). Then set the equation equal to zero because the overall charge of the chemical formula is zero or neutral.

2 (Sc) + 3 (O) = 0

2 (Sc) + 3 (2-) = 0

2 (Sc) + (6-) = 0

2 (Sc) = 6+

(Sc) = 6/2

Sc = 3+

Therefore, the oxidation state of Sc₂ is 3+; Sc⁺³

b) TiO₂

(Ti) + 2 (O) = 0

(Ti) + 2 (2-) = 0

(Ti) + (4-) =0

Ti = 4+

Therefore, the oxidation state of Ti is 4+; Ti⁴⁺

c) V₂O₅

2 (V) + 5 (O) = 0

2 (V) + 5 (2-) = 0

2 (V) + (10-) = 0

2 (V) = 10+

(V) = 10/2

V = 5+

Therefore, the oxidation state of V₂ is 5+; V⁵⁺

d) CrO₃

(Cr) + 3 (O) = 0

(Cr) + 3 (2-) = 0

(Cr) + (6-) = 0

(Cr) = 6+

Therefore, the oxidation state of Cr is 6+; Cr⁶⁺

e) MnO₂

(Mn) + 2 (O) = 0

(Mn) + 2 (2-) = 0

(Mn) + (4-) = 0

Mn = 4+

Therefore, the oxidation state of Mn is 4+; Mn⁴⁺

f) Fe₃O₄ = FeO^.Fe₂O₃

FeO:

(Fe) + (O) = 0

(Fe) + (-2) = 0

(Fe) = +2

Fe₂O₃:

2 (Fe) + 3 (O) = 0

2 (Fe) + 3 (2-) = 0

2 (Fe) + (6-) = 0

2 (Fe) = 6+

(Fe) = 6/2

Fe = 3+

Therefore, the oxidation state of Fe₃ can be is 2+ OR 3+; Fe²⁺ OR Fe³⁺

g) Co₃O₄ = CoO^.Co₂O₃

CoO:

(Co) + (O) = 0

(Co) + (2-) = 0

Co = +2

Co₂O₃:

2 (Co) + 3 (O) = 0

2 (Co) + 3 (2-) = 0

2 (Co) + (6-) = 0

2 (Co) = 6+

(Co) = 6/2

Fe = 3+

Therefore, the oxidation state of Co₃ can be is 2+ OR 3+; Co²⁺ OR Co³⁺

h) NiO

(Ni) + (O) = 0

(Ni) + (2-) = 0

Ni = 2+

Therefore, the oxidation state of Ni is 2+; Mn²⁺

i) Cu₂O

2 (Cu) + (O) = 0

2 (Cu) + (2-) = 0

2 (Cu) + (2-) = 0

2 (Cu) = 2+

(Cu) = 2/2

Cu = 1+

Therefore, the oxidation state of Cu₂ is 1+; V⁺

Q12.4.18

Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.

S12.4.18

Step 1: We use the integrated first order rate law equation to walk us through the problem:

ln(N/N₀)=−kt

Step 2: Find k using first order half life equation

t_1/2= (ln2)/k

k= (ln2)/5730

k= 0.000121

Step 3: Plug all known values in to the first order rate law equation to find out how many years ago he died.

ln(N/N₀)= -kt

N= 93.79 and N₀=100 since 93.79% is left

ln(93.79/100) = -(0.000121)t

-0.0641 = -(0.000121)t

t= 529.9 years ago

Q21.3.2

Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?

S21.3.2

Nuclei are the central core of atoms that contain BOTH protons and neutrons. With this definition in mind, alpha particles are the only production of nuclear reactions that are actually nuclei. Alpha particles are actual helium nuclei with 2 protons and 2 neutrons, creating a mass number of 4.

Q21.7.4

A scientist is studying a 2.234 g sample of thorium-229 (t_1/2 = 7340 y) in a laboratory.

1. What is its activity in Bq?
2. What is its activity in Ci?

S21.7.4

1.

Step 1: Equation to use here is (Activity) = λ N , where N is the amount of nuclides and λ is the decay constant

Step 2: Find λ in units of seconds

λ = 0.693 / 7340 years = 9.44 x 10 ^-5 years

λ = 9.44 x 10 ^-5 years x (1 year/ 365 days) = 2.586 x 10^-7 days

λ = 2.586 x 10^-7 days x (1 day / 24 hours) = 1.077 x 10^-8 hours

λ = 1.077 x 10^-8 hours x (1 hours / 3600 seconds) = 2.994 x 10^-12 seconds

Step 3: Find N in units of atoms

N = 2.234g Thorium x (1 mole / 229 g Thorium) = 9.726 x 10^-3 moles Thorium

N = 9.726 x 10^-3 moles Thorium x ( 6.022 x 10²³) = 5.87 x 10²¹ moles per atom

Step 4: To determine activity just substitute values into equation: (Activity) = λ N

Activity = 2.994 x 10^-12 seconds x 5.87 x 10²¹ moles per atom

Activity = 1.759 x 10¹⁰ seconds x mole per atom

Step 5: Convert to Bq

Activity = 1.759 x 10¹⁰ seconds x mole per atom = 1.759 x 10¹⁰ becquerel

2.

Step 1: From 1. we were able to find the Activity in seconds x moles per atom

Activity = 1.759 x 10¹⁰ seconds x mole per atom

Step 2: Convert the Activity to Curie

Activity = 1.759 x 10¹⁰ Bq ( 1 curie/ 3.7 x 10¹⁰Bq)= 0.4754 curie

Q20.4.17

The standard cell potential for the oxidation of Pb to Pb²⁺ with the concomitant reduction of Cu⁺ to Cu is 0.39 V. You know that E° for the Pb²⁺/Pb couple is −0.13 V. What is E° for the Cu⁺/Cu couple?

S20.4.17

1. Standard Cell Potential is measured by:

E^o_{Cell =} E^o_Cathode - E^o_Anode

2. Cathode is where reduction occurs. Thus, Cu⁺/Cu is located at the cathode since copper goes from a 1+ charge to a 0 charge through the gain of electrons.

Anode is where the oxidation occurs. Thus, Pb/ Pb²⁺ is located at the anode since lead goes from 0 charge to a 2+ charge through the loss of electrons.

3. You are given the E^o_Cell and E^o_Anode . After substituting all know values, we solve for E^o_Cathode:

0.39V = E^o_Cathode - (-0.13V)

E^o_Cathode = +0.26V

The E° for the Cu⁺/Cu couple is +0.26V

Q20.9.2

How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?

S20.9.2

If we know the stoichiometry of an electrolysis or electrochemical reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. In other words, we can calculate the quantity of material that is oxidized or reduced at an electrode. We can determine strength of oxidants and reductants by calculating the number of moles of electrons transferred when a known current is passed through a cell. Additionally, we can use the stoichiometry of the reaction and the total charge transferred to calculate the amount of product formed or the amount of metal deposited in an electroplating process.

Q14.1.2

If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?

S14.1.2

Studying chemical kinetics determines whether to proceed with a reaction because it measures the rate of a reaction and its corresponding mechanism. Reactions conducted in an industrial facility mix compounds together, heating and stirring them for a while, before moving to the next process. When the compounds are then moved to the next phase of the process it is important to know how long to hold the reaction at one stage before continuing, to make sure a reaction has finished before starting the next one.