Extra Credit 30

Q17.4.3

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

1. Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)
2. The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br₂(l) is the same as that of Br₂(aq).

Q17.4.3

(a) Ag is getting reduced so E*cathode= +0.8(standard potential). Hg is getting oxidized so E^oanode= +0.851(standard potential).

Therefore E^ocell=E^ocathode-E^oanode=(0.8) -(0.851)= -0.051V Non-spontaneous

Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)

Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s).

n=2 as two electrons are involved. F=96485 (contant). T=298^oK. R=8.3145.

Ecell= E⁰cell - RT/(nF) lnQ. = -0.051-RT/nF ln(1/[0.1][0.25]^2) = -0.051-0.01284(5.025) = -0.051-0.0645 = -0.116V

non-spontaneous as the electrode potential is negative.

(b) Aluminium is getting oxidized E⁰anode= -1.66. Nitrate is getting reduced so E⁰cathode= -0.23.

The E^ocell=E^ocathode-E^oanode=(-0.23)-(-1.66)=1.43v.

Therefore the reaction is spontaneous. 3Ni²⁺+ Al(s) → 2Al³⁺ + Ni(s) n=6 as six electrons are involved. F=96485 (contant). T=298^oK. R=8.3145

Ecell= E⁰cell - RT/(nF) lnQ. = 1.43- 8.315(298^oK)/(6)96485 ln[Al³⁺]²/[Ni²⁺]³ = 1.43 -(0.003921)(-4.2405) =1.4466V

spontaneous as the electrode potential is positive.

(c) 2Br^- (aq)(1M)⟶ Br₂ (0.11M) + 2e^- is getting oxidized. E⁰anode= 1.09

Al³⁺(0.023M)+3e^- ⟶ Al is getting reduced. E⁰cathode= -1.66

E^ocell=E^ocathode-E^oanode= -1.66-(1.09) = -2.75V.

Ecell= E⁰cell - RT/(nF) lnQ. = -(2.75) - 8.315(298^oK)/(6)96485 ln[Br₂]³/[Al²⁺]²[Br^-]⁶

= -2.75-(0.0036169)

= -2.6536V

nonspontaneous as the electrode potential is negative.

Q12.1.3

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Cl2(g)+3F2(g)⟶2ClF3(g)Cl2(g)+3F2(g)⟶2ClF3(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl₂ and F₂ and the formation of ClF₃.(12.1.3)

Q12.1.3

Cl2(g)+3F2(g)⟶2ClF3(g)Cl2(g)+3F2(g)⟶2ClF3(g). is the reaction.

So, we need to know the individual rate of reaction for each element.

The stoichiometry should be taken into concern as it will tell us how the rate are related to one another.

Rate of reaction for Cl₂(g) is - because its on the reactant side. so - (ΔCl₂/ΔT)

Rate of reaction for 3F₂(g) is - because its on the reactant side. so - 1/3 (ΔF₂/ΔT)

The two reactants will have negative sign because as the reaction occurs, they will be decreasing in abundance.

Rate of reaction for because its on the products side. so + 1/2 (ΔClF₃/ΔT)

Thus we can equate because all of them represent the same equations rate of reaction.

Rate= - (ΔCl₂/ΔT) = - 1/3 (ΔF₂/ΔT) = + 1/2 (ΔClF₃/ΔT)

Q12.5.1

Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

Q12.5.1

According to the collision theory the collisions should have high amounts of molecules in required orientation for it to have a high rate of reaction. However the kinetic energy with which the reactants collide contributes a lot. If the collision does not produce a chemical reaction, the reactants could be moving too slowly and did not have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring. k=Ae-Ea/RT explains a lot about the factors mentioned above and how it contributes during the chemical reaction.

Q21.3.5

Write a balanced equation for each of the following nuclear reactions:

1. the production of ¹⁷O from ¹⁴N by α particle bombardment
2. the production of ¹⁴C from ¹⁴N by neutron bombardment
3. the production of ²³³Th from ²³²Th by neutron bombardment
4. the production of ²³⁹U from ²³⁸U by H12H12 bombardment

Q21.3.5

1. ¹⁴N which has 7 protons and 7 neutrons. It can be produced by the help of the alpha particles. The alpha particles contain 2 protons and 2 neutrons. So, according to the conservation of mass the masses on the reactant side and the product side should be equal. The production of oxygen should produce a H₁¹ to balance it out. However with the production of H₁¹ reproduces the ¹⁴N molecule.

¹⁴N₇+⁴He₂⟶¹⁷O₈+H₁¹

2.¹⁴N₇+n₀¹⟶¹⁴N₆+H₁¹

When the N molecule attains a neutron its protons are reduced in order to follow the conservation of mass.

3. ²³²Th₉₀+n₀¹⟶²²³Th₉₀

When the Thallium when attains a neutron loses neutrons in order follow the conservation of mass.

4. ²³⁸U₉₂+H₁²⟶U²³⁹₉₂+H₁¹ when the uranium attains the H₁² molecule we must add H₁¹ to the right side in order to follow the conservation of mass while getting U²³⁹₉₂ as our product.

Q20.2.1

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

Q20.2.1

Good oxidents or oxidizing reagents are elements which can get reduced easily. So, to get reduced the elements should be able to acquire electrons easily which could be easily done with the non-mettallic elements on the right side of the periodic table. Specifically Elements from III-VIII need 1 or 2 electrons in their orbital shell to have a stable electron configuration. Similarly good reductants oxidize easily and those elements which want to get rid of its few electrons are on the left side of the periodic table Specifically group I-II.

Q20.4.20

Will each reaction occur spontaneously under standard conditions?

1. Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
2. Zn²⁺(aq) + Pb(s) → Zn(s) + Pb²⁺(aq)

Q20.4.20

1) Cu(s)→Cu²⁺(aq)+2e^- Oxidization so anode. E^oanode= 0.34. 2e^-+2H⁺(aq)→H₂(g). E^ocathode=0

E^ocell=E^ocathode-E^oanode=0-(0.34)= -0.34V

Its a negative answer therefore non- spontaneous reaction.

2) 2e^-+Zn²⁺(aq)→Zn(s) Zn is getting reduced. E^ocathode= -0.76

Cu(s) → Cu²⁺(aq) + 2e^- Cu is getting oxidized. E^oanode= +0.34

E^ocell=E^ocathode-E^oanode= -0.76-0.34 = -1.1V

Therefore because the voltage is a negative number it is not a spontaneous reaction.

Q20.9.5

The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?

1. AlCl₃
2. MgCl₂
3. FeCl₃

Q20.9.5

1. Al³⁺+3e^-→Al. The molecular mass of aluminum is 26.98u.

Therefore because (c/s).(s).(1e^-/1.602*10^-19C).(1mole e^-/6.022*10²³e^-).(moles metal/moles e^-).(g metal/moles of metal) = g metal

We require,

(moles metal/moles e^-).(g metal/moles of metal) for getting the grams of metal which are deposited from these salts for each mole of electrons.

=25.98u/3

= 8.66u.

2. Mg²⁺+2e^- →Mg The molecular mass is 24.3u.

Therefore because (c/s).(s).(1e^-/1.602*10^-19C).(1mole e^-/6.022*10²³e^-).(moles metal/moles e^-).(g metal/moles of metal) = g metal

We require,

(moles metal/moles e^-).(g metal/moles of metal) for getting the grams of metal which are deposited from these salts for each mole of electrons.

=24.3u/2

= 12.15u.

3. Fe³⁺+3e^-→Fe molecular mass is 55.84u.

Therefore because (c/s).(s).(1e^-/1.602*10^-19C).(1mole e^-/6.022*10²³e^-).(moles metal/moles e^-).(g metal/moles of metal) = g metal

We require,

(moles metal/moles e^-).(g metal/moles of metal) for getting the grams of metal which are deposited from these salts for each mole of electrons.

= 55.84u/3

= 18.613u

Q14.5.1

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

Q14.5.1

Arrhenius showed that the relationship between temperature and the rate constant for a reaction obeyed the following equation.

k=Ae-Ea/RT

More kinetic energy the more collision in the reaction. The rate of reaction is highly dependent on the temperature as temperature proportionally increases the kinetic energy within the molecules with the increase in the temperature. The temperature as it keeps increasing helps the molecule attain the kinetic energy and collide frequently with other molecules in the reaction even before the activation energy is attained. The collisions at one point of time will be able to attain enough activation energy after the increase in temperature.