Extra Credit 30

Last updated
Save as PDF

Page ID: 82738

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

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.

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

S17.4.3

Solution:

1. First begin splitting the overall reaction into two half reactions.The half reactions should result in one oxidized and one reduced reaction. The oxidized half reaction Hg(l) + S^2----> HgS + 2e^- has a Standard Cell potential of -.70. The reduced half-reaction is determined to be Ag⁺+ e^---> Ag(s) with a Standard cell Potential of .799. This half-reaction appears in the same order on the table of Standard Reduction Cell Potentials so the value does not need to be switched. Using these two values, plug them into the equation E^o_cell= E _cathode- E _anode to calculate Standard Cell Potential. E^o _cell= .799-(-.70)= 1.4996 V. Since this value is positive, it is spontaneous. In order to calculate Cell Potential under stated conditions you must first balance each half-reaction and combine them. This results in the overall equation that was already given in the problem. Plug in all the values into the equation E_cell= E^o_cell - (.0592/n) log Q. (where n is the total # of electrons transferred and Q is the ratio of the concentration of products raised to the power of their coeff. divided by the concentration of reactants raised to the power of their coeff.) The total number of electrons transferred is 2 because in order to cancel out the electrons from both sides of the equation in the overall reaction, they both must have a coeff. of 2. Q= 1/ [.1][.25]², where the numerator is the concentration of the products, and the denominator is the product of the concentrations of the reactants raised to their respective coeff. Plug in all values into the equation, E_cell= 1.4996 - (.0592/2) log 160. This results in E_cell = 1.43 V , and because the value is positive, it is spontaneous.

2. First begin splitting the overall reaction into two half reactions.The half reactions should result in one oxidized and one reduced reaction. The half reactions include Al³⁺ + 3e^----> Al(s) with a Standard Cell potential of -1.66 and Ni²⁺ + 2e^- --> Ni(s) with a Standard Cell potential of -.25. Using these two values, plug them into the equation E^o _cell = E _cathode - E _anode to calculate Standard Cell Potential. E^o_cell= -.25 - (-1.66) = 1.41 V. Since this value is positive, it is spontaneous. In order to calculate the Cell Potential under stated conditions, you must first balance each half reaction and combine them. This results in the overall equation 2 Al + 3 Ni(NO3)₂---> 2Al(NO3)₃+ 3 Ni. Plug in all the values into the equation E_cell= E^o_cell - (.0592/n) log Q. (where n is the total # of electrons transferred and Q is the ratio of the concentration of products raised to the power of their coeff. divided by the concentration of reactants raised to the power of their coeff.) The total number of electrons transferred is 6 because in order to cancel out the electrons from both sides of the equation in the overall reaction, they both must have a coeff. of 6. Q= [.015]²/[.25]³, where the numerator is the concentration of the products, and the denominator is the product of the concentrations of the reactants raised to their respective coeff. Plug in all values into the equation, E_cell= 1.41 - (.0592/6) log .0144. This results in E_cell = 1.43 V , and because the value is positive, it is spontaneous.

3. First begin splitting the overall reaction into two half reactions.The half reactions should result in one oxidized and one reduced reaction. The half reactions include 2Br^-(aq) ---> Br₂(aq)+ 2e^- with a Standard Cell potential of 1.087 V and Al³⁺+ 3e^---> Al(s) with a Standard Cell potential of -1.676. Using these two values, plug them into the equation E^o _cell = E _cathode - E _anode to calculate Standard Cell Potential. E^o_cell= -1.676-1.087=-2.763 V. Since this value is negative, it is nonspontaneous. In order to calculate the Cell Potential under stated conditions, you must first balance each half reaction and combine them. This results in the overall equation 6Br^-(aq) +2Al³⁺--->2Al(s) + 3Br₂ (l). Plug in all the values into the equation E_cell= E^o_cell - (.0592/n) log Q. (where n is the total # of electrons transferred and Q is the ratio of the concentration of products raised to the power of their coeff. divided by the concentration of reactants raised to the power of their coeff.) The total number of electrons transferred is 6 because in order to cancel out the electrons from both sides of the equation in the overall reaction, they both must have a coeff. of 6. Q= [.023]²/[.11]3, where the numerator is the concentration of the products, and the denominator is the product of the concentrations of the reactants raised to their respective coeff. Plug in all values into the equation, E_cell= -2.763 - (.0592/6) log .397 . This results in Ecell = -2.76 V , and because the value is negative, it is nonspontaneous.

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)Cl2g3F2g2ClF3g. 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.

S12.1.3

Solution:

Rate expressions are written in terms of delta which means "change" in something. In this case, a rate is determined as the change in concentration of a substance, with respect to the delta t (or change in time). When writing rate equations, you must include the coefficient from the balanced equation, and use it as a coefficient in the rate law equation. However, when it is used in the rate law equation it will be written as 1/coeff. and if the it is a product then it will be positive because product is being formed, but if it is a reactant it will be negative because as the reaction proceeds, reactant will disappear. The rate of disappearance for Cl₂ would be $-(\Delta\frac{[Cl_2]}{\Delta t})$ .Since it is a reactant it will disappear making it a negative value, and it has a coeff. of one. The rate of disappearance of F₂ would be $-\frac{1}{3}(\Delta\frac{[F_2]}{\Delta t})$ . Since F₂ has a coeff. of 3 in the overall equation, it will transfer as 1/3 in the rate equation, and since it is a reactant is will be disappearing as the reaction proceeds making it a negative value. ClF₃ has a rate of $\frac{1}{2}(\Delta\frac{[ClF_3]}{\Delta t})$ . Due to the fact that it has a coeff. of 2 in the balanced equation, it is written as a coeff. of 1/2 in the rate expression, and the value would be positive because it is a product meaning that it will be formed as the reaction proceeds. All of these rate reactions are equal to one another due to the fact the rate for the loss of a reactant has to equal the rate of formation for a product to follow the law that no matter can be created nor destroyed.

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

Solution:

Two factors that may prevent a collision from producing a chemical reaction include low temperature and orientation. Low temperature results in particle motion to be slow and it prevents particles from colliding with one another with enough kinetic energy to exceed activation energy and produce a reaction. Reactants have to collide at high energy to break bonds so that new bonds can be formed, to result in a reaction. Orientation also affects whether or not a reaction can occur. If molecules collide at a reactive site, then a reaction will occur. However, it is very rare for larger, more complex molecules to combine at a reactive site with another complex molecule. In some cases, the molecule is so complex that the reaction site is completely blocked off and no reaction occurs.

Q21.3.5

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

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

S21.3.5

Solution:

1. First write out the equation with the missing terms, and determine which particle when combined with $\ce{^{14}_{7}N}$ would produce a mass # closest to 17 and an atomic # closest to 8. After, add the $\ce{^{1}_{1}H}$ particle on the right side of the equation to balance it. The overall equation is $\ce{^{14}_{7}N}$+$\ce{^{4}_{2}He}$-->$\ce{^{17}_{8}O}$+ $\ce{^{1}_{1}H}$

2.First write out the equation with the missing terms, then determine the particle that when combined with$\ce{^{14}_{7}N}$would produce a mass closest to 14 and an atomic number closest to 6. After, add a $\ce{^{1}_{1}H}$ particle on the right side of the equation to balance both sides of the equation. The overall equation is $\ce{^{14}_{7}N}$+ $\ce{^{1}_{0}n}$--> $\ce{^{14}_{6}C}$+ $\ce{^{1}_{1}H}$

3. First write out the equation with the missing term. Determine what mass particle will increase the mass number by 1 and will not change the atomic number. Only the $\ce{^{1}_{0}n}$ particle will result in this. The equation is $\ce{^{232}_{90}Th}$+ $\ce{^{1}_{0}n}$-->$\ce{^{233}_{90}Th}$

4. First write out the equation with missing terms. Determine what particle will increase the mass # to be closest to 239 and what # would keep the atomic number closest to 92. By adding $\ce{^{2}_{1}H}$ to the reactant side, you are able to increase the mass number and the atomic number. In order to balance the reactants and the products the particle $\ce{^{1}_{1}H}$ must be added to make the overall mass number equal 240 and the overall atomic number equal to 93. The equation is $\ce{^{238}_{92}U}$+$\ce{^{2}_{1}H}$--> $\ce{^{239}_{92}U}$+ $\ce{^{1}_{1}H}$

Q20.2.1

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

S20.2.1

Solution:

Elements they are good oxidizing agents are nonmetals which can be found on the right side of the periodic table. This is because they tend to be highly electronegative and therefore pull electron pairs towards them thus oxidizing the other elements and reducing themselves. Elements that are good reducing agents are metals, found on the left side of the periodic table. This is because they tend to want to lose electrons to another element, thus, reducing the other element and oxidizing themselves.

Q20.4.20

Will each reaction occur spontaneously under standard conditions?

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

S20.4.20

Solution:

1. First, write out the two half reactions associated with the overall equation. The half reactions include Cu(s)-->Cu²⁺(aq)+2e^- and 2e^-+ 2H⁺--> H²(g). Next, use a Standard Reduction Table to find the Cell potentials associated with each half reaction. The values for these reactions include .34 and 0. After finding these values, put them into the equation E^o_cell= E_cathode-E_anode . E^o_cell= 0-(.34)= -.34. Since the value is negative, this reaction will not occur spontaneously.

2. First, write out the two half reactions associated with the overall equation. The half reactions include Zn²⁺(aq) +2e^--->Zn(s) and Pb(s)--> Pb²⁺+ 2e^-. Next, use a Standard reduction table to find the Cell potentials associated with each half reaction. The values for these reactions include -.76 and -.13. After finding these values, put them into the equation E^o_cell= E_cathode- E_anode . E^o_cell = (-.76)-(-.13)= -.63. Since this value is negative, this reaction is non-spontaneous.

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?

AlCl3
MgCl2
FeCl3

Q20.9.5

Solution:

1. First, begin by writing out the balanced half reactions and calculate how many moles of electrons were transferred. The balanced half reactions are 3Cl^--->Cl₃+3e^- and Al³⁺+ 3e^--->Al. Overall, 3 moles of electrons were transferred. Since no information was given about (amp) or time, use the number of moles of electrons in a stoichiometric conversion to determine how many grams of the metal are deposited. Use the Al half reaction, since it is where a metal is being formed. Using a stoichiometric conversion, it is concluded that for every 3 moles of electrons, there is one mole of Al, and for every one mole of Al, there are 26.982g. This yields to 8.994 g Al/ mole e^-.

2. First, being by writing out the balanced half reactions and calculate how many moles of electrons were transferred. The balanced half reactions are Mg²⁺+2e^--->Mg(s) and 2Cl^--->Cl₂+2e^-. Overall, 2 moles of electrons were transferred. Since no information was given about (amp) or time, use the number of moles of electrons in a stoichiometric conversion to determine how many grams of the metal are deposited. Use the Mg half reaction, since it is where a metal is being formed. Using a stoichiometric converion, it is concluded that for every 2 moles of electrons, there is 1 mole of Mg, and for every one mole of Mg, there are 24.305g. This yields to 12.153 g Mg/mole e^-.

3. First begin by writing out the balanced half reactions and calculate how many moles of electrons were transferred. The balanced have reactions are Fe³⁺+3e^--->Fe(s) and 3Cl^--->Cl₃+ 3e^-. Overall, 3 moles of electrons were transferred. Since no information was given about (amp) or time, use the number of moles of electrons in a stoichiometric conversion to determine how many grams of the metal are deposited. Use the Fe half reaction, since it is where the metal is being formed. Using a stoichiometric conversion, it is concluded that for every 3 moles of electrons, there is 1 mole of Fe, and for every 1 mole of Fe, there are 55.847g. This yields to 18.62 g Fe/mole e^-.

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?

S14.5.1

Solution:

Activation energy relates to the chemical kinetics of a reaction in showing that the the higher the kinetic energy of a system is, the proportion of collisions that can overcome the activation energy increases as the temperature increases. Raising the temperature increases the reactions rate and the amount of collisions that occur. The activation energy is the minimal amount of energy that a system must reach in order for the reaction to occur. Thus, raising the temperature of a system will only make it more likely for particles to collide with enough energy to overcome this barrier. At given temperatures, however, not all molecules are moving with the same kinetic energy. Some are moving slower, and some are moving faster, thus making the average kinetic energy somewhat of a mis- representation of the overall rate of the reaction.

Search

Text Color

Text Size

Margin Size

Font Type