Extra Credit 10

Q17.1.9

a.) Why is it not possible for hydroxide ion (OH−) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?

a.) An acidic solution has a surplus of H+ ions, meaning that it has a higher concentration of hydrogen ions than water, [H+]> \[1x10^{-7}\], so the [OH-] is basically 0.If OH- were made it would react with the huge amount of H+ to produce water. Because OH- would be the limiting reactant, all of it would be used up.

b.) Why is it not possible for hydrogen ion (H+) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?

b.) A basic solution uses OH- ions instead of H+ ions, meaning that it has a lower concentration of hydrogen ions than water. The [OH]> \[1x10^{-7}\] while the [H+]<M. The H+ concentration is basically zero. If H+ ions were made it would react with the surplus of OH- to make water, so H+ should not appear at all.

Q19.1.8

The following reactions all occur in a blast furnace. Which of these are redox reactions?

\[a.) 3Fe_{2}O_{3(s)}+ CO(g)\rightarrow 2Fe_{3}O_{4}(s)+CO_{2}(g)\]

\[b.)Fe_{3}O_{4}(s)+CO9(g)\rightarrow 3FeO(s) + CO_{2}(g)\]

\[c.)FeO(s)+CO(g)\rightarrow Fe(l)+CO_{2}(g)\]

\[d.)C(s)+O_{2}(g)\rightarrow CO_{2}(g)\]

\[e.)C(s)+CO_{2}(g)\rightarrow 2CO(g)\]

\[f.)CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)\]

\[g.)CaO(s)+SiO_{2}(s)\rightarrow CaSiO_{3}(l)\]

A redox reaction means that both oxidation and reduction are occuring.

\[A.) 3Fe_{2}O_{3(s)}+ CO(g)\rightarrow 2Fe_{3}O_{4}(s)+CO_{2}(g)\]

This reaction is a redox reaction because:

Fe goes from 3+ to 2+ oxidation state, reduction

Carbon changes from 2+ to 4+ oxidation state, oxidation

B) \[Fe_{3}O_{4}(s)+CO(g)\rightarrow 3FeO(s)+CO_{2}(g)\]

This reaction is a redox reaction because:

Fe changes from partial charge of 8/3+ to 2+ (reduction)

Carbon goes from 2+ to 4+ (oxidation)

C) \[FeO(s)+CO(g)\rightarrow Fe(l)+CO_{2}(g)\]

This reaction is a redox reaction because:

Fe changes from +2 to 0 (reduction)

Carbon changes from 2+ to 4+ (oxidation)

D) \[C(s)+O_{2}(g)\rightarrow CO_{2}(g)\]

This reaction is a redox reaction because:

Oxygen changes from 0 in the to -2 (reduction)

Carbon changes from 0 to 4+ (oxidation)

E)\[C(s)+CO_{2}(g)\rightarrow 2CO(g)\]

This reaction is a redox reaction because:

Carbon solid changes from 0 to +2 (oxidation)

Carbon dioxide gas changes from +4 to +2 (reduction)

F. \([CaCO]_{3} + [CO](g)\rightarrow 2[Fe_{3}O](s)+[CO]_{2}\)

In the reactants calcium is equal to +2 while carbon is equal to +4 where oxygen is equal to -2. In the products side calcium is equal to +2 while carbon is equal to +4 and oxygen is equal to -2. Thus the oxidation numbers for the reactants and products are equal and thus this is not an oxidation reduction reaction.

G. \([CaO](s) + [SiO]_{2}(s)\rightarrow[CaSiO]_{3}(l)\)

In the reactants calcium is equal to +2 while silicon is equal to +4 where oxygen is equal to -2. In the products side calcium is equal to +2 while silicon is equal to +4 where oxygen is equal -2. Thus the oxidation numbers for the reactants and products are equal and thus this is not an oxidation reduction reaction.

Q19.2.10

Draw the geometric, linkage, and ionization isomers for [CoCl₅CN][CN].

A geometric isomer is where two or more compounds are different from each other in the group arrangement. A linkage isomer is where the coordination compounds that have the same composition have a different connecting strand of metal to the ligand. An ionization isomer is where it is nearly identical except a ligand changes place with a neutral or anion molecule.

Q12.3.23

NO is second order

Cl2 is first order

Overall rate reaction is : rate= k[No}^{2}[Cl_2]

b) Calculate the specific rate constant for this reaction.

\[[NO]^{1} [Cl_{2}]^{1}k=rate\]

\[[.50]^{1}[.50]^{1}k=(5.1x10^{-3})\]

\[.25k=(5.1x10^{-3})\]

\[k=.0204M^{-1}s^{-1}\]

Wrong formula correct answer and formula is

\[k=\tfrac{5.1x10^{-3}}{0.5atm^{2}(0.5atm)}\]

k= \(0.0408atm^{-2}s^{-1}\)

Q12.7.3

a. A catalyst speeds up a chemical reaction without being used up in the process, or being permanently changed in anyway. Which means it should appear in the reactant and the products, meaning that it has been recovered from previously being used to speed up the reaction since it is not changed nor does it permanently disappear. In the gas phase transformation reaction 2^nd step, Cl is a reactant, and in the 3^rd step Cl is regenerated in the products. It was not used up thus it is a catalyst.

b. NO is also a catalyst because it is a reactant and acts as an intermediate in the 2^nd step and then reappears in the 3^rd step as a product.

Q21.4.26

a.\[_{13}^{26}\textrm{Al}\rightarrow _{12}^{26}\textrm{Mg}+_{+1}^{0}\textrm{B}+_{0}^{0}\textrm{v}\]

\[_{26}^{13}\textrm{Al}+_{-1}^{0}\textrm{e}\rightarrow _{12}^{26}\textrm{Mg}+_{0}^{0}\textrm{v}\]

b.

Here is the half life equation:

\[t_{1/2} = ln(2)/k\]

Plug in the half life to get k (the rate constant):

\[(7.2X10^{5})=ln(2)/k\]

solve for k:

\[k=9.627x10^{-7} years\]

Plug k, initial amount, and amount left to get time:

\[ln\tfrac{[A]_{t}}{[A]_{0}} = -kt\]

\[ln\tfrac{.000000001}{100} = -(9.627x10^{-7})t\]

solve for time:

\[t=2.8701x10^{7}\]

Subtract the time from how old the Earth is to get how old the Earth was when 99.999999% of Al-26 decayed

\[(4.7x10^{9})-(2.8701 x 10^{7})= 4.671299x10^{9} years\]

Good job ! I agree.

Q20.3.13

For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.

a.

(reduction cathode) \[2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(aq)\]

(oxidation, anode) \[Zn(s)\rightarrow Zn^{2+}(aq) + 2e^{-}\]

result \[Zn(s)+2H^{+}(aq)\rightarrow Zn^{2+}(aq)+H_{2}(aq)\]

b.

(reduction,cathode) \[AgCl(s) + e^{-}\rightarrow Ag(s)+Cl^{-}\]

\[2AgCl(s) + 2e^{-}\rightarrow 2Ag(s)+2Cl^{-}\]

(oxidation,anode) \[H_{2}(g)\rightarrow 2H^{+}(aq)+2e^{-}\]

result \[2AgCl(s) + H_{2}(g)\rightarrow 2H^{+}(aq)+2Ag(s)+2Cl^{-}(aq)\]

c.

(reduction,cathode) \[Fe^{3+}(aq) +e^{-}\rightarrow Fe^{2+}(aq)\]

\[2Fe^{3+}(aq) + 2e^{-}\rightarrow 2Fe^{2+}(aq)\]

(oxidation,anode) \[H_{2}(g)\rightarrow 2H^{+}(aq)+2e^{-}\]

result \[2Fe^{3+}(aq)+H_{2}(g)\rightarrow 2H^{+}(aq)+2Fe^{2+}(aq)\]

Good job! I agree.

Q20.5.25

Yes, FAD will be an effective oxidant for the conversion of acetaldehyde to acetate at PH 4.00, since this reaction has a negative value and the reaction is spontaneous. An oxidant is an oxidizing agent which means it gets reduced according to OIL RIG since oxidizing is losing.

\[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode}\]

\[(-0.18v)-(-0.58v)\]

\[E^{\circ} = 0.40V\]