Extra Credit 41

S17.6.1 Corrosion is the process in which the metals act as anodes and get oxidized. (Being oxidized is giving up electrons)

A. Looking at the standard reduction table, the element that is more negative (lower on the table) has a lesser chance of being reduced and a high chance of being oxidized or corroded. So Ca is below Mg so it is more likely to be corroded.

B. Au is higher (more positive) on reduction potential chart so it will most likely be reduced. Since Hg has a lower standard reduction potential value than Au, it will more likely be oxidized or corroded.

C. Fe is higher (more positive) on reduction potential chart so it will most likely be reduced. Zn is below it and more negative so it has a higher chance of being corroded (oxidized).

D. Pt is higher (more positive) on reduction potential chart so it will more then likely be reduced. Ag is below it and less positive so it has a higher chance of being corroded (oxidized).

S 12.3.4

A. \[0.5\rightarrow 0.25\] Rate=\[K{[NO_2]}^2\]

R1= \[0.5\cdot 0.5=0.25K\]

R2=\[0.25\cdot 0.25= 0.0625K\]

Compare R2 with R1 and do \[\frac{R2}{R1}=1/4\] so it reduces the rate by a factor of 4.

B. Rate is \[K{[NO_2]}^2\] , CO is not included in the overall rate law for the reaction so the rate cannot be influenced by changes to CO.

For instance, we can see that Step 1 is the rate-determining step because it's rate law matches the rate law for the overall reaction:

Step 1: \[NO_{2}+NO_{2}\overset{slow}{\rightarrow}NO_{3}+NO\]

where rate = k[NO₂]²
Step2: \[NO_{3}+CO \rightarrow NO_{2}+CO_{2}\]

Overall reaction: \[NO_{2}+CO \rightarrow NO+CO_{2}\]

S 12.5.13

\[ln(\frac{k2}{k1})=\frac{Ea}{R}\cdot (\frac{1}{T1}-\frac{1}{T2})\]

=

\[ln(\frac{2.42x10^-6}{6.23x10^-7})=\frac{Ea}{8.31}\cdot (\frac{1}{555}-\frac{1}{575})\]

=

\[1.356898=(\frac{Ea}{8.31})\cdot (\frac{1}{555}-\frac{1}{575})\]

=

\[11.2765=Ea(0.00006267)\]
=

Ea=179934 J= \[180\frac{kJ}{mol}\]

S21 4.8

even numbers of protons and neutrons prove to have more stability then odd numbers.

A. \[_{15}^{34}\textrm{P}\]
This would be expected to undergo beta decay (\[_{-1}^{0}\textrm{} \beta\])

because by looking at the band of stability and realizing the correlation between number of protons and neutrons. We also see that decay would give an even number of protons and neutrons and that would then mean more stable then odd number of both. With the number of neutrons being slightly higher than the number for protons for Phosphorus-34, we can see that the neutron-to-proton ratio is somewhat high for an element of that mass.

B. \[_{92}^{239}\textrm{U}\]

This would expect to use alpha decay \[_{2}^{4}\textrm{\alpha }\]

to reach stability because the band of stability reflects this and it would then give us an even number of protons and neutrons for Uranium. We also know that nuclei with atomic numbers > 83 are too heavy to be stable and thus, Uranium-239 will undergo alpha emission.

C. \[_{20}^{38}\textrm{Ca}\]

This element would undergo beta decay as well (\[_{-1}^{0}\textrm{} \beta\])

because the of the correlation on the band of stability. This would give us an even number of protons and neutrons and get a close to the magic numbers desired.

I disagree with this answer because we know that \[_{20}^{40}\textrm{Ca}\] has a neutron-to-proton ratio that equals to 1 and is therefore stable. However, in this case, \[_{20}^{38}\textrm{Ca}\] has a neutron-to-proton ratio that is less than 1 (where .9 < 1 and there is 18 neutrons & 20 protons) and thus, Ca-38 will either decay by positron emission or electron capture in order to convert a proton to a neutron.

D. \[_{1}^{3}\textrm{H}\]

This would be expected to undergo neutron emission \[_{0}^{1}\textrm{n}\]

because of the correlation with the band of stability. On this graph you can see where the number of protons and neutrons fall and tell what type of reaction is needed.

I also disagree with this answer because since Hydrogen-3 has 2 neutrons and 1 proton, its neutron-to-proton ratio is quite high than what we see for the neutron-to-proton ratio of Hydrogen-2 (which has a 1:1 neutron-to-proton ratio). Therefore, Hydrogen-3 is expected to undergo beta emission in order to convert a neutron to a proton.

E. \[_{94}^{245}\textrm{Pu}\]

This would expect to use alpha decay \[_{2}^{4}\textrm{\alpha }\]

because the band of stability indicates that is where this number of protons and neutrons fall and is the correct move to reach even number stabiity. Once again, we also know that nuclei with atomic numbers > 83 are too heavy to be stable and thus, Plutonium-245 will undergo alpha emission.

S20. 2.12

I disagree with his solutions (A, B, C, and D) on these problems because he did not refer to the activity series chart and his net ionic/complete ionic equations seem wrong.

Use reduction potential table to determine spontaneity for the reactions. If it proves to be spontaneous then the reaction can then occur.

A. \[NiBr_2\rightarrow Ni^{2+}+2Br\] then do cathode minus anode to see -0.23- (-0.44)= +0.21 meaning the reaction is spontaneous and the reaction can then occur.

net ionic: \[Fe^{2+}+2Br\rightarrow FeBr_2\]

Complete equation: \[NiBr_2+Fe^{2^+}+2Br\rightarrow Ni^{2+}+2Br^{-}+FeBr_2\]

I think that these are the net ionic and complete ionic equations:

net ionic: \[Fe(s) \rightarrow 2e^{-} + Fe^{2+}\]
\[Ni^{2+} + 2e^{-} \rightarrow Ni(s)\]

complete ionic: \[NiBr_{2} + Fe(s)\rightleftharpoons FeBr_{2} + Ni(s)\]

Iron will be oxidized since it is higher on the activity series chart and nickel will be reduced since it is lower than iron on the activity series chart.

B. \[HCl\rightarrow H^++Cl_2\] then do cathode minus anode to see -0.76-1.36= -2.12V which means the reaction is not spontaneous and the reaction cannot occur.

\[ZnCl_2\rightarrow Zn^{2+}+2Cl\]

I think that these are the net ionic and complete ionic equations:

net ionic: \[H^{+} + e^{-} \rightarrow H_{2}\]

\[Zn(s) \rightarrow 2e^{-} + Zn^{2+}\]

complete ionic: \[Zn(s) + 2HCl \rightleftharpoons H_{2} +ZnCl_{2}\]

Zinc will be oxidized since it is higher on the activity series chart and the hydrogen ion will be reduced since it is lower than zinc on the activity series chart.

C. Do cathode minus anode to see 1.36-0.16= +1.2V meaning the reaction is spontaneous and the reaction can then occur.

net ionic:\[Cu^{2+}+Cl^{-}\rightarrow CuCl_2\]

complete ionic:\[ZnCl_2+Cu^{2+}+Cl^{-}\rightarrow CuCl_2+Zn^{2+}+2Cl\]

I think that these are the net ionic and complete ionic equations:

net ionic: \[Cu^{2+} +2e^{-} \rightarrow Cu(s)\]

\[Zn \rightarrow 2e^{-} + Zn^{2+}\]

complete ionic: \[Zn(s) + Cu^{2+} \rightleftharpoons Zn^{2+} +Cu(s)\]

Zinc will be oxidized since it is higher on the activity series chart and the copper ion will be reduced since it is lower than zinc on the activity series chart.

D. \[Al +2Ag^{+}\rightarrow Al^{3+}+Ag\] then do cathode minus anode to see -1.66+0.80= -0.86 which means the reaction is not spontaneous and the reaction cannot occur.

I think that these are the net ionic and complete ionic equations:

net ionic: \[Ag^{2+} + e^{-} \rightarrow Ag^{+}\]

\[Al^{+} \rightarrow Al^{2+} + e^{-}\]

complete ionic: \[Ag(NO_{3}) + Al^{+} \rightleftharpoons Al(NO_{3}) +Ag^{+}\]

Aluminum will be oxidized since it is higher on the activity series chart and silver will be reduced since it is lower than aluminum on the activity series chart.

S20 5.7

Gold is a metal so it automatically is a good conductor, Gold is also very positive resulting in a position that is very high on the reduction potential chart (meaning that it has a very position reduction potential value) (See the chart at top of page) so it is very unlikely to be oxidized (corroded), elements that most likely will be corroded are lower on the table and usually more negative.

S24. 6.3

From Weak to Strong - \[I‐ < Br‐ < SCN‐ < Cl‐ < F‐ ≤ OH‐ , ONO‐ < OH2 < NCS‐ < NCCH3 < NH3 , py < NO2‐ < CN‐ , NO , CO\]

\[I^-\] is a weak field ligand so it has a very small \[\Delta o\], while \[NO^2-\] is one of the strongest field ligands meaning it has a very high \[\Delta o\]. This means the value \[\Delta o\] will increase if \[NO^2-\] ligand replaces the \[I^-\] ligand.

S14. 7.6

First, a homogenous catalyst refers to catalytic reactions where the catalyst is in the same phase as the reactants. Homogeneous catalysis applies to reactions in the gas phase and even in solids.

The main reason why homogenous catalyst would be preferred is that they are a lot easier to mix together. if you have a gas, applying a solid catalyst in would make the process grueling and more more difficult then if you add a gas or a homogenous catalyst.