Extra Credit 2
- Page ID
- 83527
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From the following observations, estimate the E∘ from the half reaction M+(aq)+e−→M(s):
- The metal M reacts with H2SO4(aq); but not with HI(aq); M displaces Au+(aq), but not Fe3+(aq).
- The metal M reacts with HI(aq), producing H2(g), but displaces neither Al3+(aq nor Na+(aq).
S19.1A
1. Since the metal does not react with HI, it has a reduction potential greater than H, thus it is greater than 0V. Because it does react with H2SO4 it has a reduction potential less than SO42- which is 0.17V. This means 0V< M< 0.17V and the metal is between 0V and 0.17V. Since it displaces Au which has a reduction potential of 1.68V, the metal has a reduction potential less than this. Since it does not displace Iron it has reduction potential greater than Iron which is 0.769V. So 1.68V> M> 0.769V.
2. Since the Metal reacts with HI it has a redcution potential less than 0V for H2 but since it doesn't displace Al or Na it has a greater reduction potential than both of them. Na has a reduction potential of -2.714V and Al has a reduction potential of -1.7676V so the range is 0V> M>-1.7676V for the Metal.
Q19.27B
Find E∘cell, ΔG∘, K, and whether reaction goes to completion at 298 K and all substances at standard states for:
3HClO2(aq)+2Cr3+(aq)+12H2O(l)→3HClO(aq)+Cr2O2−7(aq)+8H3O+(aq)
S19.27B
To Find E∘cell for standard conditions we must look at the half reactions:
Oxidation(anode): 2Cr3+(aq)+21H2O(l)→Cr2O2−7(aq)+14H3O+(aq)+6e− E∘cell=+1.33V
Reduction(cathode): 3HClO2(aq)+6H3O+(aq)+6e−→3HClO(aq)+9H2O(l) E∘cell=+1.64V
To calculate E∘cell:
E∘cell=E∘cathode−E∘anode
E∘cell= 1.64-1.33= 0.31V
To calculate ΔG∘:
ΔG∘=-nFE∘cell
n= 6 moles of electrons
F= Faraday's constant 96,485C
E∘cell=0.31V
(-6mole e-)x(96485C)x(0.31V)=179462.1 J or 179.4621 kJ
To calculate K:
lnKeq =-ΔG∘/RT
-(-179.4621)/(0.008314 kJxmol)x(298K)
lnKeq=72.4
K=2.86x1031 Since K is very large at 298 degrees K this means it will go to completion.
Q20.11C
Complete and balance the following reaction: Cr2O3(s) +Al(s) →
S20.11C
In order to complete and balance the reaction, we must use the single replacement rule where in this case Chromium goes from being in the compound to its elemental form. Aluminum does the opposite and goes from its elemental form to a compound. Then we add coefficients and subscripts to balance out the charges.
Thus the answer is:
Cr2O3(s) +2Al(s) →2Cr(s)+Al2O3(s)
Q21.21A
Predict:
- Which of the following complex ions is paramagnetic and which is diamagnetic? [CoCl2]2- and [Ni(CN)4]2−
- What is the number of unpaired electrons in the tetrahedral complex, [CoCl2]2−?
S21.21A
Since the Nickel in [Ni(CN)4]2- is 2+ this means it has 8 d-electrons, and since CN- is the ligand means it has low spin all the electrons pair with each other first so this complex is diamagnetic. Since [CoCl2]2- has 7 d-electrons it is paramagnetic and there are 3 unpaired electrons since Cl is high spin so the electrons will go to the next level before pairing up.
[CoCl2]2− has 7 d-electrons and therefore due to its high spin nature the electrons will each go to a new orbital before pairing up and this leaves 3 unpaired electrons in the end.
Q24.33A
The reaction A+B→C+D is second order in A and zero order in B. The value of k is 0.0107 M-1 min-1. What is the rate of this reaction when [A]=0.106M and [B]=3.73M?
S24.33A
(Solution was incorrect. The solution was redone and corrections have been made. -Jasmine Leung)
The rate of the reaction is given:
rate=k[A]2[B]0
Plugging in the given values we have:
rate=(0.0107 M-1min-1)[0.106M]2[3.73M]0
rate=1.2x10-4 M min-1
Q25.11A
Write equations for the following nuclear reactions:
- Bombardment of Th234 with α particles to produce U238
- Bombardment of Pa234 with hydrogen atoms to produce U235U235
- Bombardment of U238U238 with neutrons to produce U239U239 and γ rays and then ββ-decay of U239
S25.11A
234Th+4He →238U 90p+2p=92p
234 Pa+1H→235U 91p +1p=92p
238U+1n→239U+γ 92p
239U→240Np+-1e- 92p=93p+-1
Q25.42B
Which isotope is more likely to be the most abundant in nature? Explain.
- 14C vs. 12C
- 35Cl vs. 36Cl
- 96Ru vs. 95Ru
S25.45B
1. Since Carbon-14 is radioactive(decays over time) and has an unequal numbers of protons and neutrons, while Carbon-12 has an equal number of protons and neutrons, Carbon-12 is more likely to be found in nature and in fact Carbon-12 is found in all living things. It is favorable in low atomic number elements to have a 1:1 ratio of neutrons and protons.
2. Since 35Cl has an even number of neutrons and is closer to the average atomic mass of chlorine than 36Cl it is more abundant in nature.
3. 96Ru is more abundant because since this isotope has 44 protons and 52 neutrons it has an even number of neutrons and protons and therefore is more stable than95Ru in nature.
Q21.2.31
Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 141Ba, 92Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
S21.2.31
235U+1n→141Ba+92Kr+31n
E=Δmc2
Δm=mass of products - mass of reactants
Reactants =(143x1.008664)+(92x1.00727647)+(1.008664)=237.19170512 u
Products=(3x1.008664)+(92.926156)+(85x1.008664)+(56x1.00727647)=238.0960703 u
Δm=.9043652 u
(0.9043652 u)x(1.66054x10-27kg / 1 u)=1.50173459x10-27 kg
E=mc2=(1.50173459x10-27 kg) x(3.00x108 m/s)2
E=1.35x10-13 kJ/mole
1 kJ=6.242x1021
842603761.34 ev/mole x6.0223 atoms per mole =5.07x1032 electronvolts per atom
E=1.35x10-13kJ/mole or 5.07x1032 electronvolts per atom