Extra Credit 2

Q19.1A

From the following observations, estimate the E^∘ from the half reaction M⁺(aq)+e⁻→M(s):

1. The metal M reacts with H₂SO₄(aq); but not with HI(aq); M displaces Au⁺(aq), but not Fe³⁺(aq).
2. 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 SO₄^2- 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:

lnK_eq =-ΔG∘/RT

-(-179.4621)/(0.008314 kJxmol)x(298K)

lnK_eq=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:

1. Which of the following complex ions is paramagnetic and which is diamagnetic? [CoCl2]^2- and [Ni(CN)4]²⁻
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:

1. Bombardment of Th234 with α particles to produce U238
2. Bombardment of Pa234 with hydrogen atoms to produce U235U235
3. 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.

1. ¹⁴C vs. ¹²C
2. ³⁵Cl vs. ³⁶Cl
3. 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