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Extra Credit 42.

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    Q19.19C

    Write the cell reactions for the electrochemical cells provided below. Then use data from Table Table P2 calculate E°cell for each of the reactions.

    1. Ag(s)|Ag+(aq)||Hg2+(aq)|Hg(l)
    2. Al(s)|Al3+(aq)||Zn2+(aq)|Zn(s)
    3. Pt(s)|Ce4+(aq),Ce3+(aq)||I-(aq),I2(s)|C(s)
    4. Mg(s)|Mg2+(aq)||2H+(aq)|H2(g)

    S19.19C

    1. Ag(s)|Ag+(aq)||Hg2+(aq)|Hg(l)

    Step 1: Identify the reduction and oxidation of this cell.

    Left: Oxidation---Ag(s)⇌Ag+(aq) + e-

    Right: Reduction---Hg2+(aq) + 2e- ⇌ Hg(l)

    Step 2: Find the standard reduction potential of two half equation.

    E°=0.7996V Ag+(aq) + e- ⇌ Ag(s)

    E°=0.8535V Hg2+(aq) + 2e- ⇌ Hg(l)

    Step 3: Plug in values to the equation and determine cell potential.

    E°cell=E°red-E°ox

    E°cell=0.8535-0.7996=0.0539V

    2. Al(s)|Al3+(aq)||Zn2+(aq)|Zn(s)

    Step 1: Identify the reduction and oxidation of this cell.

    Left: Oxidation---Al(s) ⇌ Al3+(aq) + 3e-

    Right: Reduction---Zn2+(aq) + 2e- ⇌ Zn(s)

    Step 2: Find the standard reduction potential of two half equation.

    E°=-1.676V Al3+(aq) + 3e- ⇌ Al(s)

    E°=-0.7618V Zn2+(aq) + 2e- ⇌ Zn(s)

    Step 3: Plug in values to the equation and determine cell potential.

    E°cell=E°red-E°ox

    E°cell=(-0.7618)-(-1.676)=0.9142V

    3. Pt(s)|Ce4+(aq),Ce3+(aq)||I-(aq),I2(s)|C(s)

    Step 1: Identify the reduction and oxidation of this cell.

    Left: Oxidation---Ce3+(aq) ⇌ Ce4+(aq) + e-

    Right: Reduction---I2(s) + 2e- ⇌ 2I-(aq)

    Step 2: Find the standard reduction potential of two half equation.

    E°=1.44V Ce4+(aq) + e- ⇌ Ce3+(aq)

    E°=0.5355V I2(s) + 2e- ⇌ 2I-(aq)

    Step 3: Plug in values to the equation and determine cell potential.

    E°cell=E°red-E°ox

    E°cell=0.5355-1.44=-0.9045V

    4. Mg(s)|Mg2+(aq)||2H+(aq)|H2(g)

    Step 1: Identify the reduction and oxidation of this cell.

    Left: Oxidation--- Mg(s)⇌ Mg2+(aq) + 2e-

    Right: Reduction---2H+(aq) + 2e- ⇌H2(g)

    Step 2: Find the standard reduction potential of two half equation.

    E°=-2.356V Mg2+(aq) + 2e- ⇌ Mg(s)

    E°=0 2H+(aq) + 2e- ⇌ H2(g)

    Step 3: Plug in values to the equation and determine cell potential.

    E°cell=E°red-E°ox

    E°cell=2.356V

    Q20.3A

    Describe the following trends of the periodic table

    1. Atomic radii
    2. Ionization energy
    3. Electronegativity

    S20.3A

    1. Atomic radii decreases across the periodic table and increases down the periodic table. ( Increase below left diagonal)
    2. Ionization energy increases across the periodic table and decreases down the periodic table. (Increase above right diagonal)
    3. Electronegativity increases across the periodic table and decreases down the periodic table. ( Increase above right diagonal)

    Q21.15A

    Draw cis-dichlorobis(en)chromium (III) ion. Is it chiral? Is the trans isomer chiral?

    S21.15A

    Chiral - optically active -> no mirror plane

    Cis-Dichlorobis(ethylenediamine)cobalt(III)_chloride.pngCis: Not Chiral, Trans:Chiral

    Q24.21B

    A→product is a first order reaction. 97% of reactants decompose in 137 minutes. What is the half-life, t1/2, of this decomposition?

    S24.21B

    Step 1: We first identify that this is a first order reaction, so this will be the equation we are using:

    [A] = [A0]e−kt

    Step 2: We plug in the numbers provided in the question to get the rate constant:

    e−kt=[A]/[A0]

    k=-ln([A]/[A0])/t=-ln(0.03)/137=0.0256

    Step 3: Then we use this equation to calculate the half-life:

    t1/2=ln2/k=0.693/0.0256=27.07min

    Q25.4A

    The disintegration rate for a sample containing Co60 is 6635 dis h-1. The half-life of Co60 is 5.2 years. How many atoms of Co60 are in the sample?

    S25.4A

    Step 1: Determine the λ

    t1/2=ln2/λ

    λ=ln2/t1/2=0.693/(5.2*365*24)=1.52E-05h-1

    Step 2: Determine the number of atoms of Co-60 (N).

    A=λN

    N=A/λ=6635/1.52E-05h-1=4.36E08

    Q25.37F

    Calculate energy released in nuclear reaction: C611+42He→N714+11HC611+24He→N714+11H
    mass of nuclei:

    C611=10.9μ

    42He=3.21μ

    N714=13.104μ

    11H=1.01μ

    S25.37F

    Step 1: Determine the change in mass of this reaction

    delta mass = free neutron and proton- given mass

    (13.104μ+1.01μ)−(10.9μ+3.21μ)=0.004μ

    Step 2: Apply the equation

    E=mc2=0.004*1.66E-27*(3E08)2=5.98E-13J

    Q21.2.10

    Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules.

    1. 238Pa → ? + β−
    2. 216Fr → ? + α
    3. 199Bi → ? + β+

    S21.2.10

    1. 238Pa → ? + β−

    Step 1: Releasing a β− ( electron ) is changing a neutron in the element to a proton, so mass number remains the same but atomic number increases by 1:

    Zi=Zf=238

    Ai+1=Af=91+1=92

    Step 2: Identify the element from the information above

    U has atomic number of 92

    So reaction is 238Pa → 238U + β−

    Step 3: Calculate the mass change of the reaction:

    Initial: (238-91)*1.6749E-27+91*1.6726E-27=3.9842E-25 kg

    Final: (238-92)*1.6749E-27+92*1.6726E-27=3.9841E-25 kg

    Change: 0.0001E-25=1E-29 kg

    Step 4: Plug in numbers

    E=deltam* c2=1E-29*(3E08)2=9E-13 J=9E-16 kJ

    2. 216Fr → ? + α

    Step 1: Releasing a α is releasing a helium nucleus ( He-4) from the element , so mass number will decrease by 4 and atomic number increases by 2:

    Zi-4=Zf=216-4=212

    Ai-2=Af=87-2=85

    Step 2: Identify the element from the information above

    At has atomic number of 85

    So reaction is 216Fr → 212At + α

    Step 3: Calculate the mass change of the reaction:

    Initial: (216-87)*1.6749E-27+87*1.6726E-27=3.6158E-25 kg

    Final: (212-85)*1.6749E-27+85*1.6726E-27+6.645E-27=3.6153E-25 kg

    Change: 0.0005E-25=5E-29 kg

    Step 4: Plug in numbers

    E=deltam*c2=5E-29*(3E08)2=1.5E-12 J=1.5E-15 kJ

    3. 199Bi → ? + β+

    Step 1: Releasing a β+ (positron) is accepting a electron to the element, so mass number remains the same but atomic number decreases by 1:

    Zi=Zf=199

    Ai-1=Af=83-1=82

    Step 2: Identify the element from the information above

    Pb has atomic number of 82

    So reaction is 199Bi → 199Pb + β+

    Step 3: Calculate the mass change ( Final - Initial ) of the reaction:

    Initial: (199-83)*1.6749E-27+83*1.6726E-27=3.3311E-25 kg

    Final: (199-82)*1.6749E-27+82*1.6726E-27=3.3312E-25 kg

    Change: -0.0001E-25=-1E-29 kg

    Step 4: Plug in numbers

    E=deltam*c2=-1E-29*(3E08)2=9E-13 J=-9E-16 kJ

    Q20.3.10

    Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)6]3−, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.

    S20.3.10

    Step 1: Identify half Reactions:

    Fe(s)→Fe2+(aq)+2e-

    1/2O2(g)+H2O(l)+2e- → 2OH-(aq)

    3Fe2+(aq) + 2Fe(CN)63–(aq) → Fe3[Fe(CN)6]2(s)

    The is OH- presented in the reaction. Therefore, phenolphthalein indicator will show pink for this reaction

    Extra Credit 42. is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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