Extra Credit 26

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Q17.3.5

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

S17.3.5

The reactions for the galvanic cell are:

Anode (oxidation) : \(3Cd(s) \rightarrow 3Cd^{2+}(aq) + 6e^-\)

Cathode (reduction): \(2Al^{3+}(aq) + 6e^- \rightarrow 2Al(s)\)

Overall reaction: \(3Cd(s) + 2Al^{3+}(aq) \rightarrow 3Cd^{2+}(aq) + 2Al(s)\)

The standard cell potential for a reaction can be calculated by subtracting the standard reduction potential for half-reaction at the anode from the standard reduction potential for the half-reaction at the cathode.

\(E_{cell} = E_{cathode} - E_{anode}\)

If \(E_{cell}\) > 0, then the reaction is spontaneous

If \(E_{cell}\) < 0, then the reaction is nonspontaneous

We can find the standard reduction of each species using the Standard Reduction Potential Table found here.

\(E_{cathode}= -1.662V\)

\(E_{anode}= -0.4030V\)

\(E_{cell}= -1.662V - (-0.4030V)\)

\(E_{cell}= −1.259 V\)

Since \(E_{cell}\) < 0 , the reaction is nonspontaneous

Q19.1.24

Predict which will be more stable, \([CrO_4]^{2-}\) or \([WO_4]^{2-}\), and explain.

S19.1.24

Chromium and Tungsten are both Group 6 elements. Chromium is found in the 3d orbital while Tungsten is found in the 4d orbital.

Within a group, higher oxidation states become more stable down the group. Therefore, \([WO_4]^{2-}\)is more stable than \([CrO_4]^{2-}\).

Q12.4.17

Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for 164164 of the initial dose to remain in the athlete’s body?

S12.4.17

When a chemical reaction is a first order reaction, the half life for the starting compound is given by: \(t_{\frac{1}{2}}= \frac{0.693}{k}\)

\(t_{\frac{1}{2}}\) is the time required for half of the compound to decompose

\(k\) is the rate constant of the reaction.

Since the half life of steroids taken by the athlete is 42 days, plug in 42 to \(t_{\frac{1}{2}}\)

\(k =\frac{ 0.693 }{42}\)

\(k = 0.0165\)

The rate constant for first order reactions is: \([A] = [A]_0 e^{-kt}\)

[A]₀ is the initial concentration of the reactant (steroid)

[A] is the concentration of the reactant (steroid) left after time \(t\)

Given that [A]₀=1 and \([A]=\frac{1}{64}\) and \(k = 0.0165\)

Substitute into the equation:

\(\frac{1}{64} = e^{-0.0165t}\)

\(t = 252\) days

Therefore, it would take 252 days for \(\frac{1}{64}\) of the initial concentration of the steroid to remain in the body.

Q21.3.1

Write a brief description or definition of each of the following:

nucleon
α particle
β particle
positron
\(\gamma\) ray (gamma)
nuclide
mass number
atomic number

S21.3.1

1. Nucleons are made up of two important subatomic particles: protons and neutrons, which are the particles within the nucleus.

2. \(\alpha\) particles are made up of two protons and two neutrons like the nucleus of a helium particle (ie. \(_{2}^{4}\textrm{He}\)). They are produced through a radioactive process called alpha decay.

3. \(\beta\) particles are high energy and high speed electrons with a mass of zero and a charge of -1 (ie. \(_{-1}^{0}\textrm{e}\)). They are produced through a radioactive process called beta decay.

4. Positrons have the same mass of an electron, but instead of a negative charge, they have a positive charge (ie.\(_{1}^{0}\textrm{e}\)). They are emitted during beta decay.

5. Gamma rays are a form of electromagnetic radiation and they have the smallest wavelength as well as the highest energy.

6. Nuclide refers to a specific number of protons and neutrons in the nucleus.

7. The mass number refers to the the sum of the protons and the neutrons in an element.

8. The atomic number refers to the number of protons in the nucleus of an element.

Q21.7.3

Given specimens uranium-232 (\(t_\frac{1}{2}=68.9\gamma\)) and uranium-233 (\(t_\frac{1}{2}=159,200\gamma\)) of equal mass, which one would have greater activity and why?

S21.7.3

Half- life is the time required for a substance to decay by one-half.

The activity of a radioactive substance is: \(Activity = \lambda N\)

\(\lambda\) is the decay constant

N is the concentration of nuclides

The decay constant (\(\lambda\)) is calculated by: \(\lambda =\frac{ln2}{t_\frac{1}{2}}\)

Substituting into the activity equation, we get: \(Activity = \frac{ln2}{t_\frac{1}{2}}N\)

For uranium-232, the activity is:

\(Activity = \frac{ln2}{68.9}(232)\)

\(Activity = 2.33\)

For uranium-233, the activity is:

\(Activity = \frac{ln2}{159,200}(233)\)

\(Activity = 0.00101\)

Therefore, Uranium-232 has more activity than Uranium-233

Q20.4.13

Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.

Cl₂(g) + H₂(g) → 2Cl⁻(aq) + 2H⁺(aq)
Br₂(aq) + Fe²⁺(aq) → 2Br⁻(aq) + Fe³⁺(aq)
Fe³⁺(aq) + Cd(s) → Fe²⁺(aq) + Cd²⁺(aq)

S20.4.13

\(E_{cell} = E_{cathode} - E_{anode}\)

If \(E_{cell}> 0\), then the reaction is spontaneous

If \(E_{cell}< 0\), then the reaction is nonspontaneous

We can find the standard reduction of each species using the Standard Reduction Potential Table found here.

1. Anode (oxidation): H₂(g) →2H⁺(aq)+ \(2e^-\)

Cathode (reduction): Cl₂(g) + \(2e^-\)→ 2Cl⁻(aq)

Overall reaction: Cl₂(g) + H₂(g) → 2Cl⁻(aq) + 2H⁺(aq)

\(E_{cathode}=1.358V\)

\(E_{anode}=0V\)

\(E_{cell}= 1.358V - 0V = 1.358V\)

2. Anode (oxidation): Fe²⁺(aq) → Fe³⁺(aq) + \(e^-\)

Cathode (reduction): Br₂(aq) + \(2e^-\) → 2Br⁻(aq)

Overall reaction: Br₂(aq) + 2Fe²⁺(aq) → 2Br⁻(aq) + 2Fe³⁺(aq)

\(E_{cathode}= 1.087V\)

\(E_{anode}= 0.771V\)

\(E_{cell}= 1.087V - 0.771V = 0.316V\)

3. Anode (oxidation): Cd(s) → Cd²⁺(aq) + \(2e^-\)

Cathode (reduction): Fe³⁺(aq) + \(e^-\)→ Fe²⁺(aq)

Overall reaction: 2Fe³⁺(aq) + Cd(s) → 2Fe²⁺(aq) + Cd²⁺(aq)

\(E_{cathode}= -0.403V\)

\(E_{anode}= 0.771V\)

\(E_{cell}= -0.403V - 0.771V = -1.174V\)

Q20.4.15

Write a balanced chemical equation for each redox reaction.

H₂PO₂⁻(aq) + SbO₂⁻(aq) → HPO₃²⁻(aq) + Sb(s) in basic solution
HNO₂(aq) + I⁻(aq) → NO(g) + I₂(s) in acidic solution
N₂O(g) + ClO⁻(aq) → Cl⁻(aq) + NO₂⁻(aq) in basic solution
Br₂(l) → Br⁻(aq) + BrO₃⁻(aq) in basic solution
Cl(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → ClCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

S20.4.15

1. First, write out the half reactions:

H₂PO₂⁻(aq) → HPO₃²⁻(aq)

SbO₂⁻(aq) → Sb(s)

Balance the oxygens by adding H₂O:

H₂O(l) + H₂PO₂⁻(aq) → HPO₃²⁻(aq)

SbO₂⁻(aq) → Sb(s) + 2H₂O(l)

Balance the hydrogens by adding H⁺

H₂O(l) + H₂PO₂⁻(aq) → HPO₃²⁻(aq) + H⁺(aq)

4H⁺(aq)+ SbO₂⁻(aq) → Sb(s) + 2H₂O(l)

Balance by adding OH^-(aq) to both sides

OH^-(aq)+ H₂O(l) + H₂PO₂⁻(aq) → HPO₃²⁻(aq) + H⁺(aq) + OH^-(aq)

4H⁺(aq) + 4OH^-(aq) + SbO₂⁻(aq) → Sb(s) + 2H₂O(l) + 4OH^-(aq)

Combine to form H₂O and add electrons:

OH^-(aq) + H₂O(l) + H₂PO₂⁻(aq) → HPO₃²⁻(aq) + H₂O(l) + \(2e^-\)

\(3e^-\) + 4H₂O(l) + SbO₂⁻(aq) → Sb(s) + 2H₂O(l) + 4OH^-(aq)

Multiply the top half reaction by 3. Multiply the bottom half reaction by 2 to cancel the electrons:

3OH^-(aq) + 3H₂O(l) + 3H₂PO₂⁻(aq) → 3HPO₃²⁻(aq) + 3H₂O(l) + \(6e^-\)

\(6e^-\) + 8H₂O(l) + 2SbO₂⁻(aq) → 2Sb(s) + 4H₂O(l) + 8OH^-(aq)

Cancel like species to get the overall balanced equation:

OH^-(aq) + 3H₂PO₂⁻(aq) + 2SbO₂⁻(aq) → 3HPO₃²⁻(aq) + 2Sb(s) + 2H₂O(l)

2. Balance half equations

HNO₂(aq)→ NO(g)

2I⁻(aq) → I₂(s)

Add water to balance the oxygens:

HNO₂(aq)→ NO(g) + H₂O(l)

2I⁻(aq) → I₂(s)

Balance electrons on both sides:

\(e^-\) + HNO₂(aq)→ NO(g) + H₂O(l)

2I⁻(aq) → I₂(s) + \(2e^-\)

Multiply top half reaction by 2 to cancel electrons:

\(2e^-\) + 2HNO₂(aq)→ 2NO(g) + 2H₂O(l)

2I⁻(aq) → I₂(s) + \(2e^-\)

Cancel like species to get overall balanced equation:

2H⁺(aq) + 2HNO₂(aq) + 2I⁻(aq) → 2NO(g) + I₂(s) + 2H₂O(l)

3. Balance half reactions:

N₂O(g) → 2NO₂⁻(aq)

ClO⁻(aq) → Cl⁻(aq)

Balance oxygens with water:

3H₂O(l) + N₂O(g) → 2NO₂⁻(aq)

ClO⁻(aq) → Cl⁻(aq) + H₂O(l)

Balance with hydrogen:

3H₂O(l) + N₂O(g) → 2NO₂⁻(aq) + 6H⁺(aq)

2H⁺(aq) + ClO⁻(aq) → Cl⁻(aq) + H₂O(l)

Since reaction is in basic solution, we need to balance hydrogens with hydroxide:

6OH^-(aq) + 3H₂O(l) + N₂O(g) → 2NO₂⁻(aq) + 6H⁺(aq) + 6OH^-(aq)

2OH^-(aq) + 2H⁺(aq) + ClO⁻(aq) → Cl⁻(aq) + H₂O(l) + 2OH^-(aq)

Balance charges with electrons and combine water

6OH^-(aq) + 3H₂O(l) + N₂O(g) → 2NO₂⁻(aq) + 6H₂O(l) + \(4e^-\)

\(2e^-\) + 2H₂O(l) + ClO⁻(aq) → Cl⁻(aq) + H₂O(l) + 2OH^-(aq)

Multiply bottom half reaction with 2 to cancel electrons:

6OH^-(aq) + 3H₂O(l) + N₂O(g) → 2NO₂⁻(aq) + 6H₂O(l) + \(4e^-\)

\(4e^-\) + 4H₂O(l) + 2ClO⁻(aq) → 2Cl⁻(aq) + 2H₂O(l) + 4OH^-(aq)

Balanced overall equation:

2OH^-(aq) + N₂O(g) + 2ClO⁻(aq) → 2Cl⁻(aq) + 2NO₂⁻(aq) + H₂O(l)

4. Balance the half reactions:

Br₂(l) → 2Br⁻(aq)

Br₂(l) → 2BrO₃⁻(aq)

Balance oxygens with water:

Br₂(l) → 2Br⁻(aq)

6H₂O(l) + Br₂(l) → 2BrO₃⁻(aq)

Balance with hydrogens:

Br₂(l) → 2Br⁻(aq)

6H₂O(l) + Br₂(l) → 2BrO₃⁻(aq) + 12H⁺(aq)

Since the reaction happens in basic solution, we need to balance the hydrogens with hydroxide:

Br₂(l) → 2Br⁻(aq)

12OH^-(aq) + 6H₂O(l) + Br₂(l) → 2BrO₃⁻(aq) + 12H⁺(aq) + 12OH^-(aq)

Balance charges with electrons:

\(2e^-\) + Br₂(l) → 2Br⁻(aq)

12OH^-(aq) + 6H₂O(l) + Br₂(l) → 2BrO₃⁻(aq) + 12H₂O(l) + \(10e^-\)

Multiply top half reaction by 5 to cancel electrons:

\(10e^-\) + 5Br₂(l) → 10Br⁻(aq)

12OH^-(aq) + 6H₂O(l) + Br₂(l) → 2BrO₃⁻(aq) + 12H₂O(l) + \(10e^-\)

Cancel like species to get overall balanced equation:

6OH^-(aq) + 3Br₂(l) → 5Br⁻(aq) + BrO₃⁻(aq)+ 3H₂O(l)

5. Balance half reactions:

Cl(CH₂)₂OH(aq) → ClCH₂CO₂H(aq)

K₂Cr₂O₇(aq) → Cr³⁺(aq)

Balance second half reaction with missing elements:

K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺

Balance both with oxygens:

H₂O(l) + Cl(CH₂)₂OH(aq) → ClCH₂CO₂H(aq)

K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺+ H₂O(l)

Balance both with hydrogens:

H₂O(l) + Cl(CH₂)₂OH(aq) → ClCH₂CO₂H(aq) + 4H⁺(aq)

14H⁺(aq) + K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺+ 7H₂O(l)

Balance charges with electrons:

H₂O(l) + Cl(CH₂)₂OH(aq) → ClCH₂CO₂H(aq) + 4H⁺(aq) + \(4e^-\)

\(6e^-\) + 14H⁺(aq) + K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺+ 7H₂O(l)

Multiply top reaction by 3 and the bottom reaction by 2 to cancel electrons:

3H₂O(l) + 3Cl(CH₂)₂OH(aq) → 3ClCH₂CO₂H(aq) + 12H⁺(aq) + \(12e^-\)

\(12e^-\) + 28H⁺(aq) + 2K₂Cr₂O₇(aq) → 4Cr³⁺(aq) + 4K⁺+ 14H₂O(l)

Cancel like species to get balanced overall reaction:

3Cl(CH₂)₂OH(aq) + 2K₂Cr₂O₇(aq) + 16H⁺(aq) → 3ClCH₂CO₂H(aq) + 4Cr³⁺(aq) + 4K⁺(aq) + 11H₂O(l)

Q20.4.16

Write a balanced chemical equation for each redox reaction.

I⁻(aq) + HClO₂(aq) → IO₃⁻(aq) + Cl₂(g) in acidic solution
Cr²⁺(aq) + O₂(g) → Cr³⁺(aq) + H₂O(l) in acidic solution
CrO₂⁻(aq) + ClO⁻(aq) → CrO₄²⁻(aq) + Cl⁻(aq) in basic solution
S(s) + HNO₂(aq) → H₂SO₃(aq) + N₂O(g) in acidic solution
F(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → FCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

S20.4.16

1. Balance the half reactions:

I⁻(aq) → IO₃⁻(aq)

HClO₂(aq) → Cl₂(g)

Balance the oxygens with water:

3H₂O(l) + I⁻(aq) → IO₃⁻(aq)

HClO₂(aq) → Cl₂(g) + 2H₂O(l)

Balance with hydrogens:

3H₂O(l) + I⁻(aq) → IO₃⁻(aq) + 6H⁺(aq)

3H⁺(aq) + HClO₂(aq) → Cl₂(g) + 2H₂O(l)

Balance charges with electrons:

3H₂O(l) + I⁻(aq) → IO₃⁻(aq) + 6H⁺(aq) + \(6e^-\)

\(3e^-\) + 3H⁺(aq) + HClO₂(aq) → Cl₂(g) + 2H₂O(l)

Multiply bottom half reaction by 2 to cancel electrons:

3H₂O(l) + I⁻(aq) → IO₃⁻(aq) + 6H⁺(aq) + \(6e^-\)

\(6e^-\) + 6H⁺(aq) + 2HClO₂(aq) → 2CI₂(g) + 4H₂O(l)

Cancel like species. Overall balanced reaction:

I⁻(aq) + 2HClO₂(aq) → IO₃⁻(aq) + Cl₂(g) + H₂O(l)

2. Balance half reactions:

Cr²⁺(aq) → Cr³⁺(aq)

O₂(g) → H₂O(l)

Balance oxygens with water :

Cr²⁺(aq) → Cr³⁺(aq)

O₂(g) → H₂O(l) + H₂O(l)

Balance with hydrogen:

Cr²⁺(aq) → Cr³⁺(aq)

4H⁺(aq) + O₂(g) → H₂O(l) + H₂O(l)

Balance charges with electrons:

Cr²⁺(aq) → Cr³⁺(aq) + \(e^-\)

\(4e^-\) + 4H⁺(aq) + O₂(g) → H₂O(l) + H₂O(l)

Multiply top half reaction by 4 to cancel electrons:

4Cr²⁺(aq) → 4Cr³⁺(aq) + \(4e^-\)

\(4e^-\) + 4H⁺(aq) + O₂(g) → H₂O(l) + H₂O(l)

Cancel like species. Overall balanced equation:

4H⁺(aq) + 4Cr²⁺(aq) + O₂(g) → 4Cr³⁺(aq) + 2H₂O(l)

3.Balance half reactions:

CrO₂⁻(aq) → CrO₄²⁻(aq)

ClO⁻(aq) → Cl⁻(aq)

Balance oxygens with water:

2H₂O(l) + CrO₂⁻(aq) → CrO₄²⁻(aq)

ClO⁻(aq) → Cl⁻(aq) + H₂O(l)

Balance with Hydrogen:

2H₂O(l) + CrO₂⁻(aq) → CrO₄²⁻(aq) + 4H⁺(aq)

2H⁺(aq)n + ClO⁻(aq) → Cl⁻(aq) + H₂O(l)

Since reaction occurs in basic solution, we need to balance hydrogens with hydroxide:

4OH^-(aq) + 2H₂O(l) + CrO₂⁻(aq) → CrO₄²⁻(aq) + 4H⁺(aq) + 4OH^-(aq)

2OH^-(aq) + 2H⁺(aq)n + ClO⁻(aq) → Cl⁻(aq) + H₂O(l) + 2OH^-(aq)

Balance charges with electrons. Combine hydrogens and hydroxide into water:

4OH^-(aq) + 2H₂O(l) + CrO₂⁻(aq) → CrO₄²⁻(aq) + 4H₂O(l) + \(3e^-\)

\(2e^-\) + 2H₂O(l) + ClO⁻(aq) → Cl⁻(aq) + H₂O(l) + 2OH^-(aq)

Multiply top half reaction by 2. Multiply bottom half reaction by 3:

8OH^-(aq) + 4H₂O(l) + 2CrO₂⁻(aq) → 2CrO₄²⁻(aq) + 8H₂O(l) + \(6e^-\)

\(6e^-\) + 6H₂O(l) + 3ClO⁻(aq) → 3Cl⁻(aq) + 3H₂O(l) + 6OH^-(aq)

Cancel like species. Balanced overall equation:

2OH^-(aq) + 2CrO₂⁻(aq) + 3ClO⁻(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) +H₂O(l)

4. Balance half reactions:

S(s) → H₂SO₃(aq)

HNO₂(aq) → N₂O(g)

Balance oxygens with water:

3H₂O(l) + S(s) → H₂SO₃(aq)

HNO₂(aq) → N₂O(g) + 3H₂O(l)

Balance with hydrogen:

3H₂O(l) + S(s) → H₂SO₃(aq) + 4H⁺(aq)

4H⁺(aq) + HNO₂(aq) → N₂O(g) + 3H₂O(l)

Balance charges with electrons:

3H₂O(l) + S(s) → H₂SO₃(aq) + 4H⁺(aq) + \(4e^-\)

\(4e^-\) + 4H⁺(aq) + HNO₂(aq) → N₂O(g) + 3H₂O(l)

Cancel like species. Balanced overall equation:

S(s) + 2HNO₂(aq) → H₂SO₃(aq) + N₂O(g)

5. Balance half reactions:

F(CH₂)₂OH(aq) → FCH₂CO₂H(aq)

K₂Cr₂O₇(aq) →Cr³⁺(aq)

Balance second half reaction with missing elements:

K₂Cr₂O₇(aq) →2Cr³⁺(aq) + 2K⁺

Balance both with oxygens:

H₂O(l) + F(CH₂)₂OH(aq) → FCH₂CO₂H(aq)

K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺ + 7H₂O(l)

Balance both with hydrogens:

H₂O(l) + F(CH₂)₂OH(aq) → FCH₂CO₂H(aq) + 4H⁺

14H⁺ + K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺ + 7H₂O(l)

Balance charges with electrons:

H₂O(l) + F(CH₂)₂OH(aq) → FCH₂CO₂H(aq) + 4H⁺+ \(4e^-\)

\(6e^-\) + 14H⁺ + K₂Cr₂O₇(aq) → 2Cr³⁺(aq) + 2K⁺ + 7H₂O(l)

Multiply top reaction by 3 and the bottom reaction by 2 to cancel electrons:

3H₂O(l) + 3F(CH₂)₂OH(aq) → 3FCH₂CO₂H(aq) + 12H⁺+ \(12e^-\)

\(12e^-\) + 28H⁺ + 2K₂Cr₂O₇(aq) → 4Cr³⁺(aq) + 4K⁺ + 14H₂O(l)

Cancel like species to get balanced overall reaction:

3F(CH₂)₂OH(aq) + 2K₂Cr₂O₇(aq) + 16H⁺(aq) → 3FCH₂CO₂H(aq) + 4Cr³⁺(aq) + 4K⁺(aq) + 11H₂O(l)

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