Extra Credit 22

Q17.3.1

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

1. \[Mg(s)+Ni^{_{2+}}(aq)\rightarrow Mg^{_{2+}}(aq)+Ni(s)\]
2. \[2Ag^{_{1+}}(aq)+Cu(s)\rightarrow Cu^{_{2+}}(aq)+2Ag(s)\]
3. \[Mn(s)+Sn(NO_3)_2(aq)\rightarrow Mn(NO_3)_2(aq)+Sn(s)\]
4. \[3Fe(NO_3)_2(aq)+Au(NO_3)_3(aq)\rightarrow 3Fe(NO_3)_3(aq)+Au(s)\]

S17.3.1

(a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)

With these problems, we must separate the reaction into two reactions where one is a reduction reaction and the other is a oxidation reaction by separating the reactions based off of the elements involved. To determine the reduction and oxidation reactions:

1) balance the elements in the equation

2) add water (H2O) to balance out the oxygens (O) and balance out the hydrogens in the reaction (H+) with H+

3) balance the charge of the reaction by counting up the positive/negative charges and using electrons to balance the equation

From left to right (forward reaction), oxidation is the loss of an electron and reduction is the gain of an electron.

Oxidation occurs at the anode and reduction occurs at the cathode. We can then determine the cell potential of these individual reactions since

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

where E°cathode and E°anode is determined from the standard reduction potential table.

1. \[Mg(s)+Ni^{_{2+}}(aq)\rightarrow Mg^{_{2+}}(aq)+Ni(s)\]

\[Ni^{_{2+}}(aq)\rightarrow Ni(s)\]

\[Mg(s)\rightarrow Mg^{_{2+}}(aq)\]

Balancing the equations (following the steps listed above):

\[Ni^{_{2+}}(aq)\rightarrow Ni(s)\]

1) \[Ni^{_{2+}}(aq)\rightarrow Ni(s)\] (already balanced elements)

2) \[Ni^{_{2+}}(aq)\rightarrow Ni(s)\] (no oxygens or hydrogens to balance)

3) \[Ni^{_{2+}}(aq)+2e^{_{-}}\rightarrow Ni(s)\] (balance charges)

This is a reduction reaction since electrons are gained.

\[Mg(s)\rightarrow Mg^{_{2+}}(aq)\]

1) \[Mg(s)\rightarrow Mg^{_{2+}}(aq)\] (already balanced elements)

2) \[Mg(s)\rightarrow Mg^{_{2+}}(aq)\] (no oxygens or hydrogens to balance)

3) \[Mg(s)\rightarrow Mg^{_{2+}}(aq)+2e^{_{-}}\] (balance charges)

This is an oxidation reaction since electrons are lost.

Oxidation occurs at the anode and reduction occurs at the cathode.

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

From the standard reduction potential table E°cathode = -0.25 V, E°anode = -2.37 V

E°cell = -0.25 - (-2.37) =2.12V.

*The slight difference in E°cell is due to the use of different standard reduction potential tables*

A reaction is spontaneous if E°cell > 0, so this reaction is spontaneous.

2. \[2Ag^{_{1+}}(aq)+Cu(s)\rightarrow Cu^{_{2+}}(aq)+2Ag(s)\]

\[2Ag^{_{+}}(aq)\rightarrow 2Ag(s)\]

\[Cu(s)\rightarrow Cu^{_{2+}}(aq)\]

Balancing the equations (following the steps listed above):

\[2Ag^{_{+}}(aq)\rightarrow 2Ag(s)\]

1) \[2Ag^{_{+}}(aq)\rightarrow 2Ag(s)\] (already balanced elements)

2) \[2Ag^{_{+}}(aq)\rightarrow 2Ag(s)\] (no oxygens or hydrogens to balance)

3) \[2Ag^{_{+}}(aq)+2e^{_{-}}\rightarrow 2Ag(s)\] (balance charges)

4) simplify the equation (by dividing by 2)

\[Ag^{_{+}}(aq)+e^{_{-}}\rightarrow Ag(s)\]

This is a reduction reaction since electrons are gained.

\[Cu(s)\rightarrow Cu^{_{2+}}(aq)\]

1) \[Cu(s)\rightarrow Cu^{_{2+}}(aq)\] (already balanced elements)

2) \[Cu(s)\rightarrow Cu^{_{2+}}(aq)\] (no oxygens or hydrogens to balance)

3) \[Cu(s)\rightarrow Cu^{_{2+}}(aq)+2e^{_{-}}\] (balance charges)

This is an oxidation reaction since electrons are lost.

Oxidation occurs at the anode and reduction occurs at the cathode.

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

From the standard reduction potential table E°cathode = 0.80 V, E°anode = 0.337 V

E°cell = 0.80 - 0.337 = 0.463V

*The slight difference in E°cell is due to the use of different standard reduction potential tables*

A reaction is spontaneous if E°cell > 0, so this reaction is spontaneous.

3. \[Mn(s)+Sn(NO_3)_2(aq)\rightarrow Mn(NO_3)_2(aq)+Sn(s)\]

\[Mn(s)\rightarrow Mn(NO_3)_2(aq)\]

\[Sn(NO_3)_2(aq)\rightarrow Sn(s)\]

Instead of balancing the equations, we can just determine the charge of the elements to determine if the reaction is a reduction or oxidation.

\[Mn(s) \rightarrow Mn(NO_3)_2(aq)\]

In Mn(NO3)2(aq), NO3 has a charge of -1 and there are two of them, then Mn has a charge of +2.

Then in the reaction (focusing only on the element Mn), \[Mn(s)\rightarrow Mn^{_{2+}}(aq)\] which means that there has to be electrons added to balance the equation:

\[Mn(s)\rightarrow Mn^{_{2+}}(aq)+2e^{_{-}}\]

This is an oxidation reaction since electrons are lost.

\[Sn(NO_3)_2(aq)\rightarrow Sn(s)\]

In Sn(NO3)2(aq), NO3 has a charge of -1 and there are two of them, then Sn has a charge of +2.

Then in the reaction (focusing only on the element Sn), \[Sn^{_{2+}}(aq)\rightarrow Sn(s)\] which means that there has to be electrons added to balance the equation:

\[Sn^{_{2+}}(aq)+2e^{_{-}}\rightarrow Sn(s)\]

This is a reduction reaction since electrons are gained.

Oxidation occurs at the anode and reduction occurs at the cathode.

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

From the standard reduction potential table E°cathode = -0.14 V, E°anode = -1.18 V

E°cell = -0.14 - (-1.18) = 1.04 V.

*The slight difference in E°cell is due to the use of different standard reduction potential tables*

A reaction is spontaneous if E°cell > 0, so this reaction is spontaneous.

4. \[3Fe(NO_3)_2(aq)+Au(NO_3)_3(aq)\rightarrow 3Fe(NO_3)_3(aq)+Au(s)\]

\[3Fe(NO_3)_2(aq)\rightarrow 3Fe(NO_3)_3(aq)\]

\[Au(NO_3)_3(aq)\rightarrow Au(s)\]

Instead of balancing the equations, we can just determine the charge of the elements to determine if the reaction is a reduction or oxidation.

\[3Fe(NO_3)_2(aq)\rightarrow 3Fe(NO_3)_3(aq)\]

Simplify the reaction by dividing both sides by 3, so that \[Fe(NO_3)_2(aq)\rightarrow Fe(NO_3)_3(aq)\] .

In Fe(NO₃)₃(aq), NO₃ has a charge of -1 and there are two of them, then Fe has a charge of +2.

In Fe(NO₃)₃(aq), NO₃ has a charge of -1 and there are three of them, then Fe has a charge of +3.

Then in the reaction (focusing only on the element Fe), \[Fe^{_{2+}}\rightarrow Fe^{_{3+}}\] which means that there has to be electrons added to balance the equation:

\[Fe^{_{2+}}\rightarrow Fe^{_{3+}}+e^{_{-}}\]

This is an oxidation reaction since electrons are lost.

\[Au(NO_3)_3(aq)\rightarrow Au(s)\]

In Au(NO3)3(aq), NO3 has a charge of -1 and there are three of them, then Au has a charge of +3.

Then in the reaction (focusing only on the element Fe), \[Au^{_{3+}}\rightarrow Au\] which means that there has to be electrons added to balance the equation:

\[Au^{_{3+}}+3e^{_{-}}\rightarrow Au\]

This is a reduction reaction since electrons are gained.

Oxidation occurs at the anode and reduction occurs at the cathode.

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

From the standard reduction potential table E°cathode = +1.5 V, E°anode = +0.771 V

E°cell = 1.5 - 0.771 = 0.729 V.

*The slight difference in E°cell is due to the use of different standard reduction potential tables*

A reaction is spontaneous if E°cell > 0, so this reaction is spontaneous.

Q19.1.20

What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?

Sulfur has the chemical formula S2-, so iron(II) sulfide has the chemical formula: FeS.

A nonoxidizing acid is an acid in which there is no stronger oxidant than H+ at pH 0 and cannot dissolve any noble metal at 1 M1 M concentration, but should (in the absence of effects like overpotential) dissolve any non-noble metal at that concentration.

Metals that react with these acids produce a metal salt and hydrogen gas:
metal + non-oxidizing acid → metal salt + hydrogen gas

The metal in this case would be iron (Fe) and we can substitute dilute hydrochloric acid (HCl) as our nonoxidizing acid. By using the general reaction:

metal + non-oxidizing acid → metal salt + hydrogen gas

\[FeS+2HCl\rightarrow FeCl_2+H_2S\] (Hydrogen gas will be produced with any nonoxidizing acid)

The gas produced when iron(II) sulfide is treated with a nonoxidizing acid is hydrogen sulfide gas.

Q19.3.12

Would you expect salts of the gold ion, Au⁺, to be colored? Explain.

S19.3.12

No. Au⁺ has a complete 5d sublevel.

For this problems, I will be using the shorthand method of doing electron configurations.

I will go across each period to reach the element that I am interested in. For example, Li would be 1s²2s¹ because it goes through the first period and to reach Li on the periodic table, we would need to add another electron and it would be from the 2s shell. In another example, N would be 1s²2s²2p³ because you need to go across the p shell to reach the N element. The d orbitals are included after Ca, so that Ca would be [Ar] 4s² and Sc would be [Ar] 4s²3d¹. The order for adding electrons would be to go (from left to right) the shells: s -> p -> d -> f.

This is because generally, f requires more energy to fill than d and so on. The order for removing electrons would be to go in the same order as filling electrons because it would be easier to remove electrons from the s orbital than the p orbital.

For electron configuration of Au (uncharged):

The highest noble gas that is in the period before Au would be Xe, so to simplify the electron configuration, we can start on Xe and build off from its electron configuration, since [Xe] = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6. So moving down the period, Au would be [Xe] 6s2 4f14 5d9 if we went down the period since we would need to add 2 electrons from the s orbital (with the energy level prefix 6), 14 electrons from the f orbital (with the energy level prefix 4), and 9 electrons from the d orbital (with the energy level prefix 5) to get to Au. However, Au is a special case element that has a full d electron configuration, so Au would actually have the configuration:

[Xe] 6s¹ 4f¹⁴ 5d¹⁰

by taking one s orbital electron and putting it in the d orbital.

For Au+ (+1 charge):

This element has a charge of +1, so we would need the electron configuration of Au without any charge and remove 1 electron from it to have this positive charge. By taking what we already done in question one, Au = [Xe] 6s¹ 4f¹⁴ 5d¹⁰, we can then remove 1 electron by following the order s->p->d->f. We need to remove 1 electron and we must remove all the s orbital electrons first before the d orbital electrons, so Au+ would be:

[Xe] 4f¹⁴ 5d¹⁰

To determine the color of this complex, we must look at the d electrons. Au+ has a d orbital that is completely filled with electrons, so it would absorb all the possible wavelengths. It wouldn't absorb light because there are no free d orbitals for d electrons to promote to.

We would not expect the salts of the gold ion, Au+, to be colored.

Q12.4.12

Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 × 10⁴ g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 × 10⁻⁶ g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.

S12.4.12

The reaction is first order.

k = 1.0 × 10⁷ mol⁻¹ min⁻¹

To determine the order of the reaction, we must put put each molarity of penicilin ([P]) with its corresponding rate in the equation to find the value of n:

rate=k[P]ⁿ

rate₁ = 1.0 x 10^-10= k(2.0x10^-6)ⁿ

rate₂ = 1.5 x 10^-10= k(3.0x10^-6)ⁿ

rate₃ = 2.0 x 10^-10= k(4.0x10^-6)ⁿ

\(\displaystyle \frac{rate_1}{rate_2}= \frac{1.0 \cdot 10^{_{-10}}}{1.5 \cdot 10^{_{-10}}}=\frac{(2.0 \cdot 10^{_{-6}})^{_{n}}}{(3.0 \cdot 10^{_{-6}})^{_{n}}}\)

\(\displaystyle \frac{2}{3}= \frac{(2.0 \cdot 10^{_{-6}})^{_{n}}}{(3.0 \cdot 10^{_{-6}})^{_{n}}}\)

\(\displaystyle 2/3 = (2/3)ⁿ\)

\(\displaystyle n=1\)

The order of the reaction with respect to penicillin is 1.

So plugging n into rate₃, we get:

\(\displaystyle 2.0 x 10^-10= k(4.0x10^-6)¹\)

\(\displaystyle k=5.0x10^-5\)

Since this is a first order reaction, k's units are min⁻¹. k=5.0x10^-5 min⁻¹

Q21.2.7

What are the two principal differences between nuclear reactions and ordinary chemical changes?

S21.2.7

Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Nuclear reactions involve much larger energies than chemical reactions and have measurable mass changes.

Chemical reactions:

• atoms are rearranged by the breaking and forming of chemical bonds
• Only electrons in atomic or molecular orbitals are involved in the breaking and forming of bonds
• Reactions are accompanied by absorption or release of relatively small amounts of energy
• Rates of reaction are influenced by temperature, pressure, concentration, and catalysts

Nuclear reactions:

• Elements (or isotopes of the same elements) are converted from one to another
• protons, neutrons, electrons, and other elementary particles may be involved
• Reactions are accompanied by absorption or released of tremendous amounts of energy
• Rates of reaction normally are not affected by temperature, pressure, and catalysts

Nuclear reactions are different from ordinary chemical changes because they involve changes in the nucleus. They require larger energies than chemical changes and affect mass. Nuclear reactions occur when the nuclei emit particles/rays.

Q21.6.2

Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?

When 75% of the dose has decayed, it means that 25% is left over, so we can use the initial amount ([A]_o) of 1 and final amount ([A]) of 0.25. The half life t_1/2 of this reaction is 6.01 hours. We need to find k (the rate constant).

\(\displaystyle t_(1/2)=(ln2)/k\)

\(\displaystyle 6.01 = (ln2)/k\)

\(\displaystyle 6.01(k) = ln2\)

\(\displaystyle k = (ln2)/6.01 = 0.115/hr\)

Since k = 0.115/hr, [A]_o = 1, and [A] = 0.25, then we can use [A]=[A]_o(e^-kt) to find t.

\(\displaystyle [A]=[A]_o(e^{_{-kt}})\)

\(\displaystyle 0.25=1(e^{_{-0.115t}})\)

\(\displaystyle ln(0.25)=-0.115t\)

\(\displaystyle t = ln(0.25)/-0.115 = 12 hrs\)

The patient is allowed to leave after 12hours.

Q20.4.8

Identify the oxidants and the reductants in each redox reaction.

1. \[Br_2(l)+2I^{_{-}}(aq)\rightarrow 2Br^{_{-}}(aq)+I_2(s)\]
2. \[Cu^{_{2+}}(aq)+2Ag(s)\rightarrow Cu(s)+2Ag^{_{+}}(aq)\]
3. \[H^{_{+}}(aq)+2MnO_4^{_{-}}(aq)+5H_2SO_3(aq)\rightarrow 2Mn^{_{2+}}(aq)+3H_2O(l)+5HSO_4^{_{-}}(aq)\]
4. \[IO_3^{_{-}}(aq)+5I^{_{-}}(aq)+6H^{_{+}}(aq)\rightarrow 3I_2(s)+3H_2O(l)\]

In a redox reaction, the oxidant is the substance being reduced and the reductant is the substance being oxidized. The substance being reduced is the one gaining an electron in the reaction and the substance being oxidized is the one losing an electron in the reaction.

To determine the reduction and oxidation reactions:

1) separate the reaction based on the elements involved

2) balance the elements in the equation

3) add water (H2O) to balance out the oxygens (O) and balance out the hydrogens in the reaction (H+) with H+

4) balance the charge of the reaction by counting up the positive/negative charges and using electrons to balance the equation

1. \[Br_2(l)+2I^{_{-}}(aq)\rightarrow 2Br^{_{-}}(aq)+I_2(s)\]

1) Br₂(l) → 2Br⁻(aq) and 2I⁻(aq) → I₂(s)

For Br₂(l) → 2Br⁻(aq),

2) Br₂(l) → 2Br⁻(aq) (the elements are already balanced)

3) Br₂(l) → 2Br⁻(aq) (no oxygens or hydrogens involved in the reaction)

4) 2e^- + Br₂(l) → 2Br⁻(aq) (balance charges w/ e^-)

This is a reduction reaction since electrons are gained.

For 2I⁻(aq) → I₂(s),

2) 2I⁻(aq) → I₂(s) (the elements are already balanced)

3) 2I⁻(aq) → I₂(s) (no oxygens or hydrogens involved in the reaction)

4) 2I⁻(aq) → I₂(s) + 2e^- (balance charges w/ e^-)

This is an oxidation reaction since electrons are lost.

Since Br₂(l) is being reduced, it is the oxidant. Since I⁻(aq) is being oxidized, it is the reductant.

2. \[Cu^{_{2+}}(aq)+2Ag(s)\rightarrow Cu(s)+2Ag^{_{+}}(aq)\]

1) Cu²⁺(aq) → Cu(s) and 2Ag(s) → 2Ag⁺(aq) or Ag(s) → Ag⁺(aq)

For Cu²⁺(aq) → Cu(s),

2) Cu²⁺(aq) → Cu(s) (the elements are already balanced)

3) Cu²⁺(aq) → Cu(s) (no oxygens or hydrogens involved in the reaction)

4) 2e^- + Cu²⁺(aq) → Cu(s) (balance charges w/ e^-)

This is a reduction reaction since electrons are gained.

For Ag(s) → Ag⁺(aq)

2) Ag(s) → Ag⁺(aq) (the elements are already balanced)

3) Ag(s) → Ag⁺(aq) (no oxygens or hydrogens involved in the reaction)

4) Ag(s) → Ag⁺(aq) + e^-(balance charges w/ e^-)

This is an oxidation reaction since electrons are lost.

Since Cu²⁺(aq) is being reduced, it is the oxidant. Since Ag(s) is being oxidized, it is the reductant.

3. \[H^{_{+}}(aq)+2MnO_4^{_{-}}(aq)+5H_2SO_3(aq)\rightarrow 2Mn^{_{2+}}(aq)+3H_2O(l)+5HSO_4^{_{-}}(aq)\]

1) 5H₂SO₃(aq) → 5HSO₄⁻(aq) or H₂SO₃(aq) → HSO₄⁻(aq) and

2MnO₄⁻(aq) → 2Mn²⁺(aq) or MnO₄⁻(aq) → Mn²⁺(aq)

For H₂SO₃(aq) → HSO₄⁻(aq)

2) H₂SO₃(aq) → HSO₄⁻(aq) (the element S is already balanced)

3) H₂O(l) + H₂SO₃(aq) → HSO₄⁻(aq) (balance oxygens or hydrogens involved in the reaction; first w/ H2O, then w/ H+)

H₂O(l) + H₂SO₃(aq) → HSO₄⁻(aq) +3H⁺

4) H₂O(l) + H₂SO₃(aq) → HSO₄⁻(aq) +3H⁺ +2e^- (balance charges w/ e^-)

This is an oxidation reaction since electrons are lost.

For MnO₄⁻(aq) → Mn²⁺(aq)

2) MnO₄⁻(aq) → Mn²⁺(aq) (the element Mn is already balanced)

3) MnO₄⁻(aq) → Mn²⁺(aq) +4H₂O(l) (balance oxygens or hydrogens involved in the reaction; first w/ H2O, then w/ H+)

8H⁺ +MnO₄⁻(aq) → Mn²⁺(aq) +4H₂O(l)

4) 8H⁺ +MnO₄⁻(aq) + 5e^- → Mn²⁺(aq) +4H₂O(l) (balance charges w/ e^-)

This is a reduction reaction since electrons are gained.

Since MnO₄⁻(aq) is being reduced, it is the oxidant. Since H₂SO₃(aq) is being oxidized, it is the reductant.

4. \[IO_3^{_{-}}(aq)+5I^{_{-}}(aq)+6H^{_{+}}(aq)\rightarrow 3I_2(s)+3H_2O(l)\]

1) IO₃⁻(aq) → 3I₂(s) and 5I⁻(aq) → 3I₂(s)

For IO₃⁻(aq) → 3I₂(s)

2) 6IO₃⁻(aq) → 3I₂(s) (balance element I)

3) 6IO₃⁻(aq) → 3I₂(s) + 18H₂O(l) (balance oxygens or hydrogens involved in the reaction; first w/ H2O, then w/ H+)

36H⁺ + 6IO₃⁻(aq) → 3I₂(s) +18H₂O(l)

4) 30e^- +36H⁺ + 6IO₃⁻(aq) → 3I₂(s) +18H₂O(l) (balance charges w/ e^-)

This is a reduction reaction since electrons are gained.

For 5I⁻(aq) → 3I₂(s)

2) 30I⁻(aq) → 15I₂(s) (balance element I)

3) 30I⁻(aq) → 15I₂(s) (no oxygens or hydrogens involved in the reaction)

4) 30I⁻(aq) → 15I₂(s) + 30e^- (balance charges w/ e^-)

6I⁻(aq) → 3I₂(s) + 10e^-

This is an oxidation reaction since electrons are lost.

Since IO₃⁻(aq) is being reduced, it is the oxidant. Since I⁻(aq) is being oxidized, it is the reductant.

Q20.7.5

This reaction is characteristic of a lead storage battery:

\[Pb(s)+PbO_2(s)+2H_2SO_4(aq)\rightarrow 2PbSO_4(s)+2H_2O(l)\]

If you have a battery with an electrolyte that has a density of 1.15 g/cm³ and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?

S20.7.5

1. [H₂SO₄] = 3.52 M; E > E°

To find the standard cell potential, we must separate the reaction into two reactions where one is a reduction reaction and the other is a oxidation reaction by separating the reactions based off of the elements involved. To determine the reduction and oxidation reactions:

1) balance the elements in the equation

2) add water (H2O) to balance out the oxygens (O) and balance out the hydrogens in the reaction (H+) with H+

3) balance the charge of the reaction by counting up the positive/negative charges and using electrons to balance the equation

From left to right (forward reaction), oxidation is the loss of an electron and reduction is the gain of an electron.

Oxidation occurs at the anode and reduction occurs at the cathode. We can then determine the cell potential of these individual reactions since

E°cell = E°cathode - E°anode

where E°cathode and E°anode is determined from the standard reduction potential table.

\[Pb(s)+PbO_2(s)+2H_2SO_4(aq)\rightarrow 2PbSO_4(s)+2H_2O(l)\] can be split into:

PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) and Pb(s) + H₂SO₄(aq) → PbSO₄(s)

Balancing the equations (following the steps listed above):

PbO₂(s) + H₂SO₄(aq) → PbSO₄(s)

1) PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) (balance Pb)

2) PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) +2H₂O(l) (balance oxygens or hydrogens involved in the reaction; first w/ H2O, then w/ H+)

2H⁺ + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) +2H₂O(l)

3) 2e^- +2H⁺ + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) +2H₂O(l) (balance charges w/ e^-)

This is a reduction reaction since electrons are gained.

Pb(s) + H₂SO₄(aq) → PbSO₄(s)

1) Pb(s) + H₂SO₄(aq) → PbSO₄(s) (Pb is already balanced)

2) Pb(s) + H₂SO₄(aq) → PbSO₄(s) +2H⁺ (balance oxygens or hydrogens involved in the reaction; first w/ H2O, then w/ H+)

3) Pb(s) + H₂SO₄(aq) → PbSO₄(s) +2H⁺ + 2e^- (balance charges w/ e^-)

This is an oxidation reaction since electrons are lost.

Oxidation occurs at the anode and reduction occurs at the cathode.

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

From the standard reduction potential table E°cathode = +1.685 V, E°anode = -0.356 V

E°cell = 1.685 - (-0.356) =2.041V.

To balance the equations and get Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l):

2e^- +2H⁺ + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) +2H₂O(l)

Pb(s) + H₂SO₄(aq) → PbSO₄(s) +2H⁺ + 2e^-

___________________________________________________

Pb(s) + H₂SO₄(aq) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) +2H₂O(l) + PbSO₄(s)

Pb(s) + 2H₂SO₄(aq) + PbO₂(s) → 2PbSO₄(s) +2H₂O(l)

This means that the reduction and oxidation reactions were balanced properly and that there were 2 electrons transferred.

n=2

If the density of 1.15 g/cm³, then:

\(\displaystyle \frac{1 L soln}{1}\cdot \frac{1.15g}{1cm^{3}}\cdot \frac{1cm^{3}}{1mL}\cdot \frac{1000mL}{1L}=1150g\)

The molar mass of H₂SO₄ is 98.079 g/mol, so (\displaystyle (1150g)\cdot\frac{1 mol}{98.079 g}= 1(1.7 mol/L) H_2SO_4\)

The molarity of H₂SO₄ is 11.7 mol/L and since only the battery only contains 30.0% sulfuric acid by mass, then the molarity of H₂SO₄ is \(\displaystyle (11.7M)(.3)= 3.52M H_2SO_4\):

[H₂SO₄]=3.52 M

To find E, we have to use \(\displaystyle E=E^{0}-\frac{RT}{nF}lnQ\)

The overall reaction is: \[Pb(s)+PbO_2(s)+2H_2SO_4(aq)\rightarrow 2PbSO_4(s)+2H_2O(l)\] so Q would be equal to (1/[H₂SO₄]²) since the other compounds are in the solid or liquid state.

\(\displaystyle Q=\frac{prod}{reac}\)

\(\displaystyle Q=\frac{1}{3.52M}^{2}=0.0807\)

So for \(\displaystyle E=E^{0}-\frac{RT}{nF}lnQ\),

\(\displaystyle E=E^{0}-\frac{RT}{nF}lnQ\)

\(\displaystyle E = 2.041 - (0.0591/2)(log(0.0807))\)

\(\displaystyle E=2.07 V\)

Since E°= 0.0591V and E = 2.07 V, E > E°.

The potential is greater than that of the standard cell.