Chapter 19

Electrochemistry

19.1

Strategy:

We follow the stepwise procedure for balancing redox reactions presented in Section 19.1 of the text.

Solution:

a.	In Step 1, we separate the half-reactions. oxidation: Fe^2+ ® Fe^3+ reduction: H2O2 ® H2O Step 2 is unnecessary because the half-reactions are already balanced with respect to iron. Step 3: We balance each half-reaction for O by adding H2O. The oxidation half-reaction is already balanced in this regard (it contains no O atoms). The reduction half-reaction requires the addition of one H2O on the product side. H2O2 ® H2O + H2O or simply H2O2 ® 2H2O Step 4: We balance each half reaction for H by adding H^+. Again, the oxidation half-reaction is already balanced in this regard. The reduction half-reaction requires the addition of two H^+ ions on the reactant side. H2O2 + 2H^+ ® 2H2O Step 5: We balance both half-reactions for charge by adding electrons. In the oxidation half-reaction, there is a total charge of +2 on the left and a total charge of +3 on the right. Adding one electron to the product side makes the total charge on each side +2. Fe^2+ ® Fe^3+ + e^- In the reduction half-reaction, there is a total charge of +2 on the left and a total charge of 0 on the right. Adding two electrons to the reactant side makes the total charge on each side 0. H2O2 + 2H^+ + 2e^- ® 2H2O Step 6: Because the number of electrons is not the same in both half-reactions, we multiply the oxidation half-reaction by 2. 2´(Fe^2+ ® Fe^3+ + e^-) = 2Fe^2+ ® 2Fe^3+ + 2e^- Step 7: Finally, we add the resulting half-reactions, cancelling electrons and any other identical species to get the overall balanced equation.
b.	In Step 1, we separate the half-reactions. oxidation: Cu ® Cu^2+ reduction: HNO3 ® NO Step 2 is unnecessary because the half-reactions are already balanced with respect to copper and nitrogen. Step 3: We balance each half-reaction for O by adding H2O. The oxidation half-reaction is already balanced in this regard (it contains no O atoms). The reduction half-reaction requires the addition of two H2O molecules on the product side. HNO3 ® NO + 2H2O Step 4: We balance each half reaction for H by adding H^+. Again, the oxidation half-reaction is already balanced in this regard. The reduction half-reaction requires the addition of three H^+ ions on the reactant side. 3H^+ + HNO3 ® NO + 2H2O Step 5: We balance both half-reactions for charge by adding electrons. In the oxidation half-reaction, there is a total charge of 0 on the left and a total charge of +2 on the right. Adding two electrons to the product side makes the total charge on each side 0. Cu ® Cu^2+ + 2e^- In the reduction half-reaction, there is a total charge of +3 on the left and a total charge of 0 on the right. Adding three electrons to the reactant side makes the total charge on each side 0. 3H^+ + HNO3 + 3e^- ® NO + 2H2O Step 6: Because the number of electrons is not the same in both half-reactions, we multiply the oxidation half-reaction by 3, and the reduction half-reaction by 2. 3´(Cu ® Cu^2+ + 2e^-) = 3Cu ® 3Cu^2+ + 6e^- 2´(3H^+ + HNO3 + 3e^- ® NO + 2H2O) = 6H^+ + 2HNO3 + 6e^- ® 2NO + 4H2O Step 7: Finally, we add the resulting half-reactions, cancelling electrons and any other identical species to get the overall balanced equation.
c.	In Step 1, we separate the half-reactions. oxidation: CN^− ® CNO^− reduction: MnO ® MnO2 Step 2 is unnecessary because the half-reactions are already balanced with respect to carbon, nitrogen, and manganese. Step 3: We balance each half-reaction for O by adding H2O. The oxidation half-reaction requires the addition of one H2O molecule to the reactant side. The reduction half-reaction requires the addition of two H2O molecules on the product side. H2O + CN^- ® CNO^- MnO ® MnO2 + 2H2O Step 4: We balance each half reaction for H by adding H^+. The oxidation half-reaction requires the addition of 2 H^+ ions to the product side. The reduction half-reaction requires the addition of four H^+ ions on the reactant side. H2O + CN^- ® CNO^− + 2H^+ 4H^++ MnO ® MnO2 + 2H2O Step 5: We balance both half-reactions for charge by adding electrons. In the oxidation half-reaction, there is a total charge of -1 on the left and a total charge of +1 on the right. Adding two electrons to the product side makes the total charge on each side -1. H2O + CN^- ® CNO^− + 2H^+ + 2e^- In the reduction half-reaction, there is a total charge of +3 on the left and a total charge of 0 on the right. Adding three electrons to the reactant side makes the total charge on each side 0. 3e^- + 4H^+ + MnO® MnO2 + 2H2O Step 6: Because the number of electrons is not the same in both half-reactions, we multiply the oxidation half-reaction by 3, and the reduction half-reaction by 2. 3´( H2O + CN^- ® CNO^− + 2H^+ + 2e^-) = 3H2O + 3CN^- ® 3CNO^− + 6H^+ + 6e^- 2´( 3e^- + 4H^+ + MnO ® MnO2 + 2H2O) = 6e^- + 8H^+ 2MnO ® 2MnO2 + 4H2O Step 7: Finally, we add the resulting half-reactions, cancelling electrons and any other identical species to get the overall balanced equation. Balancing a redox reaction in basic solution requires two additional steps. Step 8: For each H^+ ion in the final equation, we add one OH^- ion to each side of the equation, combining the H^+ and OH^- ions to produce H2O. 3CN^- + 2MnO + 2H^+ ® 3CNO^- + 2MnO2 + H2O + 2OH^- + 2OH^- 2H2O + 2MnO + 3CN^-® 2MnO2 + 3CNO^- + H2O + 2OH^- Step 9: Lastly, we cancel H2O molecules that result from Step 8. 3CN^- + 2MnO + H2O ® 3CNO^- + 2MnO2 + 2OH^-
d.	Parts (d) and (e) are solved using the methods outlined in (a) through (c). 6OH^- + 3Br2 ® BrO + 3H2O + 5Br^-
e.	2S2O + I2 ® S4O+ 2I^-

19.2

a.	2OH^- + Mn^2+ + H2O2 → MnO2 + 2H2O
b.	2Bi(OH)3 + 3SnO→ 2Bi + 3SnO+ 3H2O
c.	14H^+ + 3C2O + Cr2O→ 6CO2 + 2Cr^3+ + 7H2O
d.	4H^+ + 2Cl^- + 2ClO→ Cl2 + 2ClO2 + 2H2O
e.	14H^+ + 2Mn^2+ + 5BiO → 2MnO + 5Bi^3+ + 7H2O

19.10

Half-reaction E°(V)

Mg²⁺(aq) + 2e^- ® Mg(s) -2.37

Cu²⁺(aq) + 2e^- ® Cu(s) +0.34

The overall equation is: Mg(s) + Cu²⁺(aq) ® Mg²⁺(aq) + Cu(s)

E° = 0.34 V - (-2.37 V) = 2.71 V

19.11

Strategy:	At first, it may not be clear how to assign the electrodes in the galvanic cell. From Table 19.1 of the text, we write the standard reduction potentials of Al and Ag and compare their values to determine which is the anode half-reaction and which is the cathode half-reaction.
Solution:	The standard reduction potentials are: Ag^+(1.0 M) + e^- ® Ag(s) E° = 0.80 V Al^3+(1.0 M) + 3e^- ® Al(s) E° = -1.66 V The silver half-reaction, with the more positive E° value, will occur at the cathode (as a reduction). The aluminum half-reaction will occur at the anode (as an oxidation). In order to balance the numbers of electrons in the two half-reactions, we multiply the reduction by 3 before summing the two half-reactions. Note that multiplying a half-reaction by 3 does not change its E° value because reduction potential is an intensive property. Al(s) ® Al^3+(1.0 M) + 3e^- 3Ag^+(1.0 M) + 3e^- ® 3Ag(s) Overall: Al(s) + 3Ag^+(1.0 M) ® Al^3+(1.0 M) + 3Ag(s) We find the emf of the cell using Equation 19.1. E = 0.80 V - (-1.66 V) = 2.46 V
Think About It:	The positive value of E° shows that the forward reaction is favored.

19.12

The appropriate half-reactions from Table 19.1 are

I₂(s) + 2e^- ® 2I^-(aq)

Fe³⁺(aq) + e^- ® Fe²⁺(aq)

Thus, iron(III) should oxidize iodide ion to iodine. This makes the iodide ion/iodine half-reaction the anode. The standard emf can be found using Equation 19.1.

(The emf was not required in this problem, but the fact that it is positive confirms that the reaction should favor products at equilibrium.)

19.13

The half-reaction for oxidation is:

2H₂O(l) O₂(g) + 4H⁺(aq) + 4e^-

The species that can oxidize water to molecular oxygen must have an more positive than +1.23 V. From Table 19.1 of the text we see that only Cl₂(g) and MnO(aq) in acid solution can oxidize water to oxygen.

19.14

Another way to word the problem is: “Is the reaction of NO and Mn²⁺ to produce NO and MnO spontaneous?” The overall reaction described in the problem is:

5NO(aq) + 3Mn²⁺(aq) + 2H₂O(l) ® 5NO(g) + 3MnO(aq) + 4H⁺(aq)

Using E° values from Table 19.1 and Equation 19.1, we calculate the cell emf to determine whether or not the reaction is spontaneous.

The negative emf indicates that reactants are favored at equilibrium. is positive for a spontaneous reaction. Therefore, NO will not oxidize Mn²⁺ to MnO under standard-state conditions.

19.15

Strategy:

In each case, we can calculate the standard cell emf from the potentials for the two half-reactions.

Solution:

Using E° values from Table 19.1:

a.	E° = -0.40 V - (-2.87 V) = 2.47 V. The reaction is spontaneous.
b.	E° = -0.14 V - 1.07 V = -1.21 V. The reaction is not spontaneous.
c.	E° = -0.25 V - 0.80 V = -1.05 V. The reaction is not spontaneous.
d.	E° = 0.77 V - 0.15 V = 0.62 V. The reaction is spontaneous.

19.16

From Table 19.1 of the text, we compare the standard reduction potentials for the half-reactions. The more positive the potential, the better the substance is as an oxidizing agent.

a.	Au^3+
b.	Ag^+
c.	Cd^2+
d.	O2 in acidic solution.

19.17

Strategy:

The greater the tendency for the substance to be oxidized, the stronger its tendency to act as a reducing agent. The species that has a stronger tendency to be oxidized will have a smaller reduction potential.

Solution:

In each pair, look for the one with the smaller reduction potential. This indicates a greater tendency for the substance to be oxidized.

a.	From Table 19.1, we have the following reduction potentials: Na: -2.71 V Li: -3.05 V Li has the smaller (more negative) reduction potential and is therefore more easily oxidized. The more easily oxidized substance is the better reducing agent. Li is the better reducing agent. Following the same logic for parts (b) through (d) gives
b.	H2
c.	Fe^2+
d.	Br^-

19.20

Strategy:	The Faraday constant is the electric charge contained in 1 mole of electrons. To determine the charge per mole of electrons, multiply the charge on the electron by Avogadro’s number.
Setup:	According to Table 2.1 the charge on an electron is 1.6022 × 10^−19 C.
Solution:	F = = 9.648 × 10^4 F.

19.21

Strategy:	The relationship between the equilibrium constant, K, and the standard emf is given by Equation 19.5 of the text: . Thus, knowing and n (the moles of electrons transferred) we can the determine equilibrium constant. We find the standard reduction potentials in Table 19.1 of the text.
Solution:	We must rearrange Equation 19.5 to solve for K: We see in the reaction that Mg goes to Mg^2+ and Zn^2+ goes to Zn. Therefore, two moles of electrons are transferred during the redox reaction; i.e., n = 2. K3 ´ 10^54

19.22

We use Equation 19.5:

Substitute the equilibrium constant and the moles of e^- transferred (n = 2) into the above equation to calculate E°.

E°0.368 V

19.23

In each case we use standard reduction potentials from Table 19.1 together with Equation 19.5 of the text.

a.	We break the equation into two half-reactions: Br2(l) + 2e^- 2Br^-(aq) 2I^-(aq) I2(s) + 2e^- The standard emf is Next, we can calculate K using Equation 19.5 of the text. K2 ´ 10^18
b.	We break the equation into two half-reactions: 2Ce^4+(aq) + 2e^- 2Ce^3+(aq) E= 1.61 V 2Cl^-(aq) Cl2(g) + 2e^-E= 1.36 V The standard emf is K3 ´ 10^8
c.	We break the equation into two half-reactions: MnO(aq) + 8H^+(aq) + 5e^- Mn^2+(aq) + 4H2O(l) E= 1.51 V 5Fe^2+(aq) 5Fe^3+(aq)+ 5e^- E= 0.77 V The standard emf is K3 ´ 10^62

19.24

a.	We break the equation into two half-reactions: Mg(s) Mg^2+(aq) + 2e^- Pb^2+(aq) + 2e^- Pb(s) The standard emf is given by We can calculate DG° from the standard emf using Equation 19.3. Next, we can calculate K using Equation 19.5 of the text. or and K5 ´ 10^75
b.	Parts (b) and (c) are worked in an analogous manner to part (a). DG° = -(4)(96500 J/V×mol)(0.46 V) = -178 kJ/mol K1 ´ 10^31
c.	DG° = -(6)(96500 J/V×mol)(2.19 V) = -1.27 ´ 10^3 kJ/mol K9 ´ 10^221

19.25

Strategy:	The spontaneous reaction that occurs must include one reduction and one oxidation. We examine the reduction potentials of the species present to determine which half reaction will occur as the reduction and which will occur as the oxidation. The relationship between the standard free energy change and the standard emf of the cell is given by Equation 19.3 of the text: . The relationship between the equilibrium constant, K, and the standard emf is given by Equation 19.5 of the text: . Thus, once we determine , we can calculate DG° and K.
Solution:	The half-reactions (both written as reductions) are: Fe^3+(aq) + e^- ® Fe^2+(aq) Ce^4+(aq) + e^- ® Ce^3+(aq) Because it has the larger reduction potential, Ce^4+ is the more easily reduced and will oxidize Fe^2+ to Fe^3+. This makes the Fe^2+/Fe^3+ half-reaction the anode. The spontaneous reaction is Ce^4+(aq) + Fe^2+(aq) ® Ce^3+(aq) + Fe^3+(aq) The standard cell emf is found using Equation 19.1 of the text. The values of DG° and Kc are found using Equations 19.3 and 19.5 of the text. K= 2 ´ 10^14
Think About It:	The negative value of DG° and the large positive value of K, both indicate that the reaction favors products at equilibrium. The result is consistent with the fact that E° for the galvanic cell is positive.

19.26

We can determine the of a hypothetical galvanic cell made up of two couples (Cu²⁺/Cu⁺ and Cu⁺/Cu) from the standard reduction potentials in Table 19.1 of the text.

The half-cell reactions are:

Anode (oxidation): Cu⁺(1.0 M) ® Cu²⁺(1.0 M) + e^-

Cathode (reduction): Cu⁺(1.0 M) + e^- ® Cu(s)

Overall: 2Cu⁺(1.0 M) ® Cu²⁺(1.0 M) + Cu(s)

Now, we use Equation 19.3 of the text. The overall reaction shows that n = 1.

DG° = -(1)(96500 J/V×mol)(0.37 V) = -36 kJ/mol

Next, we can calculate K using Equation 19.5 of the text.

or

and

K= 2 ´ 10⁶

19.27

Strategy:

According to Section 19.4, a simplified unbalanced equation for oxidation of tin from an amalgam filling is: Sn(s) + O2(g) ® Sn^2^+(aq) + H2O(l) Apply steps 1 through 7 for balancing redox equations to balance for mass and charge. Then use Equation 19.1 to calculate the standard cell potential for the reaction.

Solution:

Step 1. Separate the unbalanced reaction into half-reactions. oxidation: Sn ® Sn^2+ reduction: O2 ® H2O Step 2. This step is unnecessary because the half-reactions are balanced for all elements, excluding O and H. Step 3. Balance both half-reactions for O by adding H2O. Sn ® Sn^2+ O2 ® 2 H2O Step 4. Balance both half reactions for H by adding H^+. Sn ® Sn^2+ 4 H^+ + O2 ® 2 H2O Step 5. Balance the total charge of both half-reactions by adding electrons. Sn ® Sn^2+ + 2e^− 4e^− + 4 H^+ + O2 ® 2 H2O Step 6. Multiply the half-reactions to make the numbers of electrons the same in both. 2(Sn ® Sn^2+ + 2e^−) 4e^− + 4 H^+ + O2 ® 2 H2O Step 7. Add the half-reactions back together, cancelling electrons. 2Sn ® 2Sn^2+ + 4e^− 4e^− + 4 H^+ + O2 ® 2 H2O 2Sn +4 H^+ + O2 ® 2Sn^2+ + 2 H2O The standard cell potential is

19.28

Strategy:	Use Equation 19.3 to calculate the standard free-energy change. Calculate the equilibrium constant using Equation 19.5 (rearranging to solve for K).
Setup:	According to problem 19.27, E° = 1.37 V and n = 4e^−.
Solution:	Solving for the standard free-energy gives: DG° = -nFE° = −(4e^−)(96,500 J/V ∙ mol e^−)(1.37 V) DG° = −5.3 × 10^5 J/mol DG° = −5.3 × 10^2 kJ/mol Solve Equation 19.5 for the equilibrium constant:

19.30

Strategy:

The Nernst equation is given in Equation 19.6:

Setup:

Use E° values from Table 19.1 to determine E° for the reaction.

Solution:

a.	From Table 19.1, Cathode (reduction): Sn^2+(aq) + 2e^− → Sn(s) Anode (oxidation): Mg(s) → Mg^2+(aq) + 2e^− The Nernst equation for this reaction is
b.	From Table 19.1, Cathode (reduction): Pb^2+(aq) + 2e^− → Pb(s) Anode (oxidation): Cr(s) → Cr^3+(aq) + 3e^− The Nernst equation for this reaction is

19.31

Strategy:

The standard emf (E°) can be calculated using the standard reduction potentials in Table 19.1 of the text. Because the reactions are not run under standard-state conditions (concentrations are not 1 M), we need the Nernst equation (Equation 19.7) of the text to calculate the emf (E) of a hypothetical galvanic cell. Remember that solids do not appear in the reaction quotient (Q) term in the Nernst equation. We can calculate DG from E using Equation 19.2 of the text: DG = -nFEcell.

Solution:

The half-cell reactions are: . E 1.09 V

19.32

If this were a standard cell, the concentrations would all be 1.00 M, and the voltage would just be the standard emf calculated from Table 19.1 of the text. Since cell emf's depend on the concentrations of the reactants and products, we must use the Nernst equation (Equation 19.7 of the text) to find the emf of a nonstandard cell.

a.

The half-reactions are: Anode (oxidation): Mg(s) ® Mg^2+(1.0 M) + 2e^- Cathode (reduction): Sn^2+(1.0 M) + 2e^- ® Sn(s) Overall: Mg(s) + Sn^2+(1.0 M) ® Mg^2+(1.0 M) + Sn(s) From Equation 19.7 of the text, we write: E2.23 V We can now find the free energy change at the given concentrations using Equation 19.2 of the text. Note that in this reaction, n = 2. DG = -nFEcell DG = -(2)(96500 J/V×mol)(2.23 V) = -430 kJ/mol

b.

The half-cell reactions are: Anode (oxidation): 3[Zn(s) ® Zn^2+(1.0 M) + 2e^-] Cathode (reduction): 2[Cr^3+(1.0 M) + 3e^- ® Cr(s)] Overall: 3Zn(s) + 2Cr^3+(1.0 M) ® 3Zn^2+(1.0 M) + 2Cr(s) From Equation 19.7 of the text, we write: E0.04 V We can now find the free energy change at the given concentrations using Equation 19.2 of the text. Note that in this reaction, n = 6. DG = -nFEcell DG = -(6)(96500 J/V×mol)(0.04 V) = -23 kJ/mol

19.33

The overall reaction is: Zn(s) + 2H⁺(aq) ® Zn²⁺(aq) + H₂(g)

E0.78 V

19.34

We write the two half-reactions to calculate the standard cell emf. (Oxidation occurs at the Pb electrode.)

Pb(s) Pb²⁺(aq) + 2e^-

2H⁺(aq) + 2e^- H₂(g)

2H⁺(aq) + Pb(s) H₂(g) + Pb²⁺(aq)

Using the Nernst equation, we can calculate the cell emf, E.

E0.083 V

19.35

As written, the reaction is not spontaneous under standard state conditions; the cell emf is negative.

The reaction will become spontaneous when the concentrations of zinc(II) and copper(II) ions are such as to make the emf positive. The turning point is when the emf is zero. We solve the Nernst equation (Equation 19.6) for the [Cu²⁺]/[Zn²⁺] ratio at this point.

At 25°C:

6.0 ´ 10^-38

In other words for the reaction to be spontaneous, the [Cu²⁺]/[Zn²⁺] ratio must be less than 6.0 ´ 10^-38.

Think About It:

Is the reduction of zinc(II) by copper metal a practical use of copper?

19.36

All concentration cells have the same standard emf: zero volts.

Mg²⁺(aq) + 2e^- Mg(s)

Mg(s) Mg²⁺(aq) + 2e^-

We use the Nernst equation to compute the emf. There are two moles of electrons transferred from the reducing agent to the oxidizing agent in this reaction, so n = 2.

E0.010 V

19.40

a.	The total charge passing through the circuit is From the anode half-reaction we can find the amount of hydrogen. The volume can be computed using the ideal gas equation
b.	The charge passing through the circuit in one minute is We can find the amount of oxygen from the cathode half-reaction and the ideal gas equation.

19.41

We can calculate the standard free energy change, DG°, from the standard free energies of formation, using Equation 18.12 of the text. Then, we can calculate the standard cell emf, , from DG°.

The overall reaction is:

C₃H₈(g) + 5O₂(g) 3CO₂(g) + 4H₂O(l)

We can now calculate the standard emf using the following equation:

or

Check the half-reactions of the text to determine that 20 moles of electrons are transferred during this redox reaction.

Think About It:

Does this suggest that, in theory, it should be possible to construct a galvanic cell (battery) based on any conceivable spontaneous reaction?

19.44

Strategy:	A faraday is a mole of electrons. Knowing how many moles of electrons are needed to reduce a mole of copper ions, we can determine how many moles of copper will be produced.
Setup:	The half-reaction shows that two moles of e^- are required per mole of Cu. Cu^2+(aq) + 2e^- → Cu(s)
Solution:

19.45

Strategy:	A faraday is a mole of electrons. Knowing how many moles of electrons are needed to reduce a mole of magnesium ions, we can determine how many moles of magnesium will be produced. The half-reaction shows that two moles of e^- are required per mole of Mg. Mg^2+(aq) + 2e^- → Mg(s)
Solution:

19.46

a.

The only ions present in molten BaCl2 are Ba^2+ and Cl^-. The electrode reactions are: anode: 2Cl^-(aq) → Cl2(g) + 2e^- cathode: Ba^2+(aq) + 2e^- → Ba(s) This cathode half-reaction tells us that 2 moles of e^- are required to produce 1 mole of Ba(s).

b.

According to Figure 19.12 of the text, we can carry out the following conversion steps to calculate the quantity of Ba in grams. current ´ time ® coulombs ® mol e^- ® mol Ba ® g Ba First, we calculate the coulombs of electricity that pass through the cell. Then, we will continue on to calculate grams of Ba. We see that for every mole of Ba formed at the cathode, 2 moles of electrons are needed. The grams of Ba produced at the cathode are:

19.47

The half-reactions are: Na⁺ + e^- ® Na

Al³⁺ + 3e^- ® Al

As long as we are comparing equal masses, we can use any mass that is convenient. In this case, we will use 1 g.

It is cheaper to prepare 1 ton of sodium by electrolysis.

19.48

The cost for producing various metals is determined by the moles of electrons needed to produce a given amount of metal. For each reduction, we first calculate the number of tons of metal produced per 1 mole of electrons (1 ton = 9.072 ´ 10⁵ g). The reductions are:

Mg²⁺ + 2e^- → Mg

Al³⁺ + 3e^- → Al

Na⁺ + e^- → Na

Ca²⁺ + 2e^- → Ca

Now that we know the tons of each metal produced per mole of electrons, we can convert from $155/ton Mg to the cost to produce the given amount of each metal.

a.	For aluminum :
b.	For sodium:
c.	For calcium:

19.49

Find the amount of oxygen using the ideal gas equation

Since the half-reaction shows that one mole of oxygen requires four faradays of electric charge, we write

19.50

a.	The half-reaction is: 2H2O(l) → O2(g) + 4H^+(aq) + 4e^- First, we can calculate the number of moles of oxygen produced using the ideal gas equation. Since 4 moles of electrons are needed for every 1 mole of oxygen, we will need 4 F of electrical charge to produce 1 mole of oxygen.
b.	The half-reaction is: 2Cl^-(aq) → Cl2(g) + 2e^- The number of moles of chlorine produced is: Since 2 moles of electrons are needed for every 1 mole of chlorine gas, we will need 2 F of electrical charge to produce 1 mole of chlorine gas.
c.	The half-reaction is: Sn^2+(aq) + 2e^- → Sn(s) The number of moles of Sn(s) produced is Since 2 moles of electrons are needed for every 1 mole of Sn, we will need 2 F of electrical charge to reduce 1 mole of Sn^2+ ions to Sn metal.

19.51

The half-reactions are: Cu²⁺(aq) + 2e^- ® Cu(s)

2Br^-(aq) ® Br₂(l) + 2e^-

The mass of copper produced is:

The mass of bromine produced is:

19.52

a.	The half-reaction is: Ag^+(aq) + e^- → Ag(s)
b.	Since this reaction is taking place in an aqueous solution, the probable oxidation is the oxidation of water. (Neither Ag^+ nor NO3^- can be further oxidized.) 2H2O(l) → O2(g) + 4H^+(aq) + 4e^-
c.	The half-reaction tells us that 1 mole of electrons is needed to reduce 1 mol of Ag^+ to Ag metal. We can set up the following strategy to calculate the quantity of electricity (in C) needed to deposit 0.67 g of Ag. grams Ag ® mol Ag ® mol e^- ® coulombs

19.53

The half-reaction is: Co²⁺ + 2e^- ® Co

The half-reaction tells us that 2 moles of electrons are needed to reduce 1 mol of Co²⁺ to Co metal. We can set up the following strategy to calculate the quantity of electricity (in C) needed to deposit 2.35 g of Co.

19.54

a.	First find the amount of charge needed to produce 2.00 g of silver according to the half-reaction: Ag^+(aq) + e^- → Ag(s) The half-reaction for the reduction of copper(II) is: Cu^2+(aq) + 2e^- → Cu(s) From the amount of charge calculated above, we can calculate the mass of copper deposited in the second cell.
b.	We can calculate the current flowing through the cells using the following strategy. Coulombs ® Coulombs/hour ® Coulombs/second Recall that 1 C = 1 A×s The current flowing through the cells is:

19.55

The half-reaction for the oxidation of chloride ion is:

2Cl^-(aq) ® Cl₂(g) + 2e^-

First, let's calculate the moles of e^- flowing through the cell in one hour.

Next, let's calculate the hourly production rate of chlorine gas (in kg). Note that the anode efficiency is 93.0%.

19.56

Step 1: Balance the half-reaction.

Cr₂O(aq) + 14H⁺(aq) + 12e^- → 2Cr(s) + 7H₂O(l)

Step 2: Calculate the quantity of chromium metal by calculating the volume and converting this to mass using the given density.

Volume Cr = thickness ´ surface area

Converting to cm³,

Next, calculate the mass of Cr.

Mass = density ´ volume

Step 3: Find the number of moles of electrons required to electrodeposit 18 g of Cr from solution. The half-reaction is:

Cr₂O(aq) + 14H⁺(aq) + 12e^- → 2Cr(s) + 7H₂O(l)

Six moles of electrons are required to reduce 1 mol of Cr metal. But, we are electrodepositing less than 1 mole of Cr(s). We need to complete the following conversions:

g Cr ® mol Cr ® mol e^-

Step 4: Determine how long it will take for 2.1 moles of electrons to flow through the cell when the current is 25.0 C/s. We need to complete the following conversions:

mol e^- ® coulombs ® seconds ® hours

Think About It:

Would any time be saved by connecting several bumpers together in a series?

19.57

The quantity of charge passing through the solution is:

Since the charge of the copper ion is +2, the number of moles of copper formed must be:

The units of molar mass are grams per mole. The molar mass of copper is:

63.3 g/mol

19.58

Based on the half-reaction, we know that one faraday will produce half a mole of copper.

Cu²⁺(aq) + 2e^- → Cu(s)

First, we calculate the charge (in C) needed to deposit 0.300 g of Cu.

We know that one faraday will produce half a mole of copper, but we don’t have a half a mole of copper. We have:

We calculated the number of coulombs (912 C) needed to produce 4.72 ´ 10^-3 mol of Cu. How many coulombs will it take to produce 0.500 moles of Cu? This will be Faraday’s constant.

19.59

The number of faradays supplied is:

Since we need three faradays to reduce one mole of X³⁺, the molar mass of X must be:

27.0 g/mol

19.60

First we can calculate the number of moles of hydrogen produced using the ideal gas equation.

The number of faradays passed through the solution is:

19.65

a.	The half-reactions are: H2(g) ® 2H^+(aq) + 2e^- Ni^2+(aq) + 2e^- ® Ni(s) The complete balanced equation is: H2(g) + Ni^2+(aq) ® 2H^+(aq) + Ni(s) Ni(s) is below and to the right of H^+(aq) in Table 19.1 of the text (see the half-reactions at -0.25 and 0.00 V). Therefore, the spontaneous reaction is the reverse of the above reaction; therefore, the reaction will proceed to the left.
b.	The half-reactions are: 5e^- + 8H^+(aq) + MnO(aq) ® Mn^2+(aq) + 4H2O 2Cl^-(aq) ® Cl2(g) + 2e^- The complete balanced equation is: 16H^+(aq) + 2MnO(aq) + 10Cl^-(aq) ® 2Mn^2+(aq) + 8H2O + 5Cl2(g) In Table 19.1 of the text, Cl^-(aq) is below and to the right of MnO4^-(aq); therefore the spontaneous reaction is as written. The reaction will proceed to the right.
c.	The half-reactions are: Cr(s) ® Cr^3+(aq) + 3e^- Zn^2+(aq) + 2e^- ® Zn(s) The complete balanced equation is: 2Cr(s) + 3Zn^2+(aq) ® 2Cr^3+(aq) + 3Zn(s) In Table 19.1 of the text, Zn(s) is below and to the right of Cr^3+(aq); therefore the spontaneous reaction is the reverse of the reaction as written. The reaction will proceed to the left.

19.66

The balanced equation is:

Cr₂O + 6 Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

The remainder of this problem is a solution stoichiometry problem.

The number of moles of potassium dichromate in 26.0 mL of the solution is:

From the balanced equation it can be seen that 1 mole of dichromate is stoichiometrically equivalent to
6 moles of iron(II). The number of moles of iron(II) oxidized is therefore

Finally, the molar concentration of Fe²⁺ is:

19.67

The balanced equation is:

5SO₂(g) + 2MnO(aq) + 2H₂O(l) ® 5SO₄^2-(aq) + 2Mn²⁺(aq) + 4H⁺(aq)

The mass of SO₂ in the water sample is given by

0.00944 g SO₂

19.68

The balanced equation is:

MnO+ 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

First, we calculate the number of moles of potassium permanganate in 23.30 mL of solution.

From the balanced equation it can be seen that 1 mole of permanganate is stoichiometrically equivalent to
5 moles of iron(II). The number of moles of iron(II) oxidized is therefore

The mass of Fe²⁺ oxidized is:

Finally, the mass percent of iron in the ore can be calculated.

19.69

a.	The balanced equation is: 2MnO + 6H^+ + 5H2O2 ® 2Mn^2+ + 8H2O + 5O2
b.	The number of moles of potassium permanganate in 36.44 mL of the solution is From the balanced equation it can be seen that in this particular reaction 2 moles of permanganate is stoichiometrically equivalent to 5 moles of hydrogen peroxide. The number of moles of H2O2 oxidized is therefore The molar concentration of H2O2 is:

19.70

a.	The half-reactions are: (i) MnO(aq) + 8H^+(aq) + 5e^- → Mn^2+(aq) + 4H2O(l) (ii) C2O4^2^-(aq) → 2CO2(g) + 2e^- We combine the half-reactions to cancel electrons, that is, [2 ´ equation (i)] + [5 ´ equation (ii)] 2MnO(aq) + 16H^+(aq) + 5C2O^-(aq) → 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)
b.	We can calculate the moles of KMnO4 from the molarity and volume of solution. We can calculate the mass of oxalic acid from the stoichiometry of the balanced equation. The mole ratio between oxalate ion and permanganate ion is 5:2. Finally, the percent by mass of oxalic acid in the sample is:

19.71

The balanced equation is:

2MnO + 5C₂O + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Therefore, 2 mol MnO reacts with 5 mol C₂O

Recognize that the mole ratio of Ca²⁺ to C₂O is 1:1 in CaC₂O₄. The mass of Ca²⁺ in 10.0 mL is:

Finally, converting to mg/mL, we have:

0.232 mg Ca²⁺/mL blood

19.72

E DG Cell Reaction

> 0 < 0 spontaneous

< 0 > 0 nonspontaneous

= 0 = 0 at equilibrium

19.73

The solubility equilibrium of AgBr is: AgBr(s) Ag⁺(aq) + Br^-(aq)

By reversing the first given half-reaction and adding it to the first, we obtain:

Ag(s) ® Ag⁺(aq) + e^-

AgBr(s) + e^- ® Ag(s) + Br^-(aq)

AgBr(s) Ag⁺(aq) + Br^-(aq)

At equilibrium, we have:

log K_sp = -12.33

K_sp = 10^-12.33 = 5 ´ 10^-13

(Note that this value differs from that given in Table 17.4 of the text, since the data quoted here were obtained from a student's lab report.)

19.74

a.	The half-reactions are: 2H^+(aq) + 2e^- → H2(g) Ag^+(aq) + e^- → Ag(s)
b.	The spontaneous cell reaction under standard-state conditions is: 2Ag^+(aq) + H2(g) → 2Ag(s) + 2H^+(aq)
c.	Using the Nernst equation we can calculate the cell potential under nonstandard-state conditions. (i) The potential is: E= 0.92 V (ii) The potential is: E= 1.10 V
d.	From the results in part (c), we deduce that this cell is a pH meter; its potential is a sensitive function of the hydrogen ion concentration. Each 1 unit increase in pH causes a voltage increase of 0.060 V.

19.75

a.

If this were a standard cell, the concentrations would all be 1.00 M, and the voltage would just be the standard emf calculated from Table 19.1 of the text. Since cell emf's depend on the concentrations of the reactants and products, we must use the Nernst equation (Equation 19.7 of the text) to find the emf of a nonstandard cell. E3.14 V

b.

First we calculate the concentration of silver ion remaining in solution after the deposition of 1.20 g of silver metal Ag originally in solution: Ag deposited: Ag remaining in solution: (3.46 ´ 10^-2 mol Ag) - (1.11 ´ 10^-2 mol Ag) = 2.35 ´ 10^-2 mol Ag The overall reaction is: Mg(s) + 2Ag^+(aq) ® Mg^2+(aq) + 2Ag(s) We use the balanced equation to find the amount of magnesium metal suffering oxidation and dissolving. The amount of magnesium originally in solution was The new magnesium ion concentration is: The new cell emf is: E3.13 V

19.76

The overvoltage of oxygen is not large enough to prevent its formation at the anode. Applying the diagonal rule, we see that water is oxidized before fluoride ion.

F₂(g) + 2e^- → 2F^-(aq) E° = 2.87 V

O₂(g) + 4H⁺(aq) + 4e^- → 2H₂O(l) E° = 1.23 V

The very positive standard reduction potential indicates that F^- has essentially no tendency to undergo oxidation. The oxidation potential of chloride ion is much smaller (-1.36 V), and hence Cl₂(g) can be prepared by electrolyzing a solution of NaCl.

This fact was one of the major obstacles preventing the discovery of fluorine for many years. HF was usually chosen as the substance for electrolysis, but two problems interfered with the experiment. First, any water in the HF was oxidized before the fluoride ion. Second, pure HF without any water in it is a nonconductor of electricity (HF is a weak acid!). The problem was finally solved by dissolving KF in liquid HF to give a conducting solution.

19.77

The cell voltage is given by:

E_cell0.035 V

19.78

We can calculate the amount of charge that 4.0 g of MnO₂ can produce.

Since a current of one ampere represents a flow of one coulomb per second, we can find the time it takes for this amount of charge to pass.

0.0050 A = 0.0050 C/s

19.79

Since this is a concentration cell, the standard emf is zero. (Why?) Using Equation 19.6, we can write equations to calculate the cell voltage for the two cells.

(1)

(2)

In the first case, two electrons are transferred per mercury ion (n = 2), while in the second only one is transferred (n = 1). Note that the concentration ratio will be 1:10 in both cases. The voltages calculated at 18°C are:

(1)

(2)

Since the calculated cell potential for cell (1) agrees with the measured cell emf, we conclude that the mercury(I) is Hg.

19.80

According to the following standard reduction potentials:

O₂(g) + 4H⁺(aq) + 4e^- ® 2H₂O E° = 1.23 V

I₂(s) + 2e^- ® 2I^-(aq) E° = 0.53 V

we see that it is easier to oxidize the iodide ion than water (because O₂ is a stronger oxidizing agent than I₂). Therefore, the anode reaction is:

2I^-(aq) ® I₂(s) + 2e^-

The solution surrounding the anode will become brown because of the formation of the triiodide ion:

I^- + I₂(s) ® I(aq)

The cathode reaction will be the same as in the NaCl electrolysis. (Why?) Since OH^- is a product, the solution around the cathode will become basic which will cause the phenolphthalein indicator to turn red.

19.81

We begin by treating this like an ordinary stoichiometry problem (see Chapter 3).

Step 1: Calculate the number of moles of Mg and Ag⁺.

The number of moles of magnesium is:

The number of moles of silver ion in the solution is:

Step 2: Calculate the mass of Mg remaining by determining how much Mg reacts with Ag⁺.

The balanced equation for the reaction is:

2Ag⁺(aq) + Mg(s) 2Ag(s) + Mg²⁺(aq)

Since you need twice as much Ag⁺ compared to Mg for complete reaction, Ag⁺ is the limiting reagent. The amount of Mg consumed is:

The amount of magnesium remaining is:

Step 3: Assuming complete reaction, calculate the concentration of Mg²⁺ ions produced.

Since the mole ratio between Mg and Mg²⁺ is 1:1, the mol of Mg²⁺ formed will equal the mol of Mg reacted. The concentration of Mg²⁺ is:

Step 4: We can calculate the equilibrium constant for the reaction from the standard cell emf.

We can then compute the equilibrium constant.

Step 5: To find equilibrium concentrations of Mg²⁺ and Ag⁺, we have to solve an equilibrium problem.

Let x be the small amount of Mg²⁺ that reacts to achieve equilibrium. The concentration of Ag⁺ will be 2x at equilibrium. Assume that essentially all Ag⁺ has been reduced so that the initial concentration of Ag⁺ is zero.

	2Ag^+ (aq)	+ Mg(s)		2Ag(s)	+ Mg^2+ (aq)
Initial (M):	0.0000				0.0500
Change (M):	+2x				-x
Equilibrium (M):	2x				(0.0500 - x)

We can assume 0.0500 - x » 0.0500.

(2x)² = 5.00 ´ 10^-109 = 50.0 ´ 10^-110

2x = 7 ´ 10^-55 M

[Ag⁺] = 2x = 7 ´ 10^-55 M

[Mg²⁺] = 0.0500 - x = 0.0500 M

19.82

Weigh the zinc and copper electrodes before operating the cell and re-weigh afterwards. The anode (Zn) should lose mass and the cathode (Cu) should gain mass.

19.83

a.

Since this is an acidic solution, the gas must be hydrogen gas from the reduction of hydrogen ion. The two electrode reactions and the overall cell reaction are: anode: Cu(s) → Cu^2+(aq) + 2e^- cathode: 2H^+(aq) + 2e^- → H2(g) Cu(s) + 2H^+(aq) → Cu^2+(aq) + H2(g) Since 0.584 g of copper was consumed, the amount of hydrogen gas produced is: At STP, 1 mole of an ideal gas occupies a volume of 22.41 L. Thus, the volume of hydrogen gas at STP is:

b.

From the current and the time, we can calculate the amount of charge: Since we know the charge of an electron, we can compute the number of electrons. Using the amount of copper consumed in the reaction and the fact that 2 mol of e^- are produced for every 1 mole of copper consumed, we can calculate Avogadro's number. 6.09 ´ 10^23 e^-/mol e^- In practice, Avogadro's number can be determined by electrochemical experiments like this. The charge of the electron can be found independently by Millikan's experiment.

19.84

The reaction is: Al³⁺ + 3e^- ® Al

First, we calculate the number of coulombs of electricity that must pass through the cell to deposit 60.2 g of Al.

The time (in min) needed to pass this much charge is:

19.85

a.	We can calculate DG° from standard free energies of formation. DG = 0 + (6)(-237.2 kJ/mol) - [(4)(-16.6 kJ/mol) + 0] DG = -1356.8 kJ/mol
b.	The half-reactions are: 4NH3(g) → 2N2(g) + 12H^+(aq) + 12e^- 3O2(g) + 12H^+(aq) + 12e^- → 6H2O(l) The overall reaction is a 12-electron process. We can calculate the standard cell emf from the standard free energy change, DG°.

19.86

Cathode: Au³⁺(aq) + 3e^- ® Au(s)

Anode: H₂O(l) ® O₂(g) + 4H⁺(aq) + 4e^-

a.	First, from the amount of gold deposited, we can calculate the moles of O2 produced. Then, using the ideal gas equation, we can calculate the volume of O2.
b.

19.87

The reduction of Ag⁺ to Ag metal is:

Ag⁺(aq) + e^- → Ag

We can calculate both the moles of Ag deposited and the moles of Au deposited.

We do not know the oxidation state of Au ions, so we will represent the ions as Auⁿ⁺. If we divide the mol of Ag by the mol of Au, we can determine the ratio of Ag⁺ reduced compared to Auⁿ⁺ reduced.

That is, the same number of electrons that reduced the Ag⁺ ions to Ag reduced only one-third the number of moles of the Auⁿ⁺ ions to Au. Thus, each Auⁿ⁺ required three electrons per ion for every one electron for Ag⁺. The oxidation state for the gold ion is +3; the ion is Au³⁺.

Au³⁺(aq) + 3e^- → Au

19.88

Strategy:	Use Equation 19.6 to determine the value of Ecell.
Setup:	The equation represented by the diagram is balanced by applying steps 1 through 7 for balancing redox equations to give: Pb + 2AgNO3 → 2Ag + Pb(NO3)2 Use E° values from Table 19.1 to determine E° for the reaction; n = 2.
Solution:	Cathode (reduction): Ag^+(aq) + e^− → Ag(s) Anode (oxidation): Pb(s) → Pb^2+(aq) + 2e^− The Nernst equation for this reaction is: Ecell = 0.91 V If [Ag^+] were increased by a factor of 4, Ecell = 0.95 V

19.89

We reverse the first half-reaction and add it to the second to come up with the overall balanced equation

Hg₂²⁺ → 2Hg²⁺ + 2e^-

Hg₂²⁺ + 2e^- → 2Hg

2Hg → 2Hg²⁺ + 2Hg

Since the standard cell potential is an intensive property,

Hg(aq) → Hg²⁺(aq) + Hg(l)

We calculate DG° from E°.

DG° = -nFE° = -(1)(96500 J/V×mol)(-0.07 V) = 6.8 kJ/mol

The corresponding equilibrium constant is:

We calculate K from DG°.

DG° = -RTln K

K = 0.064

19.90

a.	Anode 2F^- ® F2(g) + 2e^- Cathode 2H^+ + 2e^- ® H2(g) Overall: 2H^+ + 2F^- ® H2(g) + F2(g)
b.	KF increases the electrical conductivity (what type of electrolyte is HF(l))? The K^+ is not reduced.
c.	Calculating the moles of F2 Using the ideal gas law:

19.91

The reactions for the electrolysis of NaCl(aq) are:

Anode: 2Cl^-(aq) → Cl₂(g) + 2e^-

Cathode: 2H₂O(l) + 2e^- → H₂(g) + 2OH^-(aq)

Overall: 2H₂O(l) + 2Cl^-(aq) → H₂(g) + Cl₂(g) + 2OH^-(aq)

From the pH of the solution, we can calculate the OH^- concentration. From the [OH^-], we can calculate the moles of OH^- produced. Then, from the moles of OH^- we can calculate the average current used.

pH = 12.24

pOH = 14.00 - 12.24 = 1.76

[OH^-] = 1.74 ´ 10^-2 M

The moles of OH^- produced are:

From the balanced equation, it takes 1 mole of e^- to produce 1 mole of OH^- ions.

Recall that 1 C = 1 A×s

19.92

a.	Anode: Cu(s) ® Cu^2+(aq) + 2e^- Cathode: Cu^2+(aq) + 2e^- ® Cu(s) The overall reaction is: Cu(s) ® Cu(s). Cu is transferred from the anode to cathode.
b.	Consulting Table 19.1 of the text, the Zn will be oxidized, but Zn^2+ will not be reduced at the cathode. Ag will not be oxidized at the anode.
c.	The moles of Cu: The coulombs required: The time required:

19.93

The reaction is:

Ptⁿ⁺ + ne^- → Pt

Thus, we can calculate the charge of the platinum ions by realizing that n mol of e^- are required per mol of Pt formed.

The moles of Pt formed are:

Next, calculate the charge passed in C.

Convert to moles of electrons.

We now know the number of moles of electrons (0.187 mol e^-) needed to produce 0.0466 mol of Pt metal. We can calculate the number of moles of electrons needed to produce 1 mole of Pt metal.

Since we need 4 moles of electrons to reduce 1 mole of Pt ions, the charge on the Pt ions must be +4.

19.94

Using the standard reduction potentials found in Table 19.1

Cd²⁺(aq + 2e^- ® Cd(s) E° = -0.40 V

Mg²⁺(aq) + 2e^- ® Mg(s) E° = -2.37 V

Thus Cd²⁺ will oxidize Mg so that the magnesium half-reaction occurs at the anode.

Mg(s) + Cd²⁺(aq) ® Mg²⁺(aq) + Cd(s)

19.95

The half-reaction for the oxidation of water to oxygen is:

2H₂O(l) O₂(g) + 4H⁺(aq) + 4e^-

Knowing that one mole of any gas at STP occupies a volume of 22.41 L, we find the number of moles of oxygen.

Since four electrons are required to form one oxygen molecule, the number of electrons must be:

The amount of charge passing through the solution is:

We find the electron charge by dividing the amount of charge by the number of electrons.

In actual fact, this sort of calculation can be used to find Avogadro's number, not the electron charge. The latter can be measured independently, and one can use this charge together with electrolytic data like the above to calculate the number of objects in one mole. See also Problem 19.81.

19.96

a.	Au(s) + 3HNO3(aq) + 4HCl(aq) ® HAuCl4(aq) + 3H2O(l) + 3NO2(g).
b.	The function of HCl is to increase the acidity and to form the stable complex ion, AuCl.

19.97

Cells of higher voltage require very reactive oxidizing and reducing agents, which are difficult to handle. (From Table 19.1 of the text, we see that 5.92 V is the theoretical limit of a cell made up of Li⁺/Li and F₂/F^- electrodes under standard-state conditions.) Batteries made up of several cells in series are easier to use.

19.98

The overall cell reaction is:

2Ag⁺(aq) + H₂(g) ® 2Ag(s) + 2H⁺(aq)

We write the Nernst equation for this system.

The measured voltage is 0.0589 V, and we can find the silver ion concentration as follows:

[Ag⁺] = 2.8 ´ 10^-4 M

Knowing the silver ion concentration, we can calculate the oxalate ion concentration and the solubility product constant.

K_sp = [Ag⁺]²[C₂O₄^2-] = (2.8 ´ 10^-4)²(1.4 ´ 10^-4) = 1.1 ´ 10^-11

19.99

The half-reactions are:

Zn(s) + 4OH^-(aq) ® Zn(OH)₄^2-(aq) + 2e^-

Zn²⁺(aq) +2e^- ® Zn(s)

Zn²⁺(aq) + 4OH^-(aq) ® Zn(OH)₄^2-(aq)

K_f2 ´ 10²⁰

19.100

The half reactions are:

H₂O₂(aq) ® O₂(g) + 2H⁺(aq) + 2e^-

H₂O₂(aq) + 2H⁺(aq) + 2e^- ® 2H₂O(l)

2H₂O₂(aq) ® 2H₂O(l) + O₂(g)

Thus, products are favored at equilibrium. H₂O₂ is not stable (it disproportionates).

19.101

a.	Since electrons flow from X to SHE, E for X is negative. Thus E for Y is positive.
b.	Y^2+ + 2e^- ® Y X ® X^2+ + 2e^- X + Y^2+ ® X^2+ + Y

19.102

a.	The half reactions are: Sn^4+(aq) + 2e^- ® Sn^2+(aq) 2Tl(s) ® Tl^+(aq) + e^- Sn^4+(aq) + 2Tl(s) ® Sn^2+(aq) + 2Tl^+(aq)
b.	K = 8 ´ 10^15
c.

19.103

a.	Gold does not tarnish in air because the reduction potential for oxygen is not sufficiently positive to result in the oxidation of gold. O2 + 4H^+ + 4e^- ® 2H2O That is, for either oxidation by O2 to Au^+ or Au^3+. or
b.	3(Au^+ + e^- ® Au) Au ® Au^3+ + 3e^- 3Au^+ ® 2Au + Au^3+ Calculating DG, DG° = -nFE° = -(3)(96,500 J/V×mol)(0.19 V) = -55.0 kJ/mol For spontaneous electrochemical equations, DG° must be negative. Yes, the disproportionation occurs spontaneously.
c.	Since the most stable oxidation state for gold is Au^3+, the predicted reaction is: 2Au + 3F2 ® 2AuF3

19.104

It is mercury ion in solution that is extremely hazardous. Since mercury metal does not react with hydrochloric acid (the acid in gastric juice), it does not dissolve and passes through the human body unchanged. Nitric acid (not part of human gastric juices) dissolves mercury metal (see Problem 19.114); if nitric acid were secreted by the stomach, ingestion of mercury metal would be fatal.

19.105

The balanced equation is: 5Fe²⁺ + MnO₄^- + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4 H₂O

Calculate the amount of iron(II) in the original solution using the mole ratio from the balanced equation.

The concentration of iron(II) must be:

The total iron concentration can be found by simple proportion because the same sample volume (25.0 mL) and the same KMnO₄ solution were used.

[Fe³⁺] = [Fe]_total - [Fe²⁺] = 0.0680 M

Think About It:

Why are the two titrations with permanganate necessary in this problem?

19.106

Viewed externally, the anode looks negative because of the flow of the electrons (from Zn ® Zn²⁺ + 2e^-) toward the cathode. In solution, anions move toward the anode because they are attracted by the Zn²⁺ ions surrounding the anode.

19.107

From Table 19.1 of the text.

H₂O₂(aq) + 2H⁺(aq) + 2e^- ® 2H₂O(l)

H₂O₂(aq) ® O₂(g) + 2H⁺(aq) + 2e^-

2H₂O₂(aq) ® 2H₂O(l) + O₂(g)

Because E° is positive, the decomposition is spontaneous.

19.108

a.	The overall reaction is: Pb + PbO2 + H2SO4 ® 2PbSO4 + 2H2O Initial mass of H2SO4: Final mass of H2SO4: Mass of H2SO4 reacted = 355 g - 224 g = 131 g
b.

19.109

a.	Unchanged.

b.	Unchanged.

c.

Squared.

d.

Doubled.

e.

Doubled.

19.110

a.