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Question 1

Calculate the value of the solubility product constant for PbSO4 from the half-cell potentials.

PbSO4(s) + 2e- → Pb(s) + SO42-(aq) Eº = -0.322 V
Pb2+ + 2e- → Pb(s) Eº = -0.166 V

These half-cell potentials are not accurate. You many imagine that they were determined by a poor experimentalist. Explanation

Remember, don't round until the end when exponentiating.

PbSO4(s) + 2e- → Pb(s) + SO42-(aq)     Eº = -0.322 V
Pb(s) Pb2+ + 2e-    Eº = 0.166 V

Ecell = Ec - Ea

=(-0.322) + 0.166

=-0.156

Ecell = 0.0592log(k)/n

-0.156=0.0592log(k)/2

k=5.3e-6

Question 2

Use standard reduction potentials (shown below) to calculate the potential of a nickel-cadmium cell that uses a basic electrolyte that has a 0.6 M hydroxide concentration.

Cd(OH)2(s) + 2e- → Cd(s) + 2OH-(aq) Eº = -0.83 V
NiO(OH)s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq) Eº = 0.52 V Explanation

Careful, this is easier that it looks.

Cd(s) + 2OH-(aq)→ Cd(OH)2(s) + 2e-   Eº = 0.83 V
2(NiO(OH)s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)) Eº = 0.52 V

Ecell = Ecell - 0.0592logQ/n      Q=1

Ecell= (0.52+.83)-0.0592log(1)/2

Ecell=1.35

Question 3

What happens to the voltage of this cell as the cell is discharged? Why?

Select one or more:  Question 4

Write the half-reactions and the products for the reactions that occur in the electrolysis of Molten MgCl2 using inert electrodes. (Hint, after writing the half-reactions write the net reaction to determine the products.)

Select one or more:    Explanation

2Cl-   + Mg (2+)   + 2e-    ------> Cl2   + Mg(s)   + 2e-

not an aqueous solution so no need to consider electrolysis of water

Question 5

Write the half-reactions and the products for the reactions that occur in the electrolysis of a saturated solution of magnesium sulfate using inert electrodes.

Select one or more:    Explanation

Remember to include the reduction and oxidation of water in your cathode and anode list of possible reactions.

saturated solution = aqueous

(must consider the electrolysis of water)

1) Anode can be either

2H20(l)---->O2   + 4H(+)  + 4e-        E= 1.23*** greater oxidation

2SO4(2-)(aq) ---------> S2O8(2-)(aq)  + 2e-    E=2.05

2) Cathode can be either

2H20(l)---->O2   + 4H(+)  + 4e-

2(2H2O (l) + 2e-→ H2(g) + 2OH - (aq))

[remember that OH- and H+ can combine to form one mole of H2O]

2H2O---->O2 + 2H2

Question 6

Write the half-reactions and the products for the reactions that occur in the electrolysis of a nickel chloride solution, using nickel electrodes.

Select one or more:   Explanation

Anode

2Cl- → 2Cl2(g) + 2e-   E= 1.36

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

Ni(s) → Ni2+(aq) + 2e-     E=-0.23 ***

Cathode

Ni2+(aq) + 2e- → Ni(s)   E=-0.23***

2H2O (l) + 2e-→ H2(g) + 2OH-  E=-0.83

No not reaction

Question 7

The electrolysis of molten Al2O3 at 980 ºC is used to produce metallic aluminum. A current of 408 A is used in the electrolysis cell. What is the rate of formation of aluminum, in kilograms per hour? Explanation

408 C/s * 3600s * 1mol/96500 C = 15.22

4(Al(3+)  + 3e- ---> Al(s))

3(2H2O  -----> O2   + 4H+   +4e-)

15.22 mol e- * 4 mol Al/12 mol e- * 26.98 g*1kg/1000g=0.136885 kg

Question 8

For the above current at 4.60 V, how much electrical energy (in kw-hr) is consumed in the production of 1 kg of aluminum? (A kW-hr is 3.6 x106 J.) Explanation

1 kg Al * 1000g/ 1kg * 1 mol/26.98 g * 12 mol e-/4 mol Al * 96500 C/1mol e- = 10730170.5 C

4.6 J/C* 10730170.5 C = 49358784.28 J / (3.6 *10^6) = 13.71

Question 9

Why isn't the electrolysis of an aqueous solution of aluminum chloride used to manufacture this metal?

Select one: Question 10

The electrochemical processes that occur in the rate limiting step of the corrosion of iron (oxidation of iron) are represented by the half-reactions

Fe2+(aq) + 2e-→ Fe(s) Eº = -0.44 V
O2(g) + 4H+(aq) + 4e- → 2H2O(l) Eº = 1.23 V

First, write the overall reaction. What is the standard potential for the overall chemical reaction? Explanation

Ecell = E cath = E an

=1.23-(-0.44)

=1.67

Question 11

The tendency to rust is

Select one or more:   Explanation

Hint, use LeChatelier's Principle.

Question 12

Air is 21 mole percent oxygen, that is, 0.21 atm oxygen for 1 atm air. Suppose the pH of the water is 5.16 and that the concentration of iron(II) in the water is 4.92x 10-5 M, what is the potential of the corrosion reaction under the above conditions at 298 K? Explanation

E cell = E cell - (0.0592)log Q/n

2(Fe----> Fe(2+)   + e-)

O2 + 4H+   + 4e-  -----> 2H2O

Q= (4.92e-5)^2/(0.21)(6.918e-6)^4

Q=5.03e12

E cell = 1.23-(-0.45) - 0.0592log(5.03e12)/4

Ecell = 1.492

Question 13

The standard free energy change at 25 oC, ΔGo, is equal to -97.8 kJ for

2X(CN)63-(aq) + 2I-(aq) → 2X(CN)64-(aq) + I2(s).

Calculate the standard potential for this reaction. Normally, X is Fe. Here it is some unknown metal, allowing for somewhat random values for ΔGo. Explanation

2(2X(CN)63-(aq)  → 2X(CN)64-(aq) )

2I-(aq) →  I2(s) + 2e-

ΔG= -97.8e3=-nFE

-97.83e3=-2(96500)Ecell

Ecell=.5068

Question 14

At 298 K,

Pb(IO3)2(s) ↔ Pb2+(aq) + 2IO3-(aq) Ksp = 2.6x 10-13
Pb2+(aq) + 2e- → Pb(s) Eo = -0.126 V.

Find the standard potential of the half-reaction:
Pb(IO3)2(s) + 2e- ↔ Pb(s) + 2IO3-(aq) Eo = ?

Hint given in feedback. Explanation

Notice the relation between first 2 equations and the desired equation. Convert Ksp and Eo to ΔGo. . .

E = 0.0592logK/n

=0.0592*(1/2)*log(2.6e-13)=-0.37

(-0.37+(-0.126))= -0.496

Question 15

Calculate the potential of the Pb/Pb(IO3)2 electrode in a 0.00788 M solution of NaIO3. Explanation

Q= (0.00788)^2=0.000062094

Ecell=Ecell-(0.0592)log(0.000062094)/2

Ecell = (-0.496) - (0.0592)log(0.000062094)/2

Ecell= -0.3714