CHEM126 – CHAPTER 18 – REGULAR HOMEWORK / ELECTROCHEMISTRY II

 

CHAPTER 18 REGULAR HOMEWORK:

ELECTROCHEMISTRY II

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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.
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Explanation

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

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

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Explanation

Careful, this is easier that it looks.

Question 3

 

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

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.)
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Explanation

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.
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Explanation

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

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.
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Explanation

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?

 

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Explanation

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.)
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Explanation

Question 9

 

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

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?
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Explanation

Question 11

 

The tendency to rust is
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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?
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Explanation

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.
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Explanation

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.
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Explanation

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

Question 15

 

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

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