Highlander Help




Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we'll add them to "Resources" part of this page for other students to use.


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.




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


Ecell = 0.0592log(k)/n




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




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



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:



 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:



 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:




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 ***


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?




 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.)




 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?




 Ecell = E cath = E an




Question 11

The tendency to rust is

Select one or more:



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?




 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


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.




2(2X(CN)63-(aq)  → 2X(CN)64-(aq) )

2I-(aq) →  I2(s) + 2e-

ΔG= -97.8e3=-nFE




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.




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

E = 0.0592logK/n


(-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.




 Q= (0.00788)^2=0.000062094


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

Ecell= -0.3714