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>>
View the other Key Equations and Concepts in this
chapter
Electrolysis
>> Parts of this equation/concept include:
| A. Current to Amount of Substance |
The half–reaction is used to relate moles of electrons to
moles of substance. Normally, these half–reactions are simple
and can be determined by carefully reading the question.
Current is used as a conversion factor between time and coulombs,
where 1 A = 1C/s, so a current of 1 A means 1 C of charge passes
each second.
Since the equation expresses amount of electrons in moles and current
expresses amount of electrons in coulombs, a factor to convert between
the two is needed. This factor is Faraday's constant (F)
where F = 9.65 x 104 C/mol e–
or 1 mole electrons = 9.65 x 104 C. The conversion expressed
this way has three significant figures.
With these conversions, solving electrolysis problems is simply
a matter of dimensional analysis.
>> Example 1
How many grams of chromium metal can be made from chromium(III)
ion with a current of 0.54 A for 1.00 hour?
Solution:
The reaction is that of chromium(III) ion (Cr3+) to
its metal (Cr0), so the balanced half–reaction
is
Cr3+ + 3 e–  Cr0
The superscript zero is not required for the formula of a metal
but is sometimes a convenient way to express the metal form. That
way you are reminded that the charge is zero and you don't have
to wonder if you just forgot to write it down.
From the reaction, 3 moles of electrons are required for each
mole of chromium metal. Faraday's constant relates moles of electrons
to the electrical unit coulombs (C). Amps (or amperes) are C/s,
so it's a conversion factor, not the starting place.
Using dimensional analysis and these relationships,
The information used for the process was
| time |
 |
unit conversion |
|
unit conversion |
|
current |
|
Faraday's constant |
|
chemical reaction |
|
molar mass |
= |
mass |
The current limited significant figures in the final answer to
two.
>> Example 2
What volume of hydrogen gas at STP can be produced from aqueous
hydrochloric acid with a constant current of 5.30 A for 90.0 minutes?
Solution:
The process is similar to that in the first example. For the
reaction, first recognize that hydrochloric acid in water is really
the ions H+ and Cl–. Since hydrogen
gas (H2) is being produced, it must be made from H+
ions. Therefore the reaction is
2 H+ + 2 e– H2
Rather than grams of hydrogen, the question asks for volume of
hydrogen. Recall from Chapter
8 that the volume of 1 mole of gas at STP = 22.4 L.
Using dimensional analysis to solve the problem,
>> back
to the Top of the Page
| B. Amount of Substance to Time and Current |
These problems are also solved by dimensional analysis. The same
conversion factors are used, only the order of use changes.
>> Example 3
How long does it take to make 5.56 g nickel metal from Ni2+
with a constant current of 3.78 A?
Solution:
The reaction of making nickel metal from its ion is
Ni2+ + 2 e– Ni0
The superscript zero is not required for the formula of a metal
but is sometimes convenient.
Therefore 2 moles of electrons create 1 mole of nickel metal.
The electrons are being introduced at a rate of 3.78 C/s (amps).
The relationship between coulombs (C) and moles of electrons is
Faraday's constant (9.65 x 104 C/mol). Using these
relationships in dimensional analysis,
| 4.84 x 103 s |
|
|
= |
80.6 min |
Since the question does not specify, it is appropriate to answer
in either seconds or minutes. However, the final answer should
only have three significant figures.
>> Example 4
How many minutes are required to make 1.0 g potassium metal from
potassium ion at a current of 3.53 A?
Solution:
The reaction of potassium metal from potassium ion is
K+ + e– K
So
The information used for the preceding conversions was the following:
| mass |
|
molar mass |
|
chem rxn |
|
Faraday's constant |
|
current |
|
unit conversion |
= |
time |
The two significant figures were determined by the grams of potassium.
>> Example 5
What current is required to reduce 0.21 g of zinc ion to zinc
metal in 30.0 minutes?
Solution:
The reaction is
Zn2+ + 2 e– Zn
The charge on zinc is always +2. You were supposed to have memorized
that in Chapter 4.
Current (i) in amps is coulombs/second.
Since time is given in the problem, it just needs to be converted
to seconds.
| 30.0 min |
|
|
= |
1.80 x 103 s |
The coulombs needed can be determined from the reaction and Faraday's
constant.
| 0.21 g Zn |
|
|
|
|
|
|
= |
619.91 C |
= |
6.2 x 102 C |
So
| current |
= |
|
= |
0.34 C/s |
= |
0.34 A |
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the other Key Equations and Concepts in this chapter
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