The Foundations of Chemistry

EXAMPLE 21-1 Electrolysis

Calculate the mass of copper metal produced during the passage of 2.50 amperes of current

through a solution of copper(II) sulfate for 50.0 minutes.

Plan

The half-reaction that describes the reduction of copper(II) ions tells us the number of moles

of electrons required to produce one mole of copper metal. Each mole of electrons corresponds

to 1 faraday, or 9.65 104 coulombs, of charge. The product of current and time gives the

number of coulombs.



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t

 88n co

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.

m

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m

p

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a

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 88n o

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Solution

The equation for the reduction of copper(II) ions to copper metal is

Cu^2  2 e 88n Cu (reduction, cathode)

1 mol 2(6.02 1023 )e 1 mol

63.5 g 2(9.65 104 C) 63.5 g

We see that 63.5 grams of copper “plate out” for every 2 moles of electrons, or for every

2(9.65 104 coulombs) of charge. We first calculate the number of coulombs passing through

the cell.

_?_C50.0 min

1

6

m

0

i

s

n



2.5

s

0C

7.50 103 C

We calculate the mass of copper produced by the passage of 7.50 103 coulombs.

856 CHAPTER 21: Electrochemistry

_?_g Cu7.50 103 C

9.6

1

5

m



ol

10

e

 (^4) C
6
2
3.
m
5
o
g
l
C
e
u
 2.47 g Cu (about the mass of
a copper penny)
Notice how little copper is deposited by this considerable current in 50 minutes.
You should now work Exercises 26 and 32.
EXAMPLE 21-2 Electrolysis
What volume of oxygen gas (measured at STP) is produced by the oxidation of water in the
electrolysis of copper(II) sulfate in Example 21-1?
Plan
We use the same approach as in Example 21-1. Here we relate the amount of charge passed
to the number of moles, and hence the volume of O 2 gas produced at STP.

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 88n co
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 88n 
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2

Solution
The equation for the oxidation of water and the equivalence between the number of coulombs
and the volume of oxygen produced at STP are
2H 2 O88n O 2 4H 4 e (oxidation, anode)
1 mol 4(6.02 1023 )e
22.4 LSTP 4(9.65 104 C)
The amount of electricity in Examples
21-1 and 21-2 would be sufficient to
light a 100-watt household light bulb
for about 150 minutes, or 2.5 hours.
When the number of significant
figures in the calculation warrants, the
value 96,485 coulombs is usually
rounded to 96,500 coulombs
(9.65 104 C).
2.50 A2.50 C/s