5 Steps to a 5 AP Chemistry 2019

Electrochemistry ❮ 249

First, the cell voltage can be calculated using:

EEce°°ll=−cathodeaE°node> 0

Since the cell potential must be positive (a galvanic cell), there is only one arrangement of

• 0.25 and 0.80 volts than can result in a positive value:

Ecell° =−0.80V(−=0.25V) 1.05V

This means that the Ni electrode is the anode and must be involved in oxidation, so its reduc-

tion half-reaction must be reversed, changing the sign of the standard half-cell potential, and

added to the silver half-reaction. Note that the silver half-reaction must be multiplied by

2 to equalize electron loss and gain, but the half-cell potential remains the same:

→+ °=

×+→ °=

+→+ °=

+−

+−

++

E

E

E

Ni(s)Ni(aq)2e 0.25V

2(Ag (aq) eAg(s)) 0.80V

Ni(s)2Ag (aq) Ni (aq) 2Ag(s) 1.05V

2

(^2) cell
Note that the same cell potential is obtained as using: EEce°°ll =−cathodeaE°node> 0.
If, for example, you are given the cell notation, you could use this method to determine
the cell potential. In this case, the cell notation would be: NiN i|^2 ++|Ag|Ag|.
Electrolytic Cells
Electrolytic cells use electricity from an external source to produce a desired redox reaction.
Electroplating and the recharging of an automobile battery are examples of electrolytic cells.
Figure 16.2, on the next page, shows a comparison of a galvanic cell and an electrolytic
cell for the Sn/Cu system. On the left-hand side of Figure 16.2, the galvanic cell is shown for
this system. Note that this reaction produces 0.48 V. But what if we wanted the reverse reac-
tion to occur, the nonspontaneous reaction? This can be accomplished by applying a voltage
in excess of 0.48 V from an external electrical source. This is shown on the right-hand side of
Figure 16.2. In this electrolytic cell, electricity is being used to produce the nonspontaneous
redox reaction.
Quantitative Aspects of Electrochemistry
One of the most widely used applications of electrolytic cells is in electrolysis, the decom-
position of a compound. Water may be decomposed into hydrogen and oxygen. Aluminum
oxide may be electrolyzed to produce aluminum metal. In these situations, several questions
may be asked: How long will it take; how much can be produced; what current must be used?
Given any two of these quantities, the third may be calculated. To answer these questions,
the balanced half-reaction must be known. Then the following relationships can be applied:
1 Faraday = 96,500 coulombs per mole of electrons
(F = 96,500 C/mol e– or 96,500 J/V mol)
1 ampere = 1 coulomb/second (A = C/s)
Knowing the amperage and how long it is being applied (seconds), the coulombs can
be calculated. Then the coulombs can be converted into moles of electrons, and the moles
of electrons can be related to the moles (and then grams) of material being electrolyzed
through the balanced half-reaction.
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