5 Steps to a 5 AP Chemistry

First, the cell voltage can be calculated using:

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:

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

reduction half-reaction must be reversed, changing the sign of the standard half-cell poten-

tial, and added to the silver half-reaction. Note that the silver half-reaction must be mul-

tiplied by two to equalize electron loss and gain, but the half-cell potential remains the

same:

Note that the same cell potential is obtained as using:.

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:.

Electrolytic Cells

Electrolytic cellsuse 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 shows a comparison of a galvanic and 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 reaction to occur, the nonsponta-

neous 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 elec-

trolytic 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 longwill it take; how muchcan be produced; what currentmust 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)

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.

Ni Ni^2 ++Ag Ag

EE E°° °cell=−>cathode anode 0

Ni s Ni a q e

aq e Ag s

() ( )

() ())

→+

×+→

+−

+−

(^22) E°=0.25V
2 (Ag EE°=0.80V

Ni(s)+→+ 2 Ag aq++()Ni^2 () ()aq Ag s E°cell= 1 105.V
E°cell=−− = 080 vvV 025 105.(.).
EE E°° °cell=−>cathode anode 0
Electrochemistry  247
TIP