Chemistry - A Molecular Science

(Nora) #1

Chapter 11 Electron Transfer and Electrochemistry


Example 11.4


Write the anode and cathode half-reactions

and the balanced chemical equation for

the galvanic cells constructed from the following pairs of redox couples. Determine the standard cell potential for each cell.
a) Pb/Pb

2+ and Zn/Zn

2+^

Get the reduction half-reactions and their st

andard reduction potentials from Table 11.1

Zn

2+ + 2e

1-U^

Zn

oE
= -0.76 V

Pb

2+ + 2e

1-^ U

Pb

oE
= -0.13 V

The oxidation half-reaction is the one at more

negative potential, so

Zn is oxidized, and

Pb

2+ is reduced in the cell. Both half-reacti

ons involve two electrons, so the balanced

chemical equation is the sum of the following two reactions: Zn

U

Zn

2+ + 2e

1-^

Anode (oxidation) half-reaction

Pb

2+ + 2e

1-^ U

Pb

Cathode (reduction) half-reaction

Zn + Pb

2+^

Zn

2+ + Pb Net chemical equation

oE
cell

(^) =
o (^) E
cathode



  • E


o anode

= -0.13 - (-0.76) = 0.63 V

o (^) E
cell
is positive, so the reaction is extensive.
b) Ni
2+/Ni and Ag
1+/Ag
The two relevant half-reactions and t
heir standard reduction potentials are
Ni
2+ + 2e
1-^ U
Ni
oE
= -0.23 V
Ag
1+ + e
1-U^
Ag
oE
= +0.80 V
Reverse the half-reaction at mo
re negative potential to obtain the oxidation and multiply
the Ag
1+/Ag half-reaction by 2 to make t
he number of electrons gained by Ag
1+ equal to
the number lost by the Ni. The balanced chemic
al equation for the cell is the sum of the
two reactions. Ni
U
Ni
2+ + 2e
1-^
Anode (oxidation) half-reaction
2Ag
1+ + 2e
1-^ U
2Ag
Cathode (reduction) half-reaction
Ni + 2Ag
1+^

Ni
2+ + 2Ag Net chemical equation
Standard reduction potentials indicate the free en
ergy of the electron(s), so multiplying a
half-reaction by some number does not
affect its standard reduction potential


. Thus,


multiplying the Ag

1+/Ag half-reaction by 2 does not ch

ange the potential at which the

couple is at equilibriu

m. The cell potential is

oE
cell

=

oE
cathode


  • E


oanode

= +0.80 - (-0.23) = 1.03 V
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