Example 11.5

Write net equations or “no reaction” and determine

oE

rxn

for each process.

a) Metallic sodium is added to water.

The two half-reactions and their standard

reduction potentials are as follows:

Na

1+ + e

1-^

U

Na

-2.71 V

2H

O + 2e 2

1-^

U

H

+ 2OH 2

1- -0.41 V

* Note that red arrows are used in this example to emphasize the

direction of the required electron flow. In Example 11.5a, the arrow shows that the electrons must flow from Na to H

O, which is toward 2

the couple at less negative potential and lower free energy.

The reductant (Na) is situated above the oxidant (H

O) in Table 11.1, so the reaction is 2

extensive.* Na is a reactant, not a product, so its half-reaction must be reversed. The half-reaction must also be multiplied by two in order to balance electrons lost and gained, but multiplying the reaction by a number does not

change the potential

. Rewriting the two half-

reactions and summing them, Reduction:

2H

O + 2e 2

1-^

U

H^2

+ 2OH

1-^

Oxidation:

2Na

U

2Na

1+ + 2e

1-^

Reaction:

2H

O + 2Na 2

→

2Na

1+ + H

+ 2OH 2

1-^

§ Note the reduction potential of

pure water is used, not the standard

reduction potential, which would require standard concentrations,

i.e

.,

[OH]

1- = 1.0 M, which is a strongly basic solution.

oE

rxn

=

oE

reduced

-^ E

o oxidized

= -0.41

§ - (-2.71) = +2.30 V

The very high, positive value of

o (^) E
rxn
indicates that this is an extensive reaction.
b) Copper is placed in hydrochloric acid.
The SHE is used for strong acids other than nitric acid. 2H
1+ + 2e
1-^
U
H
(^2)
0.00 V
Cu
2+ + 2e
1-^
U
Cu
+0.34 V
No reaction.
oE
rxn

oE
reduced

-^ E

ooxidized

= 0.00 - 0.34 = -0.34 V

The reductant is below the oxidant in Table 11.1, so electrons would have to flow to higher free energy, which is also indicated by the fact that

o (^) E
rxn is negative. Thus, copper does
not react with most strong acids.
†^
†
The SHE is used for strong acids only when it is the H
1+ that is the
oxidant. However, some acids are much stronger oxidants due to their anion. Nitric acid (
o E
= 0.96V) is one example. Thus, copper does not
react with most strong acids, but it does react with nitric acid because NO
1- 3
in acid is a stronger oxidant than H
1+.
c) Iron is placed into a NiSO
solution. 4
Fe
2+ + 2e
1-^
U
Fe
-0.44 V
Ni
2+ + 2e
1-
U
Ni
-0.23 V
The reductant is above the oxidant in Table
11.1, so electron transfer is extensive.
Reduction:
Ni
2+ + 2e
1-^
U
Ni
Oxidation
Fe
U
Fe
2+ + 2e
1-^
Reaction:
Ni
2+ + Fe
→
Ni + Fe
2+^
oE
rxn

oE
reduced

-^ E

ooxidized

= -0.23 - (-0.44) = +0.21 V

Chapter 11 Electron Transfer and Electrochemistry

255