278
Electrochemistry I: Batteries and Free Energy
is the number of electrons lost by the reducing agent (or gained by the oxidizing
agent) in the balanced chemical equation. Equation 17-3, on the other hand,
applies only to a balanced half-reaction. When Equation 17-3 is applied to two
different half-reactions that are to be combined to make one complete reaction,
it is essential that the half-reactions be balanced to give the same value of n, the
value that will be used in Equation 17-4 and that will refer to the complete reaction.
CAUTION: We must call your attention to a complication that may result from
the unthinking application of Equation 17-4 to a reaction such as
Hg + Hg
2+ ?± Hg 2
2 +
where one ion (in this case Hgl
+ ) is common to both half-reactions. Application
of Equation 17-4 to this overall reaction would give the erroneous result
„
_
0.0591 ,
Ecell = £?ell
— lo
because it loses track of the fact that there are two different Hgjj
+ concentrations, one in each half-cell, that can be varied independently. However, the
method of the last problem can be applied to the separate half-reactions:
(oxidizing)
2e~ + 2Hg
2+ <=» Hg|
+
(reducing)
2e~ + Hg|
+ ±5 2Hg
Using this method, we obtain a correct expression for the cell voltage:
0.0591 ,„„ [Hgn ox [Hgj+U
[Hg
2+ ]
2
£cell - £?ell
^
log
rtr-2+12
ELECTRON-TRANSFER EQUILIBRIUM
If you permit a battery to completely "run down" or become "dead," itsfcen
becomes zero, and the cell reaction reaches an equilibrium condition. Making a
battery and letting it go dead is not the only way to let an electron-transfer reaction
reach equilibrium; the components could just as well be mixed together in a
single beaker. The important thing to realize is that, no matter how you reach
equilibrium, the ion product Q at equilibrium is now equal to the equilibrium
constant. Thus, for any electron-transfer reaction at equilibrium at 25.0°C, we
could write
Seen = 0 = E c ° ell -
log K e
or
£c°eii = °'°
591 log/Ce
(17-6)
Electrochemistry I: Batteries and Free Energy
is the number of electrons lost by the reducing agent (or gained by the oxidizing
agent) in the balanced chemical equation. Equation 17-3, on the other hand,
applies only to a balanced half-reaction. When Equation 17-3 is applied to two
different half-reactions that are to be combined to make one complete reaction,
it is essential that the half-reactions be balanced to give the same value of n, the
value that will be used in Equation 17-4 and that will refer to the complete reaction.
CAUTION: We must call your attention to a complication that may result from
the unthinking application of Equation 17-4 to a reaction such as
Hg + Hg
2+ ?± Hg 2
2 +
where one ion (in this case Hgl
+ ) is common to both half-reactions. Application
of Equation 17-4 to this overall reaction would give the erroneous result
„
_
0.0591 ,
Ecell = £?ell
— lo
because it loses track of the fact that there are two different Hgjj
+ concentrations, one in each half-cell, that can be varied independently. However, the
method of the last problem can be applied to the separate half-reactions:
(oxidizing)
2e~ + 2Hg
2+ <=» Hg|
+
(reducing)
2e~ + Hg|
+ ±5 2Hg
Using this method, we obtain a correct expression for the cell voltage:
0.0591 ,„„ [Hgn ox [Hgj+U
[Hg
2+ ]
2
£cell - £?ell
^
log
rtr-2+12
ELECTRON-TRANSFER EQUILIBRIUM
If you permit a battery to completely "run down" or become "dead," itsfcen
becomes zero, and the cell reaction reaches an equilibrium condition. Making a
battery and letting it go dead is not the only way to let an electron-transfer reaction
reach equilibrium; the components could just as well be mixed together in a
single beaker. The important thing to realize is that, no matter how you reach
equilibrium, the ion product Q at equilibrium is now equal to the equilibrium
constant. Thus, for any electron-transfer reaction at equilibrium at 25.0°C, we
could write
Seen = 0 = E c ° ell -
log K e
or
£c°eii = °'°
591 log/Ce
(17-6)