DOC.
34
197
Let
these
forces
produce
an
electric double
layer
on
the
surface
of
a
piece
of metal
M
that
borders
on
a
gas,
and
a
corresponding
potential
difference
V
between
metal
and
gas,
taken
as
positive
when
the metal
has
the
higher
potential.
Let
V1
and
V2
be
the
potential
differences
between
metals
M1
and
M2
in electrostatic
equilibrium
if
they
are
insulated from
each
other. If
the
two
metals
are
brought
into contact, the electric
equilibrium
is disturbed
and
complete1
voltage
equalization
of
the metals takes
place.
Thereby,
simple
layers
will
be
superposed
on
the aforementioned double
layers at
the
metal-gas
interfaces;
to
these
corresponds
an
electrostatic field
in
the air
space
whose
line integral
equals the voltage
difference.
If
Vl1
and
Vl2
denote the electric potentials
at
points of
the
gas
space
directly
adjacent
to
the metals
in contact, and
V'
denotes
the
potential
in the interior
of
the metals,
we
have
\
-1
i
and thus
r
-
r£
= r2
,
%
-
\
=yi-v
Thus,
the electrostatically
measurable Volta
difference
is
numerically
equal
to
the difference
of the
potentials
assumed
by
the metals in
the
gas
if
they
are
insulated
from each
other.
If
one
ionizes the
gas,
the electric forces
present
in
the
gas space
will
cause a
migration
of the ions,
to which
there
corresponds
a
current
in
the metals
which,
at
the
place
of
contact of the
metals,
is directed
from
the
metal with the
higher
V
(less electropositive)
to
the metal with
the
lower
[20]
V
(more
electropositive).
[21]
Suppose a
metal
M
is insulated in
a gas.
Let
V
be
its potential
difference with
respect
to
the
gas
that
corresponds to
the double
layer. In
order
to
move a
unit of
negative
electricity
from
the metal
into the
gas,
an
amount
of
work numerically
equal to
the
potential
V
has
to be performed.
Hence,
the
greater
the
V,
i.e., the less
electropositive the
metal,
the
more [22]
1We
disregard
the effect of thermoelectric forces.