E_-_102_F11_Notes

3145jmolk t temperatureink f

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ):
 
 F
=
(6.022137
x
1023
mol‐1)
(1.6021773
x
10‐19
C)
 
 



=
96
485.31
C
/
mol
 
 Electric
current
(I)
is
the
amount
of
charge
(Q)
flowing
through
circuit
per
unit
of
time
(t):
 
 
 
 
 The
number
of
moles
of
electrons
that
flow
through
a
circuit
can
be
calculated
by
the
 equation:

 It total ch arg e pas sin g thru the circuit 
 n= = F ch arg e of a mole of electrons ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 € 
 
 
 
 
 
 
 E ‐5 
 Application
of
Faraday’s
Law
 • From
the
number
of
moles
passing
through
the
circuit
 • We
can
calculate
the
mass
of
a
given
substance
being
produced
or
consumed
at
an
 electrode.
 • For
the
previous
Cu|Cu2+||Ag+|Ag
cell,
for
each
mole
of
electrons
that
passes
through
 the
circuit:
 • ½
mole
Cu
(s)
is
oxidized
=
(63.5
g/mol)
/
(2
mol)
=
_________
g
of
Cu
(s)
are
 dissolved
at
the
anode.
 • 1
mole
of
Ag+
is
reduced
=
_______
g
of
Ag
(s)
are
...
View Full Document

Ask a homework question - tutors are online