314
Electrochemistry III: Electrolysis
faradays =
/ coulombs\
w i
( - )
(amp)(sec)
V
sec /
n, ,0-. ~~m~«""u
_,, _ coulombs
96,487 —
;
96,487 —7
faraday
faraday
(0.500
cou omb \ (2.00 hr) (60 —) (eO -^-)
_\
sec /
\
hr / \
mm/
96,487 ^Hbs
faraday
= 0.0373 faraday
We know that 0.0373 faraday will liberate 0.0373 E-T equiv. The half-reaction
(e~ + iCu
2+ -» |Cu) tells us that copper has 2 E-T equiv per mole. Therefore,
wt of Cu = (0.0373 equiv) ( '
m
°!
e
W63.5
V 2 equiv/ \ .
equiv/ \
mole/
= 1.18 g Cu plated out
The half-reaction (OH~ -» {O 2 + iH 2 O + e~) tells us that O 2 has 4 E-T equiv
per mole. Therefore,
(0.0373 equiv) ( !
m °
le ) (62.4
torr ht ^
r ') (298 K )
, . ^
rc#r
\4 equiv/ \
mole K /
vo of O 2 = —^— =
^—r
2
P
730 torr
= 0.238 liters = 238 ml at 25.0°C and 730 torr
PROBLEM:
What length of time is required to plate out 0.1000 g Ag from an AgN0 3
solution using a current of 0.200 amp?
SOLUTION:
The number of electron-transfer equivalents of Ag plated out is
E-T equiv of Ag =
°10 °°/
g _,^ = 9.27 x 10equiv/
Because 9.27 x 10~
4 equiv of Ag requires 9.27 x 10~
4 faraday of electricity for
plating out,
(9.27 x 104 faraday)
coulombs
V '
faraday
seconds =
amperCS
' 0.200
sec
= 447 sec
447 sec _ ., .
-ir
i •
= 7.45 mm required for plating
min