1.30 recall the ? of common ? in this specification
K?
Na ?
Li ?
H?
Mg ??
Ca ??
Al ??
Cl?
Br?
I?
F?
OH?
NO??
SO4 ??
CO3 ?? Answer: 1.30 recall the CHARGES of common IONS in this
specification
K+
Na +
Li +
H+
1
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,Mg 2+
Ca 2+
Al 3+
Cl-
Br-
I-
F-
OH-
NO3-
SO4 2-
CO3 2-
1.56 recall that one ? represents one ? of ?
One ? is ? ?. That is the amount of ? in ? ? of ?. Answer: 1.56 recall that one
faraday represents one mole of electrons
One FARADAY is 96500 COULOMBS. That is the amount of COULOMBS in
ONE MOLE of ELECTRONS.
1.57 calculate the amounts of the ? of the ? of ? ? and ? solutions
One ? is ? ?. It is also one ? of ?.
If ? of 0.2 ? is passed through copper(ll) sulphate for two ?, how much ? do
you get?
1) Write out the ? equation
(Cu2+) + 2e > Cu
2
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,2) Work out ? of electrons flowing
?= ? x time
?= ?T
? is 2x60x60 (times 60 makes ?, times 60 again makes it ?)
Q= 0.2 x 7200= 1440 ?
Convert ? into ? of ?
?= C/?
?= 1440/96500
?= 0.015
Work out ? factor
(Cu2+) + 2e > Cu
For every ? moles of electrons, there will be ? Cu
Sf= Moles of ?/ moles of ?
Sf= 1/2
Sf= a half
3) Work out ? of product using ? ?
so we do the ? of electrons times the ? ?
0.015x1/2= 0.0075 Moles of Cu
Convert ? into ?
Moles x Mr
3
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, 0.0075 x 63.5= 0.48g of copper Answer: 1.57 calculate the amounts of the
products of the electrolysis of molten salts and aqueous solutions
One faraday is 96500 coulombs. It is also one mole of electrons.
If current of 0.2 Apms is passed through copper(ll) sulphate for tow hours, how
much copper do you get?
Write out the HALF equation
Cu2+ + 2e > Cu
Work out COULOMBS of electrons flowing
COULOMBS= CURRENT x time
Q= IT
TIME is 2x60x60 (times 60 makes MINUTES, times 60 again makes it
SECONDS)
Q= 0.2 x 7200= 1440 COULOMBS
Convert C into MOLES of ELECTRONS
MOLES= C/FARADAY
MOLES= 1440/96500
MOLES= 0.015
Work out SCALE factor
Cu2+ + 2e > Cu
For every TWO moles of electrons, there will be ONE Cu
Sf= Moles of PRODUCT/ moles of ELECTRONS
Sf= 1/2
4
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K?
Na ?
Li ?
H?
Mg ??
Ca ??
Al ??
Cl?
Br?
I?
F?
OH?
NO??
SO4 ??
CO3 ?? Answer: 1.30 recall the CHARGES of common IONS in this
specification
K+
Na +
Li +
H+
1
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,Mg 2+
Ca 2+
Al 3+
Cl-
Br-
I-
F-
OH-
NO3-
SO4 2-
CO3 2-
1.56 recall that one ? represents one ? of ?
One ? is ? ?. That is the amount of ? in ? ? of ?. Answer: 1.56 recall that one
faraday represents one mole of electrons
One FARADAY is 96500 COULOMBS. That is the amount of COULOMBS in
ONE MOLE of ELECTRONS.
1.57 calculate the amounts of the ? of the ? of ? ? and ? solutions
One ? is ? ?. It is also one ? of ?.
If ? of 0.2 ? is passed through copper(ll) sulphate for two ?, how much ? do
you get?
1) Write out the ? equation
(Cu2+) + 2e > Cu
2
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,2) Work out ? of electrons flowing
?= ? x time
?= ?T
? is 2x60x60 (times 60 makes ?, times 60 again makes it ?)
Q= 0.2 x 7200= 1440 ?
Convert ? into ? of ?
?= C/?
?= 1440/96500
?= 0.015
Work out ? factor
(Cu2+) + 2e > Cu
For every ? moles of electrons, there will be ? Cu
Sf= Moles of ?/ moles of ?
Sf= 1/2
Sf= a half
3) Work out ? of product using ? ?
so we do the ? of electrons times the ? ?
0.015x1/2= 0.0075 Moles of Cu
Convert ? into ?
Moles x Mr
3
APPHIA - Crafted with Care and Precision for Academic Excellence.
, 0.0075 x 63.5= 0.48g of copper Answer: 1.57 calculate the amounts of the
products of the electrolysis of molten salts and aqueous solutions
One faraday is 96500 coulombs. It is also one mole of electrons.
If current of 0.2 Apms is passed through copper(ll) sulphate for tow hours, how
much copper do you get?
Write out the HALF equation
Cu2+ + 2e > Cu
Work out COULOMBS of electrons flowing
COULOMBS= CURRENT x time
Q= IT
TIME is 2x60x60 (times 60 makes MINUTES, times 60 again makes it
SECONDS)
Q= 0.2 x 7200= 1440 COULOMBS
Convert C into MOLES of ELECTRONS
MOLES= C/FARADAY
MOLES= 1440/96500
MOLES= 0.015
Work out SCALE factor
Cu2+ + 2e > Cu
For every TWO moles of electrons, there will be ONE Cu
Sf= Moles of PRODUCT/ moles of ELECTRONS
Sf= 1/2
4
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