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Transcript
```GIBB'S FREE ENERGY AND
ELECTRIC CELLS
SUPA CHEMISTRY
MRS. PAPARELLA
SPRING 2016
DG = -n F
o
(need
to
memorize)
E
• DG = Gibb’s Free Energy
• n = number of moles of electrons
exchanged in reaction
• F = Faraday’s Constant = 96,500
Coulombs/mole of e- Given on Test
• Eo
= Cell Potential measured in
Volts; Also referred to as emf or
Electromotive Force.
• Voltage = Work/Charge
• 1 V= 1Joule/Coulomb of Charge
• A Volt describes how much energy it
takes to move 1 coulomb of charge
•
These problems will use KJ for DG that
need to be changed back to Joules ( x
by 1000) It’s never easy!
HOW DOES E CELL RELATE TO GIBB'S FREE
ENERGY?
Xrzhby7bsJw
• Spontaneous reactions have
• a + E value ( + Volts) and
• a – DG value.
• Nonspontaneous Reactions have a +
DG value and a – E value (- Volts)
CALCULATE THE GIBB'S FREE ENERGY IN A ZINC /COPPER +2
ELECTROCHEMICAL CELL WHICH HAS A VOLTAGE OF 1.1 V
• DG = -n F E How many e- are transferred?
• 1.1 V = 1.1 J/C
• Plug and Chug!
• DG = -2 mol e- x 96,500 C/mol e-x 1.1J/C
• DG = - 212,300 J or -212.3 KJ
2 mol e-
YOU CAN ALSO FIND THE CELL POTENTIAL IF YOU ARE
GIVEN THE EQUATION AND THE FREE ENERGY
INFORMATION.
• 2Al(s) + 3Cu2+(aq)-->2Al3+(aq) + 3Cu(s)
• D G = -1152.2 KJ
• Question: How many moles of e- are involved?
• Answer: 6 moles of e• Now you can proceed!
REMEMBER TO CHANGE KJ TO JOULES FOR DELTA G
• DG = -n F E
• -1,152,200 J
= - 6moles of e- x 96,500 C/ mole of e- x E
Solve for E (Cell Potential)
-1,152,200 J/ -579,000 C= 1.99 J/C or 1.99 V
ALTERNATIVE METHOD
• 1 Faraday is equal to 96.5KJ / V mol e- This is also given on the test
• Compare with 96,500 C /mol e• So, if you know the D G and the # of mol of e- you can find the cell voltage by DG
= -n F E
•
•
•
•
Using the previous problem data:
1,152.2KJ = -6 mol e- x 96.5KJ/Vmol e- x E and solve for E.
1,152.2KJ / 579 KJ/V =
E = 1.99 V
E
BOOK PRACTICE: P 764 20.41
• You will first have to calculate the Cell Potential using the Eo Red half-reaction
information given. Remember to change the sign if the substance is oxidized!
For this set of problems, remember if something is oxidized its charge goes up!
So Fe2+ is going to be oxidized to Fe 3+.
You will be doing 3 separate problems.
```
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