Electrochemistry Part IV:
Spontaneity & Nernst Equation
Jespersen Chap. 20 Sec 4 & 5
Skipping Sec 6 & 8
Dr. C. Yau
Spring 2014
1
Spontaneity of Reaction
We know to have a spontaneous rxn…
E>0
ΔG < 0
How are these two related?
ΔG = - n FEcell where n = moles of eF = Faraday's constant
ΔGo = - n FEocell
4 C/mol e9.65x10
(under standard
(remember 1 V = 1J/C)
conditions)
2
Example 20.7 p. 937
Calculate ΔGo for the reaction, given that its
standard cell potential is 0.320 V at 25oC.
NiO2 (s) + 2Cl(aq) + 4H+ (aq)
 Cl2 (g) + Ni2+ (aq) + 2H2O (l)
ΔGo = - n FEocell
F = 9.65x104 C/mol e(1 V = 1J/C, so Eo = 0.320 J/C)
How do we figure out what n is?
Ans. -61.8 kJ
Do Pract Exer 13 & 14 p. 938
3
Calculating K from Cell Potential
We know
ΔGo = - n FEocell
We also know ΔGo = - RT ln K (Chap. 19)
so
- n FEo = - RT ln K
Example 20.8 p. 789
Calculate K for the reaction in Example 20.8.
NiO2 (s) + 2Cl(aq) + 4H+ (aq)
 2Cl2 (g) + Ni2+ (aq) + 2H2O (l)
Collect all the constants we need.
Do Pract Exer 15 & 16 p. 939
4
Derivation of the Nernst Eqn
What happens when it is not under standard
conditions?
In the previous chapter on therm odynam ic s, w e know ...
G = G
o
+ R T lnQ
ΔG = - n FEcell
ΔGo = - n FEocell
0
-n F E cell = -n F E cell + R T ln Q
Divide both sides of eqn by (-nF) we get...
E cell = E
0
cell

RT
nF
ln Q
Nernst Equation
5
Common simplified version of the Nernst Equation for
25.0 oC:
E cell = E cell 
RT

RT
0
E cell = E
0
cell
0
E cell = E
E cell = E
0
c ell
E cell = E
0
cell
(2.303 log) Q
1  RT

2.303

 log Q
n F

-1
-1
1  (8 .314 J m ol K )(298.15 K )(2.303) 
 
 log Q
4
-1
n
9.6 5x10 C m ol


1
0.0 592 J C
-1
log Q
n

E cell = E cell 
0
nF
nF
E cell = E cell 
0
cell
ln Q
1
0.0592 V lo g Q
n
0.0592 V
n
This version of Nernst
Eqn will be given also,
but remember it’s only
for 25.0oC
log Q
6
Nernst Equation
Eo is the cell potential under standard conditions (for
aqueous soln, 1M)
What if it is not 1 M?
E cell = E
0
cell

0.0592 V
log Q
This type of quest will
be on your final exam.
n
Example 20.9 p. 940
Suppose a galvanic cell employs the following:
Ni2+ + 2e-  Ni
Eo = - 0.25 V
Cr3+ + 3e-  Cr
Eo = - 0.74 V
Calculate the cell potential when [Ni2+] = 4.87x10-4M
and [Cr3+] = 2.48x10-3 M
Ans. +0.44 V
7
Example 20.10 p. 941
The rxn of tin metal with acid can be written as
Sn (s) + 2H+ (aq)  Sn2+ (aq) + H2 (g)
Calculate the cell potential
(a) when the system is at standard state.
Ans. +0.02 V
(b) when the pH = 2.00
Ans. -0.16V
(c) when the pH is 5.00.
Assume that [Sn2+] = 1.00 M and the partial
pressure of H2 is also 1.00 atm.
Do Pract Exer 17, 18, 20 p. 942
8
What we are skipping in Chap. 20:
pp. 943-951
Concentration from E Measurements
Sec 20.6 Electricity
Batteries: Lead Storage Batteries
Zinc-Manganese Dioxide Cells (LeClanche cell)
Nickel-Cadmium Rechargeable Batteries
Nickel-Metal Hydride Batteries
Lithium Batteries
Lithium Ion Cells
Fuel Cells
Photovoltaic Cells
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