[ Visit http://www.wewwchemistry.com ] This example uses the Nernst equation to illustrate how changes in reactant or product concentration (effected by dilution) affect cell potentials.

1. I The Nernst Equation
 Reported cell potentials are typically measured under standard conditions:
 Solutes in aqueous solutions have a concentration of 1.0 mol dm–3
 Gaseous reactants or products have a pressure of 1 atm
 Measurements are taken at 298 K
 Galvanic cells seldom operate under standard conditions. Ecell measured under non-
standard conditions is not equal to (E— )cell, the cell potential measured under
o
standard conditions.
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2.  It has been determined that cell potentials are related to concentrations of reactants
and products, and to temperature via the Nernst equation, as follows:
Ecell = (E— )cell – ⎜ RT ⎟ ln Q
o ⎛ ⎞
⎜ nF ⎟
⎝ ⎠
where R is the gas constant (8.31 J K–1 mol–1)
T is the temperature (K)
€
n is the number of moles of electrons
transferred between oxidising and reducing
agents
F is the Faraday’s constant (96500 C mol–1)
Q is the reaction quotient, which is based on
the cell reaction
 This equation gives the Ecell measured under non-standard conditions.
 The Nernst equation is sometimes expressed in terms of base 10 logarithm. For a
cell at 298 K, the above equation becomes:
Ecell = (E— )cell – ⎜ 0.0592 ⎟ log10 Q
o ⎛ ⎞
⎜
⎝ n ⎟⎠
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€

3. II Using the Nernst Equation to Illustrate the Effect of Dilution on (E— )cell
o
 Consider the following galvanic cell set up at 298 K:
Al(s) | Al3+(aq) || Ni2+(aq) | Ni(s)
Anode : Al(s) → Al3+(aq) + 3e–
Cathode : Ni2+(aq) + 2e– → Ni(s)
Overall cell reaction : 2Al(s) + 3Ni2+(aq) → 2Al3+(aq) + 3Ni(s)
 Under standard conditions, [Al3+] and [Ni2+] are both 1.00 mol dm–3,
(E— )cell = (E— )reduction – (E— )oxidation
o o o
= –0.25 – (–1.66)
= +1.41 V
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4. Scenario 1:
 How does (E— )cell change when water is added to the Ni2+/Ni half cell such that [Ni2+]
o
is decreased ten times? (All measurements are taken at 298 K.)
— )cell – ⎛ 0.0592 ⎞ log10 Q
o
Ecell = (E ⎜
⎜ n ⎟
⎟
⎝ ⎠
3+ 2
+1.41 – ⎜ 0.0592 ⎟ log10 [Al ]
⎛ ⎞
= 6 ⎟
€
⎜
⎝ ⎠ [Ni2+ ]3
+1.41 – ⎜ 0.0592 ⎟ log10 1
⎛ ⎞
=
⎜ 6 ⎟ ⎞3
⎜ 1 ⎟
⎝ ⎠ ⎛
€ €
⎜ 10 ⎟
⎝ ⎠
= +1.41 – 0.0296
€
= +1.38 V €
Note:
Six electrons are tranferred during the oxidation of Al and the reduction Ni2+.
Thus, n = 6.
 (Q > 1) Ecell decreases.
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5. Scenario 2:
 How does (E— )cell change when water is added to the Al3+/Al half cell such that [Al3+]
o
is decreased ten times? (All measurements are taken at 298 K.)
— )cell – ⎛ 0.0592 ⎞ log10 Q
o
Ecell = (E ⎜
⎜ n ⎟
⎟
⎝ ⎠
3+ 2
+1.41 – ⎜ 0.0592 ⎟ log10 [Al ]
⎛ ⎞
= 6 ⎟
€
⎜
⎝ ⎠ [Ni2+ ]3
2
⎛
⎜ 1 ⎞
⎟
= € 0.0592 ⎞ €
⎛ ⎜ 10 ⎟
+1.41 – ⎜ ⎟ log10 ⎝ ⎠
⎜
⎝6 ⎟⎠ 1
= +1.41 – (–0.0197)
= +1.43 V
€ €
 (Q < 1) Ecell increases.
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6. Scenario 3:
 How does (E— )cell change when water is added to both half cells such that [Al3+] and
o
[Ni2+] are each decreased by ten times? (All measurements are taken at 298 K.)
— )cell – ⎛ 0.0592 ⎞ log10 Q
o
Ecell = (E ⎜
⎜ n ⎟
⎟
⎝ ⎠
3+ 2
+1.41 – ⎜ 0.0592 ⎟ log10 [Al ]
⎛ ⎞
= 6 ⎟
€
⎜
⎝ ⎠ [Ni2+ ]3
2
⎛
⎜ 1 ⎞
⎟
€ 0.0592 ⎞ €
⎛ ⎜ 10 ⎟
= +1.41 – ⎟ log10
⎜ ⎝ ⎠
6 ⎟
⎜
⎝ ⎠ ⎛
1 ⎞
3
⎜ ⎟
⎜ 10 ⎟
⎝ ⎠
= +1.41 – 0.00987
€
= +1.40 V €
 (Resulting Q > 1) Ecell decreases, albeit slightly.
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7. Conclusion
In general, any change to the cell that increases Q decreases Ecell, while any change
that decreases Q will increase Ecell. Thus, adding reactant or removing product
increases Ecell. while removing reactant or adding product decreases Ecell.
In the above example,
 when reactant concentration ([Ni2+]) is smaller than 1 mol dm–3, Ecell is less than
(E— )cell (Scenario 1).
o
 when product concentration ([Al3+]) is smaller than 1 mol dm–3, Ecell is more than
(E— )cell (Scenario 2).
o
 If concentrations of both products and reactants are decreased in the cell, whether
Ecell is more or less than (E— )cell depends on the resulting Q:
o
 Q > 1, Ecell is less than (E— )cell (Scenario 3)
o
 Q = 1, there is no change to (E— )cell, i.e. Ecell = (E— )cell
o o
 Q < 1, Ecell is more than (E— )cell
o
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