Nernst Equation

Learning Objectives
Explain and distinguish the cell potential and standard cell potential.
Calculate cell potentials from known conditions (Nernst Equation).
Calculate the equilibrium constant from cell potentials.

Nernst Equation
Electrochemistry deals with cell potential as well as energy of chemical reactions. The energy of a chemical system drives the
charges to move, and the driving force gives rise to the cell potential of a system called galvanic cell. The energy aspect is also
related to the chemical equilibrium. All these relationships are tied together in the concept of the Nernst equation.
Walther H. Nernst (1864-1941) received the Nobel prize in 1920 "in recognition of his work in thermochemistry". His
contribution to chemical thermodynamics led to the well known equation correlating chemical energy and the electric potential of a
galvanic cell or battery.

Electric Work and Gibb's Free Energy

Energy takes many forms: mechanical work (potential and kinetic energy), heat, radiation (photons), chemical energy, nuclear
energy (mass), and electric energy. A summary is given regarding the evaluation of electric energy, as this is related to
electrochemistry.
Electric Work
Energy drives all changes including chemical reactions. In a redox reaction, the energy released in a reaction due to movement of
charged particles gives rise to a potential difference. The maximum potential difference is called the electromotive force (EMF),
E, and the maximum electric work W is the product of charge q in Coulomb (C), and the potential DE in Volt (= J / C) or EMF.
W J = q DE C J/C (units)

Note that the EMF DE is determined by the nature of the reactants and electrolytes, not by the size of the cell or amounts of
material in it. The amount of reactants is proportional to the charge and available energy of the galvanic cell.
Gibb's Free Energy
The Gibb's free energy DG is the negative value of maximum electric work,
ΔG = −W (1)

= −q ΔE (2)

A redox reaction equation represents definite amounts of reactants in the formation of also definite amounts of products. The
number (n) of electrons in such a reaction equation is related to the amount of charge transferred when the reaction is completed.
Since each mole of electron has a charge of 96485 C (known as the Faraday's constant, F),
q = nF

and,
ΔG = −nF ΔE

At standard conditions,
∘ ∘
ΔG = −nF ΔE

The General Nernst Equation

The general Nernst equation correlates the Gibb's Free Energy DG and the EMF of a chemical system known as the galvanic cell.
For the reaction
a A+bB ⇌ c C +dD

and

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 c d
[C ] [D]
Q =
a b
[A] [B]

It has been shown that

∘
DG = DG + RT ln Q

and
DG = −nF DE

Therefore
∘
−nF DE = −nF DE + RT ln Q

where R, T, Q and F are the gas constant (8.314 J mol-1 K-1), temperature (in K), reaction quotient, and Faraday constant (96485 C)
respectively. Thus, we have
c d
RT [C ] [D]
∘
DE = DE − ln
a b
nF [A] [B]

This is known as the Nernst equation. The equation allows us to calculate the cell potential of any galvanic cell for any
concentrations. Some examples are given in the next section to illustrate its application.
It is interesting to note the relationship between equilibrium and the Gibb's free energy at this point. When a system is at
equilibrium, DE = 0, and Qeq = K. Therefore, we have,
c d
RT [C ] [D]
∘
DE = ln , (for equilibrium concentrations)
a b
nF [A] [B]

Thus, the equilibrium constant and DE° are related.

The Nernst Equation at 298 K
At any specific temperature, the Nernst equation derived above can be reduced into a simple form. For example, at the standard
condition of 298 K (25°), the Nernst equation becomes
c d
0.0592 V [C ] [D]
∘
DE = DE − log
a b
n [A] [B]

Please note that log is the logarithm function based 10, and ln, the natural logarithm function.
For the cell
2+ +
Zn | Zn || H |H | Pt
2

we have a net chemical reaction of

+ 2+
Zn +2 H → Zn +H
(s) 2 (g)

and the standard cell potential DE° = 0.763.

If the concentrations of the ions are not 1.0 M, and the H
2
pressure is not 1.0 atm, then the cell potential DE may be calculated
using the Nernst equation:
2+
0.0592 V P(H )[ Zn ]
∘ 2
DE = DE − log
n + 2
[H ]

with n = 2 in this case, because the reaction involves 2 electrons. The numerical value is 0.0592 only when T = 298 K. This
constant is temperature dependent. Note that the reactivity of the solid Zn is taken as 1. If the H pressure is 1 atm, the term P(H ) 2 2

may also be omitted. The expression for the argument of the log function follows the same rules as those for the expression of
equilibrium constants and reaction quotients.
Indeed, the argument for the log function is the expression for the equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, DE = 0.00 and the expression becomes an equilibrium constant K, which bears the following
relationship:

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 ∘
nDE
log K =
0.0592

where DE° is the difference of standard potentials of the half cells involved. A battery containing any voltage is not at equilibrium.
The Nernst equation also indicates that you can build a battery simply by using the same material for both cells, but by using
different concentrations. Cells of this type are called concentration cells.

Example 1
Calculate the EMF of the cell
2+ 2+
Zn | Zn (0.024 M) || Zn (2.4 M) | Zn
(s) (s)

Solution
2+ −
Zn (2.4 M) + 2 e → Zn Reduction

2+ −
Zn → Zn (0.024 M) + 2 e Oxidation
–––––––––––––––––––––––––––––––––––––––––––
2+ 2+ ∘
Zn (2.4 M) → Zn (0.024 M), DE = 0.00 ← Net reaction

Using the Nernst equation:

0.0592 0.024
DE = 0.00 − log (3)
2 2.4

= (−0.296)(−2.0) (4)

= 0.0592 V (5)

DISCUSSION
Understandably, the Zn ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to
2+

0.0592 V. From here, you can also calculate the energy of dilution.
If you write the equation in the reverse direction,
Zn
2+
(0.024 M) → Zn
2+
(2.4 M) ,
its voltage will be -0.0592 V. At equilibrium concentrations in the two half cells will have to be equal, in which case the
voltage will be zero.

Example 2
Show that the voltage of an electric cell is unaffected by multiplying the reaction equation by a positive number.
Solution
Assume that you have the cell
2+ +
Mg | Mg || Ag | Ag

and the reaction is:

+ 2+
Mg + 2 Ag → Mg + 2 Ag

Using the Nernst equation

2+
0.0592 [ Mg ]
∘
DE = DE − log
2 + 2
[ Ag ]

If you multiply the equation of reaction by 2, you will have

+ 2+
2 Mg + 4 Ag → 2 Mg + 4 Ag

Note that there are 4 electrons involved in this equation, and n = 4 in the Nernst equation:
2+ 2
0.0592 [ Mg ]
∘
DE = DE − log
4 + 4
[ Ag ]

which can be simplified as

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 2+
0.0592 [ Mg ]
∘
DE = DE − log
2 + 2
[ Ag ]

Thus, the cell potential DE is not affected.

Example 3
The standard cell potential dE° for the reaction
2+ 2+
Fe + Zn → Zn + Fe

is -0.353 V. If a piece of iron is placed in a 1 M Zn 2+

solution, what is the equilibrium concentration of Fe 2+
?
Solution
The equilibrium constant K may be calculated using
∘
(nD E )/0.0592
K = 10 (6)

−11.93
= 10 (7)
−12
= 1.2 × 10 (8)
2+ 2+
= [F e ]/[Zn ] (9)

Since [Zn 2+
, it is evident that
] =1 M

] = 1.2E-12 M.
2+
[ Fe

Example 4

From the standard cell potentials, calculate the solubility product for the following reaction:
+ −
AgCl → Ag + Cl

Solution
There are Ag and AgCl involved in the reaction, and from the table of standard reduction potentials, you will find:
+

− − ∘
AgCl + e → Ag + Cl , E = 0.2223 V (1)

Since this equation does not contain the species Ag , you need, +

+ − ∘
Ag +e → Ag, E = 0.799 V (2)

Subtracting (2) from (1) leads to,

+ − ∘
AgCl → Ag + Cl DE = −0.577

Let Ksp be the solubility product, and employ the Nernst equation,
−0.577
log Ksp = = −9.75 (10)
0.0592
−9.75 −10
Ksp = 10 = 1.8 × 10 (11)

This is the value that you have been using in past tutorials. Now, you know that Ksp is not always measured from its solubility.

Questions
1. In the lead storage battery,
Pb | PbSO | H SO | PbSO , PbO | Pb
4 2 4 4 2

would the voltage change if you changed the concentration of H 2

SO
4
? (yes/no)
2. Choose the correct Nernst equation for the cell
Zn
(s)
| Zn
2+
|| Cu
2+
| Cu
(s)
.
2+
[ Zn ]
A. DE = DE ∘
− 0.0296 log(
2+
)
[ Cu ]

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 2+
[ Cu ]
B. DE = DE
∘
− 0.0296 log(
2+
)
[ Zn ]

Zn
C. DE = DE
∘
− 0.0296 log( )
Cu

Cu
D. DE = DE ∘
− 0.0296 log( )
Zn

3. The standard cell potential DE° is 1.100 V for the cell,

Zn | Zn
2+
|| Cu
2+
| Cu .
(s) (s)

If [Zn ] = 0.01 M, and [C u ] = 1.0 M, what is DE or EMF?

2+ 2+

4. The logarithm of the equilibrium constant, log K, of the net cell reaction of the cell
2+ 2+ ∘
Zn | Zn || Cu | Cu DE = 1.100 V
(s) (s)

is
A. 1.100 / 0.0291
B. -1.10 / 0.0291
C. 0.0291 / 1.100
D. -0.0291 / 1.100
E. 1.100 / 0.0592

Solutions
1. Answer ... Yes!
Hint...
The net cell reaction is
− +
Pb + PbO + 2 HSO +2 H → 2 PbSO +2 H O
2 4 4 2

and the Nernst equation

0.0592 1
DE = DE
∘
−( ) log .
2 − 2 + 2
[ HSO ] [H ]
4

2. Answer ... A
Hint...
The cell as written has
Reduction on the right: Cu + 2 e → Cu
2+ −

Oxidation on the left: Zn → Zn + 2 e

2+ −

Net reaction of cell is Zn(s) + Cu → Cu(s) + Zn

2+ 2+

3. Answer ... 1.159 V

Hint...
A likely wrong result is 1.041 V.
2+
0.059 [ Zn ]
The term that modifies DE is − ( ) log
2+
(n = 2 in this case).
n [ Cu ]

Understandably, if the concentration of Zn 2+

is low, there is more tendency for the reaction,
Zn → Zn
2+
+2 e
−
.
4. Answer ... A
Hint...
Use the Nernst equation in the form
2+
[ Zn ]
0 = 1.100 − 0.0296 log( )
2+
[ Cu ]

The Nernst equation is useful for the determination of equilibrium constants.

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Understanding is the key. Take time to understand it; there is no point in rushing.

Contributors and Attributions

Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)

Nernst Equation is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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