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UNIT I
ENERGY
F. EFFECT OF CONCENTRATION ON CELL EMF: THE NERNST EQUATION
Introduction:
When a voltaic cell operates, supplying electrical energy, the concentration of reactants decrease
and of the products increases. As time passes, the voltage drops steadily. Eventually it becomes zero, and
we say that the cell is “dead” At that point, the redox reaction taking place within the cell is at
equilibrium, and there is no driving force to produce a voltage
In the past lesson, we have dealt only with “standard” voltages, that is voltages when all gases are at
1 atm pressure and all species in aqueous solution are at a concentration of 1 M. When the concentration
of a reactant or product changes, the voltage changes as well. Qualitatively, the direction in which the
voltage shifts I readily predicted when you realize that cell voltage is directly related to reaction
spontaneity.
Learning Outcome:
1. To discuss the driving force in concentration cells.
2. To quantify the relationship between cell potential and cell concentration.
3. To Calculate the emf of a cell under a a non-standard state condition using Nernst’s equation
Cell Potentials and Free-Energy Changes for Cell Reactions
Review the following concepts:
 A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire
from a reducing agent in the other compartment.
 The “pull,” or driving force, on the electrons is called the cell potential (Ɛcell ), or the electromotive
force (emf) of the cell.
 The unit of electrical potential is the volt (abbreviated V), which is defined as 1 joule of work
per
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coulomb of charge transferred.
 Electromotive Force (emf): The force or electrical potential that pushes the negatively charged
electrons away from the anode ( electrode) and pulls them toward the cathode (+ electrode). It is also
called the cell potential (E) or the cell voltage.
The electric charge on 1 mol of electrons and is equal to 96,500 C/mol e is called faraday or
Faraday constant.
ΔG = nFE
where:
ΔG = Free-energy
n = number of moles of electrons transferred in the reaction
E = Cell potential
F = Faraday constant, 95, 46,500 Coulomb/mol e_
the electric charge on 1 mol of electron
ΔG0 = nFE0 where the redox reaction is at standard condition, T = 250; P = 1atm;
1 M, molar concentration
Effect of Concentration on Voltage
1. Voltage will increase if the concentration of a reactant is increased or that the product is
decreased. Either of these changes increases the driving behind the redox reaction, making it more
spontaneous.
2. Voltage will decrease if the concentration of a reactant is decreased or that the emf of a product is
increased. Either of these changes makes the redox less spontaneous.
The Nernst Equation
To obtain a quantitative relation between cell voltage and concentration, it is convenient to start
with the general expression for free energy change: ΔG = ΔG0 + RTln Q
Consider a redox reaction of the type:
aA(s) + bB(aq)  cC(aq) + dD(g)
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ΔG = ΔG0 + RT ln Q , where
Q
[C ]c ( PD ) d
, Q is the reaction quotient
[ A]c [ B]b
[C ]c ( PD ) d
NOTE:
where reactant A is solid phase
Q
[ B ]b
Remember that gases enter Q as their partial pressures in atmospheres. Species in water solution enter
as molar concentration. Pure liquids and pure solids do not appear in the expression for Q
where Q is the reaction quotient, this equation is known as Nernst Equation . At T = 298K, then the
equation can be rewritten as
E =
E =
E0 - 0.0257 x 2.303V log Q
n
E0 -- 0.0592 log Q
n
In this equation, Ecell is the cell voltage, Eocell is the standard voltage or standard cell potential, n is
the number of moles of electrons exchange in the reaction, and Q is the reaction quotient.
Notice that:
If Q>1, which means that the concentrations of products are high relative to those of

reactants, ln Q is positive and Ecell < Eocell.
If Q< 1(concentrations of products low relative to reactants), lnQ is negative and Ecell > Eocell.

• If Q = 1, as is the case under standard conditions, ln Q = 0 and Ecell = Eocell.
EXAMPLE No. 1:
1. Predict whether the following reaction would proceed spontaneously as written at 298 K:
Co(s) + Fe2+(aq)
Co2+(aq) + Co2+(aq) + Fe(s) given that
[Co2+] = 1.5 M and [Fe2+] = 0.68 M.
Strategy:
1. Because the reaction is not under standard-state conditions (concentrations are not 1 M).
2. Use the Nernst’s equation to calculate the emf, E of the hypothetical galvanic cell and determine the
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spontaneity of the reaction. The standard emf, (E0) can be calculated using the standard reduction
potential in Table 20.1.
3. Remember that solids do not appear in the reaction quotient (Q) term in the Nernst equation.
4. Note that 2 moles of electrons are transferred per mole of reaction, that is, n = 2
Solution:
The half-cell reaction are:
Anode(oxidation):
Cathode(reduction):
Co(s)
Fe2+ + 2e-
Co2+(aq) + 2eFe(s)
Because E is negative, the reaction is not spontaneous in the direction written
EXAMPLE No. 2:
In a certain experiment, considering it is a galvanic cell is found to have emf (Ecell) 0.54 V at 250C.
Suppose that [Zn2+] = 1.0 M and PH2 = 1.0 ATM. Calculate the molar concentration of H+.
Strategy:
1. The equilibrium that relates standard emf and nonstandard emf is the Nernst equation. The overall
cell reaction is Zn(s) + 2H+(M)
Zn2+(M) + H2(1.0 atm)
2. Given the emf of the cell, ( E), we apply the Nernst equation, to solve for [H+}.
3. Note that 2 mole of electrons are transferred per mole of reaction; that n = 2
Solution:
As we saw earlier, the standard emf E0 for the cell is 0.76V . From the
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Check:
The fact that the nonstandard-state emf (E) is given in the problem means that not all the reacting
species are in their standard-state concentration. Thus, because both Z n2+ ions and H2 gas are in their
standard states, [H+] is not 1M.
EXAMPLE No. 3
Consider a voltaic cell in which the following reaction occurs:
O2(g, 0.98atm) + 4H+(aq, pH =1.24) + 4Br-(aq, 0.15M)  2H2O + 2Br2(l)
a. Calculate E for the cell at 25oC.
Strategy:
1. Change pH to [H+] and find Q
2. Assign oxidation numbers, write the oxidation and reduction half-reactions, and
cancel electrons to find n
3. Find Eo ,( Eored, Eoox)
4. Substitute in the Nernst Equation
Solution:
1. pH = -log [H+]; 1.24 = -log [H+]
Q
[H+] = 10 -1.24 = 0.058 M
1
1

 1.8 x10 8
 4
4
4
( PO )[ H ] [ Br ]
(0.98)(0.058) (0.15)
 4
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2. O: 0  -2 (reduction); Br-: -1  0 (oxidation)
Half-reactions: O2 (g) + 4H+(aq) + 4e-  2H2O
2Br- (aq)  Br2 (l) + 2eThe oxidation half-reaction must be multiplied by 2 to cancel out the four electrons in
the reduction half-reaction. n =4
Eored, for O2 = 1. 23 V; Eoox for Br- = -1.09 V
Eocell = 1.23V + (-1.09V) = 0.14V
Ecell =
0.14V 
0.0257
ln( 1.8 x10 8 )
4
Ecell = 0.018V
b. When the voltaic cell is at 35oC, E is measured to be 0.039 V. What is Eo at 35oC?
Strategy:
Substitute the following into the Nernst Equation
T = 350C ; ECELL = 0.039 V
T = 35 + 273 = 308 K
Solution:
RT
ln Q
nF
(8.31J / mol.K )(308K )
0.039V  E 0 
ln( 1.8 x108 )
4
4(9.648 x10 J / mol.V )
E cell  E o cell 
E 0  0.039V  0.126V  0.165V
Use of the Nernst Equation to Determine Ion Concentrations
In chemistry, the most important use of the Nernst equation lies in the experimental
determination of the concentrations of ions in solution. Suppose you measure the cell voltage E
and know the concentration of all but one species in the two half-cells. It should then be
possible to calculate the concentrations of that species by using the Nernst equation.
EXAMPLE No. 4:
Consider the voltaic cell at 25oC in which the reaction is
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
It is found that the cell voltage is +0.56V when [Zn2+] = 0.85M and PH2 = 0.98 atm. What is
the pH in the H2-H half-cell?
Strategy:
1. Assign oxidation numbers, write oxidation and reduction half-reactions, and
cancel electrons to find n.
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2. Find Eo
3. Substitute into the Nernst equation for T= 25oC and find Q.
4. Write the Q expression and substitute given concentrations and pressures to find
[H+]. Change [H+] to pH.
Solution:
1. Oxidation numbers
Half-reaction
n
2. Eo
3. Q
Zn: 0  +2 oxidation; H+: +1  0
2H+ (aq) + 2e-  H2(g)
reduction
Zn(s)  2e- + Zn2+ (aq)
2 electrons cancel out so n =2
Eo = EooxZn + EoredH+ = 0.76V + 0 V = 0.76V or
Eo = EoredH+ - EooxZn = 0 V - (-0.76V) = 0.76V
E  E0 
0.0257
0.0257
ln Q;0.56V  0.76V 
ln Q
n
2
0.0257
ln Q  0.76V  0.56V ; ln Q  15.56
2
Q  5.72 x10 6
4. [H+]
Zn ( P
Q
H 
2
H2
)
 2
5. pH
 
;H

 (0.85)(0.988) 

6

 5.72 x10 
1/ 2
 3.83x10  4
pH   log 10 (3.83x10 4 )  3.42
ENRICHMENT ACTIVITY 4: (TO BE PASSED)
1. Consider the voltaic cell at 25oC in which the following reaction takes place:
MnO2(s) + 4H+(aq) + 2Br-(aq)  Br2(l) + Mn2+(aq) + 2H2O
a. Calculate Eo
b. Write the Nernst equation for the cell.
c. Calculate E at 25oC under the following conditions: [Mn2+] =0.60M; [Br-] =
0.83M; pH = 3.17
2. What is the emf of a galvanic cell consisting of a Cd2+/Cd half-cell and a Pt/H+/H2 half-cell if [Cd2+]
= 0.20 M, [H+] = 0.16 M and PH 2 = 0.80 atm?
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Electrolytic Cells
Learning Objective:
1. To describe the stoichiometry of electrolysis reactions.
In an electrolytic cell, a nonspontaneous redox reaction is made to occur by pumping electrical
energy into the system.
Quantitative Relationships
There is a simple relationship between the amount of electricity passed through an electrolytic
cell and the amounts of substances produced by oxidation or reduction at the electrodes. From
the balanced half-reactions
Ag+(aq) + e-  Ag (s)
Cu+ (aq) + 2e-  Cu (s)
Au3+(aq) + 3e-  Au (s)
You can deduce that
1 mol of e-  1 mol of Ag (107.9 g of Ag)
2 mol of e-  1 mol of Cu (63.55 g of Cu)
3 mol of e-  1 mol of Au (197.0 g Au)
Relation of this type, obtained from balanced half-equations, can be used in many
practical calculations involving electrolytic cells.
Table 3.0 Electrical Units
Quantity
Charge
Current
Potential
Power
Energy
Example:
Unit
Coulomb (C)
Ampere (A)
Volt (V)
Watt (W)
Joule (J)
Defining Relation
1C = 1 A.s = 1 J/V
1 A = 1 C/s
1 V = 1 J/C
1 W = 1 J/s
1 J = 1 V.C
Conversion Factors
1 mol e- = 9.648x104 C
1 kWh = 3.6x106 J
Chromium metal can be electroplated from an aqueous solution of potassium dichromate. The
reduction half-reaction is
Cr2O72- (aq) + 14 H+(aq) + 12 e-  2 Cr (s) + 7H2O
A current of 6.0 A and a voltage of 4.5V are used in the electroplating.
a. How many grams of chromium can be plated if the current is run for 48 minutes?
b. How long will it take to completely convert 215 mL of 1.25 M K 2Cr2O7 to elemental
chromium?
c. How many kilowatt-hours of electrical energy are required to plate 1.0 g of
chromium?
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Solution:
a. How many grams of chromium can be plated if the current is run for 48 minutes?
1. Since mass is asked for, you may assume that you have all the information to
convert the amount of electricity to moles of electrons. Moles of electrons
provide the bridge that connects the amount of electricity to the stoichiometry
of the chemical reaction. Use the following plan:
Amperes
coulombs
6.0 A = 6.0 C/s
C
1mole 
6.0 x(48 x60) sx
 0.179mole 
4
s
9.648 x10 C
mol e-
2. Convert mol e- to mass of Cr using the stoichiometry of the reaction
mol e-
mol Cr
mass Cr
2molCr 52 gCr
x
 1.55 g
12mole  1molCr
b. How long will it take to completely convert 215 mL of 1.25 M K2Cr2O7 to elemental
chromium?
0.179mole  x
1. Since V and M are given for K2Cr2O7 (thus for Cr2O72-), you have enough information to
convert to mol e- using the stoichiometry of the reaction
VxM
mol Cr2O72mol e(0.215 L) (1.25 mol/L) (12 mol e-/mol Cr2O72-) = 3.225 mol e2. Convert mol e- to coulombs
3.225mol e- x 9.648x104 C/ mol e- = 3.11x105 C
3. Convert coulomb to time. Recall 1 A = 1 C/s, 1 A = 1 C/s, Give: A =6.0 = 6.0 C/s
time 
3.11x10 5 C
 5.16 x10 4 s  14.4h
6.0C / s
c. How many kilowatt-hours of electrical energy are required to plate 1.0 g of
chromium?
1. Convert the mass of Cr to mol e- using Stoichiometry
massl Cr
mol Cr
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1molCr 12mole 
1.0 gCrx
x
 0.115mole 
52 g
2molCr
2. Find kWh by finding the energy in joules ( C x V) and then convert to kWh
mol e-
0.115mole  x
coulomb
kWh
9.468 x10 4 C
1J
1kWh
x 4.5Vx
x
 0.014kWh

1C.V 3.60 x10 6 J
1mole
Exercises for your practice:
1. An electrolytic cell produces aluminium from Al2O3 at the rate ten kilograms a day.
Assuming a yield of 100%,
a. How many moles of electrons must pass through the cell in one day?
b. How many amperes are passing though the cell?
c. How many moles of oxygen (O2) are being produced simultaneously?
2. The electrolysis of an aqueous solution of NaCl has the overall equation
2H2O + 2Cl
H2(g) + Cl2 (g) + 2OH- (aq)
During the electrolysis, 0.228 mol of electrons pass through the cell.
a. How many electrons does this represent?
b. How many coulombs does this represent?
c. Assuming 100% yield, what masses of H2 and Cl2 are produced
References:
1. Brown, L. S & Holmes, T. A. (2015). Chemistry for Engineering Students. CENGAGE Learning
2. Chang, R. (2010). Chemistry. McGraw-Hill Companies, Inc.
3. Masterton, William L, et al. (2018).Principles and Reactions: Chemistry for Engineering
Students. C&E Publishing, Inc.
4. Zumdahl, S.S, Zumdahl, S.A.& DeCoste, D.J. (2018) Chemistry. CENGAGE Learning
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