ELECTROCHEMISTRY

Thermodynamics has many applications other than expanding gases inside

steam engines. One of the most important areas of applied
thermodynamics is ELECTROCHEMISTRY.

Electrochemistry is an area of chemistry that deals with the inter-

conversion of electrical energy and chemical energy. It is typically investigated
through the use of electrochemical cells, systems that incorporate a redox reaction to
produce or utilize electrical energy.

There are two types of electrochemical cells:

• One type of cell does work by releasing free energy from a spontaneous
reaction (∆G < 0) to produce electricity. A battery houses this type of cell
(Voltaic or galvanic cell)
• The other type of cell does work by absorbing free energy from a source of
electricity to drive a nonspontaneous (∆G > 0). (Electrolytic cells).

Redox Reactions and Electrochemical Cells

Electrochemical processes are redox (oxidation-reduction) reactions in which

the energy released by a spontaneous reaction is (converted to electricity in which
electricity is used to drive a nonspontaneous chemical reaction (electrolysis) to redox
reactions, electrons are transferred from one substance to another.

The reaction between zinc metal and sulfur is an example of a redox reaction:
0 0 +2 -2 Zn + S → ZnS
RA OA

GEROA (Gain Electron, Reduction, Oxidizing Agent)

LEORA (Loss Electron, Oxidation, Reducing Agent)

Recall that the numbers above the elements are the oxidation numbers of
the elements. By assigning an oxidation number to each atom, we can see which
species was oxidized and which reduced and, from that, which is the oxidizing agent
and which is the reducing agent. The loss of electrons by an element during
oxidation is marked by an increase in the element's oxidation number. In reduction,
there is decrease in oxidation number resulting from a gain of electrons by an
element. In the preceding reaction Zn metal is oxidized and S is reduced. Keep in
mind three key points:
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 • Oxidation (electron loss) always accompanies reduction (electron gain). The
oxidizing agent is reduced, and the reducing agent is oxidized.
• The total number of electrons gained by the atoms/ ions of the oxidizing agent
always equals the total number lost by the atoms/ions of the reducing agent.
1

• Redox reaction can be defined as one in which the oxidation numbers of the
species change.

Rules for Assigning an Oxidation Number (O.N.)

1. The oxidation number of an element in the free or uncombined state is
always zero (including
2. The algebraic sum of the oxidation numbers of all the atoms in the formula of
a compound is zero.
3. The oxidation number of an ion is the same as the charge on the ion.
4. The sum of the oxidation numbers of the atoms in a polyatomic ion must be
equal to the charge on the ion.

Rules for Specific Atoms or Periodic Table Groups

1. Group IA elements are always +1
2. Group Il A are always + 2.
3. Hydrogen is usually +1 except in hydrides. 4. Oxygen is usually -2, except in
peroxides 5. Group VII A elements are -1.

GEROA (Gain Electron, Reduction, Oxidizing Agent)

LEORA (Loss Electron, Oxidation, Reducing Agent)

Example:

Use oxidation numbers to decide which of the following are redox reactions.
For redox reaction, identify the species that undergo the reduction and oxidation
reaction; and the oxidizing and reducing agent.
x =+4 x + 2( -2) = 0; x = x + +4 3( -2) = -2;

+2 -2 +4 -2 +2 -2 +4 -2

1. CaO(s) + CO2 (g) → CaCO3(s) not redox reaction CO3

+1 +5 -2 x = x + 5 3 (-2) = -1;
+1 -1 +1 -2 0 0 +5 -2

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 2. 4KNO3(s) → 2K2O(s) + 2N2(g) + 5O2 (g) redox reaction NO3
OA RA

Balancing Redox Reactions

We balance a redox reaction by making sure that the number of electrons lost by
the reducing agent equals the number of electrons gained by the oxidizing agent.
There are two methods used to balance redox equations: the oxidation number
method and the half-reaction

method — but only the second method will be discussed in this section. The
following steps are used in balancing a redox reaction by the half-reaction method.

1. Write the unbalanced equation for the reaction in ionic form.

2. Separate the equation into two half-reactions.
3. Balance each half-reaction for number and type of atoms and charges.
Balance the atoms other than O and H in each half-reaction separately.
4. For reactions in an acidic medium, add H2O to balance the O atoms and H+
to balance the H atom.
5. Add electrons (e-) on one side of each half-reaction to balance the charges. If
necessary, equalize the number of electrons in the two half-reactions by
multiplying one or both half-reactions by appropriate coefficients,
6. Add the two half-reactions and balance the final equation by inspection. The
electrons on both sides must cancel.
7. Verify that the equation contains the same type and numbers of atoms and
the same charges on both sides of the equation.

Example:

1. Balance the equation showing the oxidation of Fe +2 ions to Fe+3 ions by dichromate
ions in an acidic medium. The dichromate ions are reduced to Cr3+ ions.

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Step 1) Fe+2
6Fe +2++Cr
Cr2O
2O-2 →Fe+3 + 6Fe
+ 14H+→
7 7-2 Cr+3+3 + 2Cr+3 + 7H2O Overall Reaction

Step 2 - 3) Fe+2 →Fe+3

Cr2O7-2→ 2Cr+3

Step 4) Cr2O7-2 + 14H+→ 2Cr+3 + 7H2O

2x + 7(-2) = -2;
Fe+2 →Fe+3 + e- 6Fe+2 →6Fe+3 + 6e- x=6
Step 5) +6 -2 +1 -2 +6 -2
- -2 + +3
6e + Cr2O7 + 14H → 2Cr + 7H2O Cr2O7-2
(-2 + 14(+1) = 12) (2(3) + 7(2)(1) + 7(-2) = 6)
(2(6) + 7(-2)+ 14(+1) = 12)

6Fe+2 →6Fe+3 + 6e- Oxidation

Step 6) - -2
6e + Cr2O7 + 14H → 2Cr + 7H2O + +3
Reduction

Electrochemical Cells

Electrochemical cell is the experimental apparatus for generating electricity

through the use of a spontaneous redox reaction (∆G < 0). An electrochemical cell
is sometimes referred to as a galvanic cell or voltaic cell, after tile scientists Luigi
Galvani and Alessandro Volta, who constructed early version of this device.
The separation of half-reactions is the essential idea behind a voltaic ceil. The
components of each half-reaction are placed in a separate container, or half-cell,
which consist of one electrode dipping into an electrolyte solution. The two half-cells
are joined by the circuit, which consists of a wire and a salt bridge (the inverted U
tube in the figure). In order to measure the voltage generated by the cell, a
voltmeter is inserted in the path of the wire connecting the electrodes. A switch
closes or opens the circuit. The oxidation half-cell (anode compartment) is shown
on the left and the reduction half-cell (cathode compartment) on the right.

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Here are the key points about the Zn/Cu2+ voltaic cell:

1. The oxidation half-cell. The anode compartment consists of a zinc bar

(the anode) immersed in a Zn 2+ electrolyte (solution of ZnS0 4). The zinc
bar conducts the released electrons out of its half-cell.

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 2. The reduction half-cell. The cathode compartment consists of a copper bar
(the cathode) immersed in a Cu2+ electrolyte (solution of CuS04). Copper
metal conducts electrons into its half-cell.
3. Relative charges on the electrodes. The electrode charges are determined by
the source of electrons and the direction of electron flow through the circuit.
The electrons flow left to right through the wire to the cathode. In any voltaic
cell, the anode is negative and the cathode is positive.
4. The purpose of the salt bridge. The oxidation half-cell and the reduction
half-cell originally contain neutral solutions. If the half-cells do not remain
neutral, the resulting charge imbalance would stop cell operation. To avoid
this situation and enable the cell to operate, the two half- cells are joined by a
salt ridge, which act as a "liquid wire," allowing ions to flow through both
compartments and complete the circuit. Therefore, salt bridge maintains the
neutrality of the solutions.

The conventional notation for representing electrochemical cells is the cell

diagram (cell notation). If we assume that the concentrations of Zn 2+ and Cu2+
ions are

Zn(s) | Zn2+ (1M) || Cu2+ (1M) | Cu(s)

• The anode is written first, to the left of the double lines, while cathode is on
right.
• The single vertical line represents a phase boundary.
• The double vertical lines denote the salt bridge.
We can write the half- cell reactions as follows:

Anode (oxidation): Zn(s) Zn2+(1M) + 2e- Eᵒox = 0.76V

Cathode (reduction): Cu2+ (1M) + 2e- Cu(s) Eᵒred = 0.34V

Overall: Zn(s) + Cu2+ (1M) Zn2+ (1M) + Cu(s) Eᵒcell= 1.10V

Standard Reduction Potentials

For a reduction reaction at an electrode when all solutes are 1M and all gases
are at 1 atm, the voltage is called the standard reduction potential (EO). The larger
(more positive) the EO value, the greater the tendency for the substance to be
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reduced, and therefore the stronger its tendency to act as an oxidizing agent (gains
electrons more readily). The smaller (more negative) E O value, the greater the
tendency for the substance to be oxidized and act as reducing agent (loses electrons
more readily).
Let us consider Cu2+, H+, and Zn2+, three oxidizing agent present in the
voltaic cell. We can rank their relative oxidizing strengths by writing each half-
reaction as gain of electrons (reduction), with its corresponding standard electrode
potential:

Cu2+ gains two electrons more readily than I-1+, which gains them more readily
than Zn2+. In terms of strength as an oxidizing agent, therefore, Cu +2> H+ > Zn2+.
Therefore Cu2+ is the strongest oxidizing agents, and Zn is the strongest reducing
agent.

Example:

Arrange the following species in order of increasing strength as oxidizing agents

under standard-state conditions: Ce4+, O2, H2O2

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*For all half-reactions the concentration is 1M for dissolved species and the pressure is
atm for gases. These are standard state values.

Please refer to the links for the Table of Standard Reduction Potential

https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/
Reference_Tables
/Electrochemistry_Tables/P2%3A_Standard_Reduction_Potentials_by_Value

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Calculating the standard emf (Eo) of an Electrochemical Cell

In an electrochemical cell, electric current flows from one electrode to the other
(from anode to cathode) because there is a difference in electrical potential energy
between the electrodes. The difference in electrical potential between the anode and
cathode is measured by a voltmeter, and the reading (in volts) is called cell voltage.
However, two other terms, electromotive force or emf (E) and cell potential are also used
to denote cell voltage. If all solutes have a concentration of 1 M and all gases have a
pressure of 1 atm (standard conditions), the voltage difference between the two
electrodes of the cell is called the standard emf (EOcell). The standard emf is the sum of
the standard oxidation potential and the standard reduction potential for each half-
reaction.

E0cell = E0ox + E0red

We can use the sign of the emf of the cell to predict the spontaneity of redox
reaction. Under standard-state conditions for reactants and products, the redox
reaction is spontaneous in the forward direction if the standard emf of the cell is
positive. If it is negative, the reaction is spontaneous in opposite direction.

Example:

We can calculate the standard emf of the cell using a table of reduction
Potentials. As an example, consider the galvanic cell represented by the following
reaction:

Cu(s) + 2Ag+(aq) Cu2+(aq) + Ag(s)

Let’s break this reaction down into its half-reaction.

We can calculate the standard cell emf by adding the oxidation and reduction potentials.

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The standard oxidation potential for an anode half-reaction is equal in magnitude,
but of opposite sign to that of the reduction potential for the reverse reaction. For
example, [lie reverse reaction of the copper oxidation is:

reduction
2+ -
Cu (aq) + 2e Cu(s)

You can look up the standard reduction potential for this half-reaction:

Therefore, the standard oxidation potential for the oxidation of Cu Cu 2+ has the
opposite of the standard reduction potential above.

We can also look up the standard reduction potential for silver:

Finally, we can calculate the standard cell emf.

Spontaneity of Redox Reactions

There is a relationship between free energy change and cell emf:

Where, n = the number of moles of electrons transferred during the

redox reaction

F = the Faraday, which is the electrical charge contained in 1 mole of electrons or the
amount of energy required for the flow of one mole of electron.

1F = 96500 C/mol = 96500 J/V.mol

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Both n and F are positive quantifies, and we know that ∆G is negative for a spontaneous
process. Therefore, ecell is positive for a spontaneous process.
For reaction in which reactants and products are in their standard states, equation
becomes:

Example:

Predict whether a spontaneous reaction will occur when the following reactants
and products are in their standard states:

Fe2+ + Cr2O72- Fe3+ + Cr3+

6Fe+2 →6Fe+3 + 6e- Oxidation Eᵒox = -0.77V

6e- + Cr2O7-2 + 14H+→ 2Cr+3 + 7H2O Reduction Eᵒred = 1.33V
6Fe+2+ Cr2O7-2 + 14H+→ 6Fe+3 + 2Cr+3 + 7H2O Overall Reaction

Eᵒcell = Eᵒox + Eᵒred = -0.77V + 1.33V = 0.56 V (spontaneous, YES)

n = 6; F = 96500 J/V-mol
∆Gᵒ = -nFEᵒcell = -(6 mol)(96500 J/Vmol)(0.56V) = -324 240 J (spontaneous, YES)

Standard Cell Potential and the Equilibrium Constant

This equation shows that the relationship between standard free energy change
and standard emf is :

Using this relationship, we can relate the standard cell potential to the equilibrium
constant of the redox reaction.

Therefore, from the two equations we obtain

Solving for gives

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 We can simplify the above equation by
• Substituting the known value of 8.314 J/ mol.K for constant R
• Substituting the known value of 96 500 J/V.mol for constant F
• Substituting the standard temperature of 298 K for T
• Multiplying by 2.303 to convert from natural to common (base — 10) logarithms

Example:

Calculate ∆GO and the equilibrium constant at 250C for the reaction

2 Br ¯ (aq) + I2 (s) Br2 (l) + 2I ¯(aq)

Oxidation Eᵒox = -
2 Br ¯ (aq) Br2 (l) + 2e- 1.07V Reduction
2e- + I2 (s) 2I ¯(aq) Eᵒred= 0.53V

Eᵒcell = Eᵒox + Eᵒred = -1.07 + 0.53 = -

0.54V n = 2; F = 96500 J/V-mol

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∆Gᵒ = -nFEᵒcell = -(2)(96500)(-0.54) = 104220 J (non spontaneous)

0.0591 𝑉
log 𝐾 Eᵒcell =
𝑛

-0.54 V = 0 .0591 𝑉 log 𝐾

2

= log K
log K = -18.274
10 log K = 10 -18.274

K = 5.321 x 10-19

The Nernst Equation

We have only focused on redox reactions in which reactants and products are in
their standard states, but standard-state conditions are often difficult, and sometimes
impossible to maintain. Therefore. we need a relationship between cell emf and
concentrations of reactants and products under nonstandard-state conditions.

We encountered a relationship between the free energy change (AG) and the standard
free energy change (∆GO)

where Q is the reaction quotient.

We also know that

and

Substituting for ∆G and ∆GO in the first equation, we find'.

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Dividing both sides of the equation by —nF

The above equation is known as the Nernst equation. At 298 K, this equation can be
simplified by substituting R, T, and F into the equation

0.0257
lnQ
n

or, expressing Equation using the base-10 logarithm of Q

0.059 2
logQ
n

At equilibrium, there is no net transfer of electrons, so e = 0 and Q=K, where K is the

equilibrium constant.

Using the Nernst Equation to Predict the Spontaneity of a Redox Reaction

To predict the spontaneity of a redox reaction, we can use the Nernst equation to
calculate the cell emf, e. If the cell emf is positive, the redox reaction is spontaneous.
Conversely, if e is negative, the reaction is not spontaneous.

Example:

Calculate the cell emf and predict whether the following reaction is spontaneous at
25oC.

Zn(s) + 2H+ (1 x 10-4 M) Zn2+ (1.5 M) + H2 (1 atm)

Zn(s) Zn2+ (1.5 M) + 2e- Oxidation Eᵒox = 0.76V

2e- + 2H+ (1 x 10-4 M) H2 (1 atm) Reduction Eᵒred = 0 V

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Eᵒcell = Eᵒox + Eᵒred = 0.76 + 0 = 0.76 V
n= 2

1.51
Q= = 1.5 x 10 8
(1 ×10 −4 )2

0.059 2
logQ
n

E = 0.76V - 0 .
log (1.5 x 108) = 0.518 V
Using the Nernst equation to Calculate Concentration

Recall that the reaction quotient Q equals the concentrations of the product raised
to the power of their stoichiometric coefficients divided by the concentrations of the
reactants raised to the power of their stoichiometric coefficients.

Since Q is in the Nernst equation, if the cell emf is measured and the standard cell
emf is calculated, we can determine the concentration of one of the components if
the concentrations of the other components are known.

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