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Chapter 18

Electrochemistry

Section 18.1

Balancing Oxidation–Reduction Equations

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(18.1) Balancing oxidation–reduction equations
(18.2) Galvanic cells
(18.3) Standard reduction potentials
(18.4) Cell potential, electrical work, and free energy
(18.5) Dependence of cell potential on concentration
(18.6) Batteries
(18.7) Corrosion
(18.8) Electrolysis
(18.9) Commercial electrolytic processes

Chapter 18

Table of Contents

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Review of Terms

Oxidation - Loss of electrons
Reduction - Gain of electrons
Reducing agent - Electron donor
Oxidizing agent - Electron acceptor

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Section 18.1

Balancing Oxidation–Reduction Equations

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Review of Terms (Continued)

Electrochemistry: Study of the interchange of chemical and electrical energy
Oxidation–reduction (redox) reaction

Involves a transfer of electrons from the reducing agent to the oxidizing agent

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Section 18.1

Balancing Oxidation–Reduction Equations

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Half-Reaction Method

Overall reaction is split into two half-reactions

One involves oxidation, and the other involves reduction

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Unbalanced equation for the oxidation–reduction reaction between cerium(IV) ion and tin(II) ion

Reduction

Oxidation

Section 18.1

Balancing Oxidation–Reduction Equations

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Problem-Solving Strategy - Half-Reaction Method (Acidic Solution)

Write separate equations for the oxidation and reduction half-reactions
For each half-reaction:

Balance all the elements except H and O
Balance O using H₂O
Balance H using H⁺
Balance the charge using electrons

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Section 18.1

Balancing Oxidation–Reduction Equations

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Problem-Solving Strategy - Half-Reaction Method (Acidic Solution) (Continued)

If necessary, multiply one or both balanced half-reactions by an integer

Helps equalize the number of electrons transferred in the two half-reactions

Add the half-reactions, and cancel identical species
Check that the elements and charges are balanced

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Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Balancing Oxidation–Reduction Reactions (Acidic)

Potassium dichromate (K₂Cr₂O₇) is a bright orange compound that can be reduced to a blue-violet solution of Cr³⁺ ions

Under certain conditions, K₂Cr₂O₇ reacts with ethanol (C₂H₅OH) as follows:

Balance this equation using the half-reaction method

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution

The reduction half-reaction is:

Chromium is reduced from an oxidation state of +6 in Cr₂O₇^2– to one of the +3 in Cr³⁺

The oxidation half-reaction is:

Carbon is oxidized from an oxidation state of –2 in C₂H₅OH to +4 in CO₂

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 1)

Balancing all elements except hydrogen and oxygen in the first half-reaction

Balancing oxygen using H₂O

Balancing hydrogen using H⁺

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 2)

Balancing the charge using electrons

Next, we turn to the oxidation half-reaction

Balancing carbon

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 3)

Balancing oxygen using H₂O

Balancing hydrogen using H⁺

We then balance the charge by adding 12e^– to the right side

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 4)

In the reduction half-reaction, there are 6 electrons on the left-hand side, and there are 12 electrons on the right-hand side of the oxidation half-reaction

Thus, we multiply the reduction half-reaction by 2 to give:

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 5)

Adding the half-reactions and canceling identical species

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.1 - Solution (Continued 6)

Check that elements and charges are balanced

Section 18.1

Balancing Oxidation–Reduction Equations

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Exercise

Balance the following oxidation–reduction reaction that occurs in acidic solution

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Section 18.1

Balancing Oxidation–Reduction Equations

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Problem-Solving Strategy - Half-Reaction Method (Basic Solution)

Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H⁺ ions were present
To both sides of the equation obtained above, add a number of OH^– ions that is equal to the number of H⁺ ions

We want to eliminate H⁺ by forming H₂O

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Section 18.1

Balancing Oxidation–Reduction Equations

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Problem-Solving Strategy - Half-Reaction Method (Basic Solution) (Continued)

Form H₂O on the side containing both H⁺ and OH^– ions, and eliminate the number of H₂O molecules that appear on both sides of the equation
Check that elements and charges are balanced

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Section 18.1

Balancing Oxidation–Reduction Equations

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Critical Thinking

When balancing redox reactions occurring in basic solutions, the text instructs you to first use the half-reaction method as specified for acidic solutions

What if you started by adding OH^– first instead of H⁺?
What potential problem could there be with this approach?

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Balancing Oxidation–Reduction Reactions (Basic)

Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores

An aqueous solution containing cyanide ion is often used to extract the silver using the following reaction that occurs in basic solution:

Balance this equation using the half-reaction method

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Solution

Balance the equation as if H⁺ ions were present

Balance the oxidation half-reaction

Balance carbon and nitrogen

Balance the charge

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Solution (Continued 1)

Balance the reduction half-reaction:

Balance oxygen

Balance hydrogen

Balance the charge

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Solution (Continued 2)

Multiply the balanced oxidation half-reaction by 4

Add the half-reactions, and cancel identical species

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Solution (Continued 3)

Add OH^– ions to both sides of the balanced equation to eliminate the H⁺ ions

We need to add 4OH^– to each side

4H₂O(l)

Section 18.1

Balancing Oxidation–Reduction Equations

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Interactive Example 18.2 - Solution (Continued 4)

Eliminate as many H₂O molecules as possible

Check that elements and charges are balanced

Section 18.1

Balancing Oxidation–Reduction Equations

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Galvanic Cell

Device in which chemical energy is changed to electrical energy
Uses a spontaneous redox reaction to produce a current that can be used to do work

Reaction occurs at the interface between the electrode and the solution where electron transfer is facilitated

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Section 18.2

Galvanic Cells

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Figure 18.2 - Galvanic Cells

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Galvanic cells can contain a salt bridge

Galvanic cells can contain a porous-disk connection

Section 18.2

Galvanic Cells

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Galvanic Cell - Components

A salt bridge or a porous disk is used to permit ions to flow without extensive mixing of the solutions

Salt bridge: Contains a strong electrolyte in a U-tube
Porous disk: Contains tiny passages that allow hindered flow of ions

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Section 18.2

Galvanic Cells

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Galvanic Cell - Components (Continued)

Anode: Electrode compartment where oxidation occurs
Cathode: Electrode compartment where reduction occurs

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Section 18.2

Galvanic Cells

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Cell Potential (E_cell)

Pull or driving force on electrons
Termed as the electromotive force (emf) of the cell
Volt (V): Unit of electrical potential

Defined as 1 joule of work per coulomb of charge transferred

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Section 18.2

Galvanic Cells

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Measuring Cell Potential

Use a voltmeter

Voltmeter: Draws current via a known resistance
Maximum cell potential can be ascertained by measuring it under zero current

Potentiometer: Variable voltage device, which is powered by an external source, inserted in opposition to the cell potential

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Section 18.2

Galvanic Cells

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Figure 18.4 - A Digital Voltmeter

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Section 18.2

Galvanic Cells

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Standard Reduction Potentials

E°values corresponding to reduction half-reactions with all solutes at 1 M and all gases at 1 atm
All half-reactions are given as reduction processes in standard tables

Section 18.3

Standard Reduction Potentials

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Table 18.1 - Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions

Section 18.3

Standard Reduction Potentials

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Table 18.1 - Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions (Continued)

Section 18.3

Standard Reduction Potentials

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Obtaining a Balanced Oxidation–Reduction Reaction - Manipulations

When a half-reaction is reversed, the sign of E° is reversed

When a half-reaction is multiplied by an integer, E° remains the same

Standard reduction potential is an intensive property

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Section 18.3

Standard Reduction Potentials

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Standard Reduction Potentials - Example

Redox reaction

Half-reactions

(1)

(2)

To balance the cell reaction and calculate the standard cell potential, reverse reaction (2)

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Section 18.3

Standard Reduction Potentials

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Standard Reduction Potentials - Example (Continued 1)

Each Cu atom produces two electrons but each Fe³⁺ ion accepts only one electron

Therefore, reaction (1) must be multiplied by 2

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Section 18.3

Standard Reduction Potentials

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Standard Reduction Potentials - Example (Continued 2)

Balanced cell reaction

Cell potential

= E°(cathode) – E°(anode)
= 0.77 V – 0.34 V
= 0.43 V

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Section 18.3

Standard Reduction Potentials

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Critical Thinking

What if you want to “plate out” copper metal from an aqueous Cu²⁺ solution?

Use Table 18.1 to determine several metals you can place in the solution to plate copper metal from the solution
Defend your choices
Why can Zn not be plated out from an aqueous solution of Zn²⁺ using the choices in Table 18.1?

Section 18.3

Standard Reduction Potentials

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Interactive Example 18.3 - Galvanic Cells

Consider a galvanic cell based on the following reaction:

The half-reactions are:

Give the balanced cell reaction, and calculate E°for the cell

Section 18.3

Standard Reduction Potentials

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Interactive Example 18.3 - Solution

The half-reaction involving magnesium must be reversed and since this is the oxidation process, it is the anode

Also, since the two half-reactions involve different numbers of electrons, they must be multiplied by integers

Section 18.3

Standard Reduction Potentials

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Interactive Example 18.3 - Solution (Continued)

Section 18.3

Standard Reduction Potentials

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Line Notations

Describe electrochemical cells
Anode components are listed on the left
Cathode components are listed on the right
Anode and cathode components are separated by double vertical lines

The lines indicate a salt bridge or a porous disk

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Section 18.3

Standard Reduction Potentials

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Line Notations (Continued)

Phase difference (boundary) is indicated by a single vertical line
Example

Mg(s)|Mg²⁺(aq)||Al³⁺(aq)|Al(s)

Mg → Mg²⁺ + 2e^– (anode)
Al³⁺ + 3e^– → Al (cathode)

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Section 18.3

Standard Reduction Potentials

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Components in the Description of a Galvanic Cell

Cell potential and the balanced cell reaction

Cell potential is always positive for a galvanic cell

= E°(cathode) – E°(anode)

Direction of electron flow

Obtained by inspecting the half-reactions and using the direction that gives a positive

Designation of the anode and cathode

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Section 18.3

Standard Reduction Potentials

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Components in the Description of a Galvanic Cell (Continued)

Nature of each electrode and the ions present in each compartment

If none of the substances participating in the half–reaction are conducting solids, a chemically inert conductor is required

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Section 18.3

Standard Reduction Potentials

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Example 17-4 - Description of a Galvanic Cell

Describe completely the galvanic cell based on the following half-reactions under standard conditions:

Ag⁺ + e^–→ Ag E° = 0.80 V (1)

Fe³⁺ + e^– → Fe²⁺ E° = 0.77 V (2)

Section 18.3

Standard Reduction Potentials

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Example 17-4 - Solution

Since a positive value is required, reaction (2) must run in reverse

Ag⁺ + e^–→ Ag E°(cathode) = 0.80 V

Fe²⁺ → Fe³⁺ + e^– –E° (anode) = – 0.77 V

Ag⁺(aq) + Fe²⁺(aq) → Fe³⁺(aq) + Ag(s) = 0.03 V

Section 18.3

Standard Reduction Potentials

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Example 17-4 - Solution (Continued 1)

Ag⁺ receives electrons, and Fe²⁺ loses electrons in the cell reaction

The electrons will flow from the compartment containing Fe²⁺ to the compartment containing Ag⁺

Oxidation occurs in the compartment containing Fe²⁺ (electrons flow from Fe²⁺ to Ag⁺)

Hence this compartment functions as the anode
Reduction occurs in the compartment containing Ag⁺, so this compartment functions as the cathode

Section 18.3

Standard Reduction Potentials

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Example 17-4 - Solution (Continued 2)

The electrode in the Ag/Ag⁺ compartment is silver metal, and an inert conductor, such as platinum, must be used in the Fe²⁺/Fe³⁺ compartment

Appropriate counterions are assumed to be present
Line notation

Pt(s)| Fe²⁺(aq), Fe³⁺(aq)|| Ag⁺(aq)|Ag(s)

Section 18.3

Standard Reduction Potentials

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Work and Cell Potential

Cell potential (E) and work (w) have opposite signs

In any real, spontaneous process some energy is always wasted

Actual work realized is always less than the calculated maximum

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Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Work and Cell Potential (Continued)

Actual work done is given by the following equation:

E - Actual potential difference at which current flows
q - Quantity of charge in coulombs transferred

Faraday (F): Charge on 1 mole of electrons

F = 96,485 C/mol e^–

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Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Cell Potential and Free Energy

Change in free energy equals the maximum useful work that can be obtained from a process

w_max = ΔG

For a galvanic cell

w_max= – qE_max = ΔG

Since q = nF

Therefore, ΔG = – qE_max = – nFE_max

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Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Maximum Cell Potential

Directly related to the free energy difference between the reactants and the products in the cell

ΔG° = –nFE°

Galvanic cell will run in the direction that provides a positive E_cell value

Positive E_cell value implies negative ΔG value, which is the condition for spontaneity

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Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Interactive Example 18.5 - Calculating ΔG°for a Cell Reaction

Using the data in Table 18.1, calculate ΔG°for the following reaction:

Is this reaction spontaneous?

Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Interactive Example 18.5 - Solution

The half-reactions are:

We can calculate ΔG° from the equation

ΔG° = –nFE°

Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Interactive Example 18.5 - Solution (Continued 1)

Since two electrons are transferred per atom in the reaction, 2 moles of electrons are required per mole of reactants and products

Thus, n = 2 mol e^–, F = 96,485 C/mol e^–, and E° = 0.78 V = 0.78 J/C

Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Interactive Example 18.5 - Solution (Continued 2)

The process is spontaneous, as indicated by both the negative sign of ΔG° and the positive sign of

This reaction is used industrially to deposit copper metal from solutions resulting from the dissolving of copper ores

Section 18.4

Cell Potential, Electrical Work, and Free Energy

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Interactive Example 18.7 - The Effects of Concentration on E

For the cell reaction

Predict whether E_cell is larger or smaller than for the following case:
[Al³⁺] = 2.0 M, [Mn²⁺] = 1.0 M
[Al³⁺] = 1.0 M, [Mn²⁺] = 3.0 M

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Section 18.5

Dependence of Cell Potential on Concentration

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Interactive Example 18.7 - Solution

A product concentration has been raised above 1.0 M

This will oppose the cell reaction and will cause E_cell to be less than (E_cell < 0.48 V)

A reactant concentration has been increased above 1.0 M

E_cell will be greater than (E_cell > 0.48 V)

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Section 18.5

Dependence of Cell Potential on Concentration

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Concentration Cells

Cells in which both compartments have the same components at different concentrations

Cell potential is a factor of difference in concentration
Voltages are small

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Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.8 - Concentration Cells

Determine the direction of electron flow, and designate the anode and cathode for the cell represented in the figure on the right:

Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.8 - Solution

The concentrations of Fe²⁺ ion in the two compartments can (eventually) be equalized by transferring electrons from the left compartment to the right

This will cause Fe²⁺to be formed in the left compartment, and iron metal will be deposited (by reducing Fe²⁺ ions to Fe) on the right electrode

Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.8 - Solution (Continued)

Since electron flow is from left to right, oxidation occurs in the left compartment (the anode) and reduction occurs in the right (the cathode)

Section 18.5

Dependence of Cell Potential on Concentration

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Nernst Equation

Explains the relationship between cell potential and concentrations of cell components

Given in a form that is valid at 25°C

Resulting potential calculated from this equation is the maximum potential before any current flow

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Section 18.5

Dependence of Cell Potential on Concentration

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Nernst Equation and Equilibrium

A cell will spontaneously discharge until it reaches equilibrium

Q = K (equilibrium constant)
E_cell = 0

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Section 18.5

Dependence of Cell Potential on Concentration

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Nernst Equation and Equilibrium (Continued)

Dead battery

Battery in which the cell reaction has reached equilibrium

At equilibrium:

Both cell compartments have the same free energy
ΔG = 0
Cell can no longer work

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Section 18.5

Dependence of Cell Potential on Concentration

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Critical Thinking

What if you are told that E°= 0 for an electrolytic cell?

Does this mean the cell is “dead”?
What if E = 0?
Explain your answer in each case

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Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.9 - The Nernst Equation

Describe the cell based on the following half-reactions:

Where,

T = 25°C
[VO₂⁺] = 2.0 M [H⁺] = 0.50 M
[VO²⁺] = 1.0 ×10^–2 M [Zn²⁺] = 1.0 × 10^–1 M

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Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.9 - Solution

The balanced cell reaction is obtained by reversing reaction (2) and multiplying reaction (1) by 2

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2 × reaction (1)

Reaction (2) reversed

Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.9 - Solution (Continued 1)

Since the cell contains components at concentrations other than 1 M, we must use the Nernst equation, where n = 2 (since two electrons are transferred), to calculate the cell potential
At 25°C we can use the following equation:

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Section 18.5

Dependence of Cell Potential on Concentration

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Example 18.9 - Solution (Continued 2)

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Section 18.5

Dependence of Cell Potential on Concentration

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Exercise

Consider the cell described below:

Zn|Zn²⁺(1.00 M)||Cu²⁺(1.00 M)|Cu

Calculate the cell potential after the reaction has operated long enough for the [Zn²⁺] to have changed by 0.20 mol/L
Assume T = 25°C

1.09 V

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Section 18.5

Dependence of Cell Potential on Concentration

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Determining Ion Concentration

Cell potentials can help determine concentration of an ion
pH meter - Device used to measure concentration using observed potential

Composed of:

Standard electrode of known potential
Glass electrode: Changes potentials based on H⁺ion concentration in a solution
Potentiometer

Section 18.5

Dependence of Cell Potential on Concentration

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Figure 18.12 - A Glass Electrode

Electrical potential

Depends on the difference in [H⁺] between the reference solution and the solution into which the electrode is dipped
Varies with the pH of the solution being tested

Section 18.5

Dependence of Cell Potential on Concentration

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Ion-Selective Electrodes

Electrodes that are sensitive to the concentration of any specific ion
Examples

Using a crystal of lanthanum(III) fluoride (LaF₃) in an electrode to measure [F ^–]
Using solid sliver sulfide (Ag₂S) to measure [Ag⁺] and [S^2–]

Section 18.5

Dependence of Cell Potential on Concentration

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Table 18.2 - Some Ions Whose Concentrations Can Be Detected by Ion-Selective Electrodes

Section 18.5

Dependence of Cell Potential on Concentration

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Calculation of Equilibrium Constants for Redox Reactions

E°and ΔG°have a quantitative relationship

At equilibrium, E_cell = 0 and Q = K

Section 18.5

Dependence of Cell Potential on Concentration

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Interactive Example 18.10 - Equilibrium Constants from Cell Potentials

For the oxidation–reduction reaction

The appropriate half-reactions are

Balance the redox reaction, and calculate E°and K (at 25°C)

Section 18.5

Dependence of Cell Potential on Concentration

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Interactive Example 18.10 - Solution

To obtain the balanced reaction, we must reverse reaction (2), multiply it by 2, and add it to reaction (1)

Section 18.5

Dependence of Cell Potential on Concentration

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Interactive Example 18.10 - Solution (Continued 1)

In this reaction, 2 moles of electrons are transferred for every unit of reaction

For every 2 moles of Cr²⁺ reacting with 1 mole of S₄O₆^2– to form 2 moles of Cr³⁺ and 2 moles of S₂O₃^2–

Thus n = 2

Section 18.5

Dependence of Cell Potential on Concentration

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Interactive Example 18.10 - Solution (Continued 2)

The value of K is found by taking the antilog of 22.6

K = 10^22.6 = 4 ×10²²

This very large equilibrium constant is not unusual for a redox reaction

Section 18.5

Dependence of Cell Potential on Concentration

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Battery

Group of galvanic cells that are connected in series

Potentials of each individual cell add up to give the total battery potential

Essential source of portable power in today’s society

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Section 18.6

Batteries

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Types of Battery

Lead storage battery
Dry cell battery
Silver cell
Mercury cell
Nickel–cadmium battery
Lithium-ion battery

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Section 18.6

Batteries

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Lead Storage Battery

Components

Anode - Lead
Cathode - Lead coated with lead dioxide
Electrolyte solution - Sulfuric acid

Electrode reactions

Section 18.6

Batteries

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Figure 18.13 - One of the Six Cells in a 12–V Lead Storage Battery

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Section 18.6

Batteries

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Dry Cell Battery - Acid Version

Components

Zinc inner case that acts as the anode
Carbon rod that is in contact with a moist plate of solid MnO₂
Solid NH₄Cl and carbon that acts as the cathode

Section 18.6

Batteries

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Dry Cell Battery - Alkaline Version

Components

Solid NH₄Cl is replaced with NaOH or KOH

Lasts for a longer period of time

Zinc corrodes less rapidly under basic conditions

Section 18.6

Batteries

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Figure 18.14 - A Common Dry Cell Battery

Section 18.6

Batteries

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Silver Cell and Mercury Cell Batteries

Silver cell

Contains a Zn anode and a cathode that uses Ag₂O as the oxidizing agent

Mercury cell

Contains a Zn anode and a cathode that uses HgO as the oxidizing agent

Section 18.6

Batteries

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Figure 18.15 - A Mercury Battery

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Section 18.6

Batteries

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Nickel–Cadmium and Lithium-Ion Batteries

Nickel–cadmium battery

Lithium-ion batteries

Li⁺ ions migrate from the cathode to the anode and enter the interior as the battery is charged
Charge-balancing electrons travel to the anode through the external circuit in the charger

The opposite process occurs on discharge

Section 18.6

Batteries

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Fuel Cells

Galvanic cells for which reactants are continuously supplied
The image depicts a hydrogen–oxygen fuel cell

Anode reaction

2H₂ + 4OH^– → 4H₂O + 4e^–

Cathode reaction

4e^– + O₂ + 2H₂O → 4OH^–

Section 18.6

Batteries

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Process of Corrosion

Can be viewed as the method of returning metals to their natural state

Natural state - Ores from which metals are originally obtained

Involves the oxidation of a metal

Spontaneous process

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Section 18.7

Corrosion

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Process of Corrosion (Continued)

Some metals tend to develop a thin oxide coating to protect against further oxidation

Aluminum forms a thin layer of aluminum oxide (Al₂(OH)₆)
Copper forms an external layer of greenish copper carbonate called patina
Silver sulfide forms silver tarnish

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Section 18.7

Corrosion

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Figure 18.17 - The Electrochemical Corrosion of Iron

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Section 18.7

Corrosion

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Corrosion Prevention

Application of paint or metal plating

Galvanizing: Process in which steel is coated with zinc to prevent corrosion

Alloying
Cathodic protection

Protects steel in buried fuel tanks and pipelines

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Section 18.7

Corrosion

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Figure 18.18 - Cathodic Protection of an Underground Pipe

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Section 18.7

Corrosion

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Electrolysis

Forcing a current through a cell to produce a chemical change for which the cell potential is negative
Electrolytic cell: Device that uses electrical energy to produce a chemical change

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Section 18.8

Electrolysis

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Figure 18.19 - A Galvanic Cell and an Electrolytic Cell

Section 18.8

Electrolysis

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Stoichiometry of Electrolysis

Refers to determining how much chemical change occurs with the flow of a given current for a specified time
Steps to solve a stoichiometry problem

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Section 18.8

Electrolysis

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Interactive Example 18.11 - Electroplating

How long must a current of 5.00 A be applied to a solution of Ag⁺ to produce 10.5 g silver metal?

Section 18.8

Electrolysis

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Interactive Example 18.11 - Solution

In this case, we must use the steps given earlier in reverse

Section 18.8

Electrolysis

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Interactive Example 18.11 - Solution (Continued 1)

Each Ag⁺ ion requires one electron to become a silver atom

Thus 9.73 ×10^–2 mole of electrons is required, and we can calculate the quantity of charge carried by these electrons

Section 18.8

Electrolysis

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Interactive Example 18.11 - Solution (Continued 2)

The 5.00 A (5.00 C/s) of current must produce 9.39 × 10³ C of charge

Thus,

Section 18.8

Electrolysis

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Electrolysis of Water - Nonspontaneous Reaction

4H₂O

Section 18.8

Electrolysis

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Interactive Example 18.12 - Relative Oxidizing Abilities

An acidic solution contains the ions Ce⁴⁺, VO₂⁺, and Fe³⁺

Using the E°values listed in Table 18.1, give the order of oxidizing ability of these species, and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage

Section 18.8

Electrolysis

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Interactive Example 18.12 - Solution

The half-reactions and E°values are as follows:

The order of oxidizing ability is therefore:

Ce⁴⁺ > VO₂⁺> Fe³⁺

The Ce⁴⁺ ion will be reduced at the lowest voltage in an electrolytic cell

Section 18.8

Electrolysis

110 of 120

Production of Aluminum

Aluminum is extracted from bauxite

Not pure aluminum oxide
Bauxite is treated with aqueous sodium hydroxide to obtain pure hydrated alumina

Alumina dissolves in the basic solution
Other metals remain as solids

Extraction of purified alumina

2CO₂(g) + 2AlO₂^–(aq) + (n + 1)H₂O(l) →

2HCO₃^–(aq) + Al₂O₃ · nH₂O(s)

Section 18.8

Electrolysis

111 of 120

Production of Aluminum (Continued)

Purified alumina is mixed with cryolite and melted

Aluminum ion is reduced to aluminum metal in an electrolytic cell
Alumina reacts with the cryolite anion

Al₂O₃ + 4AlF₆^3–→ 3Al₂OF₆^2– + 6F^–

Overall cell reaction

2Al₂O₃ + 3C → 4Al + 3CO₂

Aluminum produced is 99.5% pure

Section 18.8

Electrolysis

112 of 120

Figure 18.22 - Electrolytic Cell for Producing Aluminum by the Hall–Heroult Process

112

Section 18.9

Commercial Electrolytic Processes

113 of 120

Electrorefining of Metals - Example

Impure copper that is derived from the reduction of copper ore is cast into large slabs that serve as anodes for electrolytic cells

Electrolyte - Aqueous copper sulfate
Cathodes - Thin sheets of ultrapure copper

Section 18.9

Commercial Electrolytic Processes

114 of 120

Metal Plating

Protects metals that corrode easily
Can be done by making the object as the cathode in a tank containing ions of the plating metal

Example - Electroplating of a spoon

Section 18.9

Commercial Electrolytic Processes

115 of 120

Electrolysis of Sodium Chloride (Brine)

Process used for the production of sodium metal
Steps

Mix solid NaCl with solid CaCl₂ to lower the melting point
Electrolyze the mixture in a Downs cell

Section 18.9

Commercial Electrolytic Processes

116 of 120

Figure 18.25 - The Downs Cell

116

Section 18.9

Commercial Electrolytic Processes

117 of 120

Electrolysis of Sodium Chloride (Brine) (continued)

The sodium is drained off in the Downs cell

Cooled and cast into blocks
Stored in an inert solvent, such as mineral oil, to prevent its oxidation

Produces hydrogen and chlorine gas

Leaves a solution containing dissolved NaOH and NaCl

Section 18.9

Commercial Electrolytic Processes

118 of 120

Figure 18.26 - Mercury Cell for Production of Chlorine and Sodium Hydroxide

Contamination of the sodium hydroxide by NaCl

Can be eliminated by electrolyzing the brine in a mercury cell
Reaction of resulting sodium metal

2Na(s) + 2H₂O(l) → 2Na⁺(aq) + 2OH^–(aq) + H₂(g)

Section 18.9

Commercial Electrolytic Processes

119 of 120

Chlor–Alkali Process

Involves:

Recovering pure NaOH from an aqueous solution
Pumping the re-generated mercury into the electrolysis cell

Currently carried out in diaphragm cells

Cathode and anode are separated by a diaphragm

Allows passage of water molecules, Na⁺ ions, and Cl^– ions
Blocks passage of OH^– ions

Section 18.9

Commercial Electrolytic Processes

120 of 120

Chlor–Alkali Process (Continued)

Disadvantage

The aqueous effluent from the cathode compartment contains a mixture of NaOH and untreated NaCl

Needs to be purified to derive pure NaOH

Advancement in the chlor–alkali industry

Uses a membrane to separate the anode and cathode compartments in brine electrolysis cells

Membrane is impermeable to anions

Section 18.9

Commercial Electrolytic Processes