Chapter 21  Electrochemistry: Fundamentals
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Chapter 21 Electrochemistry: Fundamentals. Key Points About Redox Reactions. Oxidation (electron loss ) always accompanies reduction (electron gain). The oxidizing agent is reduced , and the reducing agent is oxidized.

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Chapter 21 Electrochemistry: Fundamentals

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Chapter 21 electrochemistry fundamentals

Chapter 21 Electrochemistry: Fundamentals

Key Points About Redox Reactions

Oxidation (electron loss) always accompanies reduction (electron gain).

The oxidizing agent is reduced, and the reducing agent is oxidized.

The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.


Chapter 21 electrochemistry fundamentals

Zn(s) + 2H+(aq)Zn2+(aq) + H2(g)

A summary of redox terminology.

OXIDATION

Zn loses electrons.

One reactant loses electrons.

Zn is the reducing agent and becomes oxidized.

Reducing agent is oxidized.

The oxidation number of Zn increases from 0 to + 2.

Oxidation number increases.

REDUCTION

Other reactant gains electrons.

Hydrogen ion gains electrons.

Hydrogen ion is the oxidizing agent and becomes reduced.

Oxidizing agent is reduced.

The oxidation number of H decreases from +1 to 0.

Oxidation number decreases.


Chapter 21 electrochemistry fundamentals

Electrochemical cell

Oxidation half-reaction

X X+ + e-

Oxidation half-reaction

A- A + e-

Reduction half-reaction

Y++ e- Y

Reduction half-reaction

B++ e- B

Overall (cell) reaction

X + Y+ X+ + Y; DG < 0

Overall (cell) reaction

A- + B+ A + B; DG > 0

General characteristics of voltaic and electrolytic cells.

VOLTAIC / GALVANIC CELL

ELECTROLYTIC CELL

System does work on its surroundings

Energy is released from spontaneous redox reaction

Energy is absorbed to drive a nonspontaneous redox reaction

Surroundings(power supply)

do work on system(cell)


Chapter 21 electrochemistry fundamentals

Oxidation half-reaction

Zn(s) Zn2+(aq) + 2e-

Reduction half-reaction

Cu2+(aq) + 2e- Cu(s)

Overall (cell) reaction

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

A voltaic cell based on the zinc-copper reaction.


Chapter 21 electrochemistry fundamentals

Oxidation half-reaction

2I-(aq) I2(s) + 2e-

Reduction half-reaction

MnO4-(aq) + 8H+(aq) + 5e-

Mn2+(aq) + 4H2O(l)

Overall (cell) reaction

2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)

A voltaic cell using inactive electrodes.


Chapter 21 electrochemistry fundamentals

Zn(s) Zn2+(aq) + 2e-

Cu2+(aq) + 2e- Cu(s)

inert electrode

Notation for a Voltaic Cell

components of anode compartment

(oxidation half-cell)

components of cathode compartment

(reduction half-cell)

phase of lower oxidation state

phase of lower oxidation state

phase of higher oxidation state

phase of higher oxidation state

phase boundary between half-cells

Examples:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)

graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) , Mn2+(aq) | graphite


Chapter 21 electrochemistry fundamentals

PROBLEM:

Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.

e-

Oxidation half-reaction

Cr(s) Cr3+(aq) + 3e-

Cr

Ag

K+

NO3-

Reduction half-reaction

Ag+(aq) + e- Ag(s)

Cr3+

Ag+

Overall (cell) reaction

Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s)

Diagramming Voltaic Cells

PLAN:

Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).

SOLUTION:

Voltmeter

salt bridge

Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)


Chapter 21 electrochemistry fundamentals

Oxidation half-reaction

Zn(s) Zn2+(aq) + 2e-

Reduction half-reaction

2H3O+(aq) + 2e- H2(g) + 2H2O(l)

Overall (cell) reaction

Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)

Determining an unknown E0half-cell with the standard reference (hydrogen) electrode.


Chapter 21 electrochemistry fundamentals

PROBLEM:

A voltaic cell houses the reaction between aqueous bromine and zinc metal:

Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V

anode: Zn(s) Zn2+(aq) + 2e- E = +0.76

E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V

Calculating an Unknown E0half-cell from E0cell

Calculate E0bromine given E0zinc = -0.76V

PLAN:

The reaction is spontaneous as written since the E0cell is (+). Zinc is being oxidized and is the anode. Therefore the E0bromine can be found using E0cell = E0cathode - E0anode.

SOLUTION:

E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76)

E0bromine = 1.86 - 0.76 = 1.07 V


Chapter 21 electrochemistry fundamentals

F2(g) + 2e- 2F-(aq)

Cl2(g) + 2e- 2Cl-(aq)

MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

Ag+(aq) + e- Ag(s)

Fe3+(g) + e- Fe2+(aq)

O2(g) + 2H2O(l) + 4e- 4OH-(aq)

strength of reducing agent

Cu2+(aq) + 2e- Cu(s)

strength of oxidizing agent

N2(g) + 5H+(aq) + 4e- N2H5+(aq)

Fe2+(aq) + 2e- Fe(s)

2H2O(l) + 2e- H2(g) + 2OH-(aq)

Na+(aq) + e- Na(s)

Li+(aq) + e- Li(s)

Selected Standard Electrode Potentials (298K)

Half-Reaction

E0(V)

+2.87

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

2H+(aq) + 2e- H2(g)

-0.23

-0.44

-0.83

-2.71

-3.05


Chapter 21 electrochemistry fundamentals

Example:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Writing Spontaneous Redox Reactions

  • By convention, electrode potentials are written as reductions.

  • When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential.

  • The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell.

  • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents.

stronger reducing agent

stronger oxidizing agent

weaker oxidizing agent

weaker reducing agent


Chapter 21 electrochemistry fundamentals

PROBLEM:

(a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0cell for each.

(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

E0 = 0.96V

(2) N2(g) + 5H+(aq) + 4e- N2H5+(aq)

E0 = -0.23V

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

E0 = 1.23V

Writing Spontaneous Redox Reactions and Ranking

Oxidizing and Reducing Agents by Strength

(b) Rank the relative strengths of the oxidizing and reducing agents:

PLAN:

Put the equations together in varying combinations so as to produce (+) E0cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0.

In ranking the strengths, compare the combinations in terms of E0cell.


Chapter 21 electrochemistry fundamentals

E0 = 0.96V

Rev

(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-

E0 = +0.23V

(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

(A)

4NO3-(aq) + 3N2H5+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l)

(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-

Rev

(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-

(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-

E0 = -0.96V

E0 = 1.23V

X2

(B)

2NO(g) + 3MnO2(s) + 4H+(aq) 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l)

Writing Spontaneous Redox Reactions and Ranking

Oxidizing and Reducing Agents by Strength

continued (2 of 4)

SOLUTION:

(a)

E0cell = 1.19V

X4

X3

E0cell = 0.27V

X3


Chapter 21 electrochemistry fundamentals

Rev

(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-

(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-

E0 = +0.23V

E0 = 1.23V

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

(C)

N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l)

Writing Spontaneous Redox Reactions and Ranking

Oxidizing and Reducing Agents by Strength

continued (3 of 4)

E0cell = 1.46V

X2

(b) Ranking oxidizing and reducing agents within each equation:

(A): oxidizing agents: NO3- > N2

reducing agents: N2H5+ > NO

(B): oxidizing agents: MnO2 >NO3-

reducing agents: NO > Mn2+

(C): oxidizing agents: MnO2 > N2

reducing agents: N2H5+ > Mn2+


Chapter 21 electrochemistry fundamentals

Writing Spontaneous Redox Reactions and Ranking

Oxidizing and Reducing Agents by Strength

continued (4 of 4)

A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of

Oxidizing agents: MnO2 > NO3- > N2

Reducing agents: N2H5+ > NO > Mn2+


Summary

Summary

  • A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge.

  • The salt bridge provides ions to maintain the charge balance when the cell operates.

  • Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell.

  • The output of a cell is called cell potential (Ecell) and is measured in volts.

  • When all substances are in standard states, the cell potential is the standard cell potential (Eocell).

  • Ecell equals Ecathode minus Eanode,Ecell = Ecathode - Eanode.

  • Conventionally, the half cell potential refers to its reduction half-reaction.

  • Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking the oxidizing agent or reducing agent.

  • Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones.

  • Spontaneous reaction is indicated negative ∆G and positive ∆E,

    ∆G = - nF∆E.

  • We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.


Chapter 21 electrochemistry fundamentals

strength as reducing agents

Relative Reactivities (Activities) of Metals

Li

K

Ba

Ca

Na

can displace H

from water

1. Metals that can displace H from acid

Mg

Al

Mn

Zn

Cr

Fe

Cd

2. Metals that cannot displace H from acid

can displace H

from steam

3. Metals that can displace H from water

Co

Ni

Sn

Pb

can displace H

from acid

4. Metals that can displace other metals from solution

H2

Cu

Hg

Ag

Au

cannot displace H from any source


Chapter 21 electrochemistry fundamentals

By substituting standard state values into E0cell, we get

E0cell = (0.0592V/n) log K (at 298 K)

The interrelationship of DG0, E0, and K.

Reaction at standard-state conditions

DG0

K

E0cell

DG0

< 0

> 1

> 0

spontaneous

0

1

0

at equilibrium

> 0

< 1

< 0

nonspontaneous

DG0 = -nFEocell

DG0 = -RT lnK

E0cell

K

E0cell = -RT lnK

nF


Chapter 21 electrochemistry fundamentals

RT

Ecell = E0cell -

ln Q

nF

0.0592

log Q

Ecell = E0cell -

n

The Effect of Concentration on Cell Potential

DG = DG0 + RT ln Q

-nF Ecell = -nF Ecell + RT ln Q

Nernst equation

  • When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell

  • When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell

  • When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell


Chapter 21 electrochemistry fundamentals

Calculating K and DG0 from E0cell

PROBLEM:

Lead can displace silver from solution:

Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)

As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DG0 at 298 K for this reaction.

Pb2+(aq) + 2e- Pb(s)

Ag+(aq) + e- Ag(s)

Ag+(aq) + e- Ag(s)

E0cell =

0.592V

log K

n

n x E0cell

(2)(0.93V)

=

0.592V

0.592V

PLAN:

Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide

SOLUTION:

E0 = -0.13V Anode

E0 = 0.80V Cathode

E0cell = E0cathode – E0anode = 0.93V

2X

E0cell = - (RT/n F) ln K

DG0 = -nFE0cell

= -(2)(96.5kJ/mol*V)(0.93V)

log K =

DG0 = -1.8x102kJ

K = 2.6x1031


Chapter 21 electrochemistry fundamentals

PROBLEM:

In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm

Calculate Ecell at 298 K.

H2

E0 = 0.00V

P x [Zn2+]

2H+(aq) + 2e- H2(g)

Q =

H2

E0 = -0.76V

Zn2+(aq) + 2e- Zn(s)

[H+]2

E0 = +0.76V

Zn(s) Zn2+(aq) + 2e-

(0.30)(0.010)

Q =

0.0592V

Ecell = E0cell -

(2.5)2

log Q

n

Using the Nernst Equation to Calculate Ecell

PLAN:

Find E0cell and Q in order to use the Nernst equation.

SOLUTION:

Determining E0cell :

Q = 4.8x10-4

Ecell = 0.76 - (0.0592/2)log(4.8x10-4) =

0.86V


Chapter 21 electrochemistry fundamentals

Diagramming Voltaic Cells

Ag

K+

NO3-

Reduction half-reaction

Ag+(aq) + e- Ag(s)

Ag+

Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Zn bar in a Zn(NO3)2 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Zn electrode is negative relative to the Ag electrode.

PROBLEM:

Identify the redox reactions

Write each half-reaction.

Associate the (-)(Zn) pole with the anode (oxidation) and the (+) (Ag) pole with the cathode (reduction).

PLAN:

SOLUTION:

e-

Voltmeter

Oxidation half-reaction

Zn2+(aq) + 2e- Zn(s)

salt bridge

Zn

Zn2+

Anode

Cathode

Overall (cell) reaction

Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)


Chapter 21 electrochemistry fundamentals

-wmax

-Ecell =

charge

charge = n F

n = # mols e-

F = Faraday constant

Free Energy and Electrical Work

If there is no current flows, the potential represents the maximum work the cell can do.

If there is no current flows, no energy is lost to heat the cell component.

DG a -Ecell

DG = wmax = charge x (-Ecell)

DG = - n F Ecell

In the standard state

All components are at standard state.

DG0 = - n F E0cell

F = 96,485 C/mol

DG0 = - RT ln K

1V = 1J/C

E0cell = - (RT/n F) ln K

F = 9.65x104J/V*mol

E0cell = - (0.05916/n) log K at RT


Chapter 21 electrochemistry fundamentals

RT

Ecell = E0cell -

ln Q

nF

0.0592

log Q

Ecell = E0cell -

n

The Effect of Concentration on Cell Potential

Cell operates with all components at standard states. Most cells are starting at Non-standard state.

DG = DG0 + RT ln Q

-nF Ecell = -nF Ecell + RT ln Q

Nernst equation

  • When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell forward reaction

  • When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell Equilibrium

  • When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reversereaction


Chapter 21 electrochemistry fundamentals

PROBLEM:

In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm

H2

RT

nF

P x [Zn2+]

2H+(aq) + 2e- H2(g)

Q =

H2

Zn2+(aq) + 2e- Zn(s)

[H+]2

(0.30)(0.010)

Q =

0.0592

Ecell = E0cell -

0.0592

(2.5)2

log Q

log Q

Ecell = E0cell -

n

n

Sample Problem

Using the Nernst Equation to Calculate Ecell

2H+(aq) + Zn (s) H2(g) + Zn2+ (aq)

Calculate Ecell at 298 K.

PLAN:

Find E0cell and Q in order to use the Nernst equation.

Ecell = E0cell -

ln Q

SOLUTION:

Determining E0cell :

E0 = 0.00V cathode

E0 = -0.76V anode

Ecell0 = E0c-E0a = 0.00-(-0.76)V = 0.76 V

Q = 4.8x10-4

Ecell = 0.76 - (0.0592/2)log(4.8x10-4) =

0.86V


Summary1

Summary

  • A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge.

  • The salt bridge provides ions to maintain the charge balance when the cell operates.

  • Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell.

  • The output of a cell is called cell potential (Ecell) and is measured in volts.

  • When all substances are in standard states, the cell potential is the standard cell potential (Eocell).

  • Ecell equals Ecathode minus Eanode,Ecell = Ecathode - Eanode.

  • Conventionally, the half cell potential refers to its reduction half-reaction.

  • Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking the oxidizing agent or reducing agent.

  • Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones.

  • Spontaneous reaction is indicated negative ∆G and positive ∆E,

    ∆G = -nF∆E.

  • We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.


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