Chapter 21 Electrochemistry: Fundamentals

Chapter 21 Electrochemistry: Fundamentals

Key Points About Redox Reactions

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

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

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

A summary of redox terminology.

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

OXIDATION

Zn loses electrons.

Zn is the reducing agent and becomes oxidized.

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

REDUCTION

One reactant loses electrons.

Reducing agent is oxidized.

Oxidation number increases.

Hydrogen ion gains electrons.

Hydrogen ion is the oxidizing agent and becomes reduced.

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

Other reactant gains electrons.

Oxidizing agent is reduced.

Oxidation number decreases.

Energy is absorbed to drive a nonspontaneous redox reaction

General characteristics of voltaic and electrolytic cells.

VOLTAIC / GALVANIC CELL ELECTROLYTIC CELLEnergy is released from

spontaneous redox reaction

Reduction half-reactionY++ e- Y

Oxidation half-reactionX X+ + e-

System does work on its surroundings

Reduction half-reactionB++ e- B

Oxidation half-reactionA- A + e-

Surroundings(power supply)do work on system(cell)

Overall (cell) reactionX + Y+ X+ + Y; G < 0

Overall (cell) reactionA- + B+ A + B; G > 0

Electrochemical cell

A voltaic cell based on the zinc-copper reaction.

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- Cu(s)

Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

A voltaic cell using inactive electrodes.

Reduction half-reactionMnO4

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

Mn2+(aq) + 4H2O(l)

Oxidation half-reaction2I-(aq) I2(s) + 2e-

Overall (cell) reaction2MnO4

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

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 higher oxidation state

phase of higher oxidation state

phase of lower oxidation state

phase boundary between half-cells

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

Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

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

inert electrode

Diagramming Voltaic Cells

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.

PLAN:

SOLUTION:

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).

Voltmeter

Oxidation half-reactionCr(s) Cr3+(aq) + 3e-

Reduction half-reactionAg+(aq) + e- Ag(s)

Overall (cell) reactionCr(s) + Ag+(aq) Cr3+(aq) + Ag(s)

Cr

Cr3+

Ag

Ag+

K+

NO3-

salt bridge

e-

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

Determining an unknown E0half-cell with the standard

reference (hydrogen) electrode.

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reaction2H3O+(aq) + 2e- H2(g) + 2H2O(l)

Overall (cell) reactionZn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)

Calculating an Unknown E0half-cell from E0

cell

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

PLAN:

SOLUTION:

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

Calculate E0bromine given E0

zinc = -0.76V

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 = E0

cathode - E0anode.

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

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

E0cell = E0

cathode - E0anode = 1.83 = E0

bromine - (-0.76)

E0bromine = 1.86 - 0.76 = 1.07 V

Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)

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

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)

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

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)

+2.87

-3.05

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

-0.23

-0.44

-0.83

-2.71

strength

of reducin

g agent

stre

ngt

h o

f ox

idiz

ing

agen

t

• 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 E0

cell.

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

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

stronger reducing agent

weaker oxidizing agent

stronger oxidizing agent

weaker reducing agent

Writing Spontaneous Redox Reactions

Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength

PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0

cell for each.

PLAN:

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

E0 = 0.96V(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

E0 = -0.23V(2) N2(g) + 5H+(aq) + 4e- N2H5+(aq)

E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

Put the equations together in varying combinations so as to produce (+) E0

cell 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.


continued (2 of 4)

SOLUTION: (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = 0.96V

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

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

X4

X3

E0cell = 1.19V

(a)


(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23VRev

E0 = -0.96V(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-Rev

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

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

X2

X3

E0cell = 0.27V

4NO3-(aq) + 3N2H5

+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l)(A)

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


continued (3 of 4)


E0 = +0.23V(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-Rev

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

(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) X2

E0cell = 1.46V

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

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

(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+


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

• 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 (Eo

cell).

• 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.

Relative Reactivities (Activities) of Metals

1. Metals that can displace H from acid

2. Metals that cannot displace H from acid

3. Metals that can displace H from water

4. Metals that can displace other metals from solution

can displace Hfrom water

LiKBaCaNa

stre

ngt

h as

red

ucin

g ag

ents

can displace Hfrom steam

MgAlMnZnCrFeCd

H2

cannot displace H from any source

CuHg AgAu

can displace Hfrom acid

CoNiSnPb

G0

E0cell K

G0 K

Reaction at standard-state

conditionsE0cell

The interrelationship of G0, E0, and K.

< 0 spontaneous

at equilibrium

nonspontaneous

0

> 0

> 0

0

< 0

> 1

1

< 1

G0 = -RT lnKG0 = -nFEo

cell

E0cell = -RT lnK

nF

By substituting standard state values into E0

cell, we get

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

The Effect of Concentration on Cell Potential

G = G0 + RT ln Q

-nF Ecell = -nF Ecell + RT ln Q

Ecell = E0cell - ln Q

RT

nF

•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

Ecell = E0cell -

log Q0.0592

n

Nernst equation

Calculating K and G0 from E0cell

PLAN:

SOLUTION:

PROBLEM: Lead can displace silver from solution:

As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and G0 at 298 K for this reaction.

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

Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0

cell. Substitute into the equations found on slide

E0 = -0.13V Anode

E0 = 0.80V Cathode

E0cell = E0

cathode – E0anode = 0.93V

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

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

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

E0cell = log K

0.592V

n

log K =

K = 2.6x1031

n x E0cell

0.592V(2)(0.93V)

0.592V=

G0 = -nFE0cell

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

G0 = -1.8x102kJ

2X

E0cell = - (RT/n F) ln K

Using the Nernst Equation to Calculate Ecell

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:

PLAN:

SOLUTION:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2

Calculate Ecell at 298 K.

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

Determining E0cell :

E0 = 0.00V2H+(aq) + 2e- H2(g)

E0 = -0.76VZn2+(aq) + 2e- Zn(s)

Zn(s) Zn2+(aq) + 2e- E0 = +0.76V

Q = P x [Zn2+]

H2

[H+]2

Q = 4.8x10-4

Q = (0.30)(0.010)

(2.5)2

Ecell = E0cell -

0.0592V

nlog Q

Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V

Diagramming Voltaic Cells

PROBLEM: 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.

PLAN:

SOLUTION:

Identify the redox reactions

Write each half-reaction.

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

Voltmeter

Oxidation half-reactionZn2+(aq) + 2e- Zn(s)

Reduction half-reactionAg+(aq) + e- Ag(s)

Overall (cell) reactionZn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

Zn

Zn2+

Ag

Ag+

K+

NO3-

salt bridge

e-

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

CathodeAnode

Free Energy and Electrical Work

G -Ecell

-Ecell =-wmax

charge

charge = n F

n = # mols e-

F = Faraday constant

F = 96,485 C/mol

1V = 1J/C

F = 9.65x104J/V*mol

G = wmax = charge x (-Ecell)

G = - n F Ecell

In the standard state

G0 = - n F E0cell

G0 = - RT ln K

E0cell = - (RT/n F) ln K

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.

All components are at standard state.

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

The Effect of Concentration on Cell Potential

G = G0 + RT ln Q

-nF Ecell = -nF Ecell + RT ln Q


RT

nF

• 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 Reverse reaction

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

Ecell = E0cell -

log Q0.0592

n

Nernst equation

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

Sample Problem Using the Nernst Equation to Calculate Ecell

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:

PLAN:

SOLUTION:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2

Calculate Ecell at 298 K.

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

Determining E0cell :

E0 = 0.00V cathode2H+(aq) + 2e- H2(g)

E0 = -0.76V anodeZn2+(aq) + 2e- Zn(s)Q =

P x [Zn2+]H2

[H+]2

Q = 4.8x10-4

Q = (0.30)(0.010)

(2.5)2

Ecell = E0cell -

0.0592

nlog Q

Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V


RT

nF

Ecell0 = E0

c-E0a = 0.00-(-0.76)V = 0.76 V

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

Ecell = E0cell -

log Q0.0592

n

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 (Eo

cell).

• 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.

Documents

Chapter 21 Electrochemistry: Fundamentals