Chapter 13 Electrochemistry and Cell...

OFB Chapter 13 111/1/2004

Chapter 13Electrochemistry and Cell Voltage

• 13-1 The Gibbs Function and Cell Voltage

• 13-2 Half Cell Potentials• 13-3 Oxidizing and Reducing

Agents• 13-4 Concentrations and the Nernst

Equations• 13-5 Equilibrium Constants from

Electrochemistry• 13-6 Batteries and Fuel Cells• 13-7 Corrosion and Its Prevention


• Important Chapter - ties together several concepts from previous Chapters– Oxidation-Reduction– Electrochemical Cells– Acid-Base– Solubility– Equilibrium constants

• Practical Importance– Energy storage – batteries– Energy conversion – solar cells– Chemical conversion

• E.g., printed circuits – metal deposition• Corrosion• Corrosion prevention



The Gibbs Function and Cell Voltage

Anode

(-)Oxidation

Cathode

(+)Reduction

e- flow →

Q = charge

There exists “potential” or potential difference , called ∆E, between the cells


For a Galvanic Cell, a spontaneous reaction takes place.

For the Galvanic cell the cell performs electrical work on the surroundings (acts as a battery)

For an Electrolytic Cell, a non-spontaneous reaction takes place.

For the electrolytic cell, the cell performs electrical work on the system


Welec = - Q ∆E

Type Cell ∆E Welec Performs work:

Galvanic on surroundings

Electrolytic on system


Welec = - Q ∆E

Welec = - Q ∆E

recall Q = I x t

Substituting gives

W units is Joules ∆E units is Joules/coulomb


Consider a Galvanic Cell (recall that this is a spontaneous reaction)

Let’s related ∆G (The Gibbs Function or Gibbs free energy) to Welec

∆G = Welec,max (at constant T,P)

Max. means theoretical max, some energy is lost as heat

Remember Faradays constant?

F = 96, 485 coulombs/moleIf n moles of electrons pass through our Galvanic Cell

∆G = = (at constant T,P)


Standard States and Cell Voltages

Standard States usually at 25°C, all reactants and products at standard states.

Example 13-2 and Exercise 13-2 give a straightforward example of calculating ∆G°for a chemical reaction in a cell

TRY THEM


Gains ElectronsDecreaseSubstance Reduced

Loses ElectronsIncreaseSubstance Oxidized

Supplies ElectronsIncrease

Reducing Agent, does the reducing and becomes oxidized

Picks Up electronsDecrease

Oxidizing Agent, does the oxidizing and becomes reduced

Gain of ElectronsDecreaseReduction

Loss of ElectronsIncreaseOxidation

Electron Change

Oxidation Number ChangeTerm

MnO4– + SO4

2– → Mn2+ + S2O82–

Ox Agent Red Agent

+7 +6 +2 +7

OFB Chapter 13 1011/1/2004

OFB Chapter 13 1111/1/2004

13-2 Half Cell Potentials

This section introduces Standard half cell reduction potentials

Tables that list half cell reactions as reductions. (This table from 3rd Ed OFB is slightly different than 4th edition)

OFB Chapter 13 1211/1/2004

Standard Reduction potentials E °

Ni2+ + 2e- → Ni (s) -0.257 Volts

Cu2+ + 2e- → Cu (s) +0.345 Volts

Note used OFB 4th edition corrected values

Now if we combine these as two separate cells (as we did in chapter 12)

Cu2+ reduction potential is more positive (+0.345) therefore it is Reduced(cathode) and Ni is oxidized (anode)

Notice that we want the opposite of the Nickel reduction potential, therefore the minus sign

Ni (s) → Ni2+ + 2e– – (–0.257) Volts

Cu2+ + 2e- → Cu (s) +0.345 Volts

Cu2+ + Ni → Cu (s) + Ni2+ Net Equation

OFB Chapter 13 1311/1/2004

By international convention

2H+ + 2e- → H2 (g)E = 0.0 volts

All reduction potentials can be determined relative to H+

reduction.

OFB Chapter 13 1411/1/2004

∆E° = E° (cathode) - E° (anode)

Note: The number of electrons appearing in half-equations do not

figure in to computation of the potential difference of a cell

voltCoulomb

Joules are units

ratio a is E

Why?

=

°

OFB Chapter 13 1511/1/2004


• Typical problem (Chapter 13 #10 end of chapter)A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.(a) By referring to Appendix E, write balanced chemical equations for the half reaction at the anode and the cathode and for the overall reaction(b) Calculate the cell voltage, assuming that all reactions and products are in standard state at 298.15°K.

OFB Chapter 13 1611/1/2004

A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.

(a) By referring to Appendix E, write balanced chemical equations for the half reaction at the anode and the cathode and for the overall reaction

Fe3+ + e- → Fe2+

Cd+2 + 2e- → Cd

Look up Reduction Potentials in Appendix E

For Fe3+ to Fe2+ and Cd+2 to Cd

Need to balance electrons and stoichiometry

2Fe3+ + 2e- → 2Fe2+

Cd → Cd+2 +2e-

(Overall) 2Fe3+ + Cd → 2Fe2+ + Cd+2

OFB Chapter 13 1711/1/2004

A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.

(b) Calculate the cell voltage, assuming that all reactions and products are in standard state at 298.15°K.

Fe3+ reduction potential is more positive (+0.771) therefore it is reduced (cathode) and Cd (-0.403) is oxidized (anode)

Notice that the numbers of electrons appearing in half-equations do not figure in the computation of the potential difference of a cell

Cd | Cd 2+ || Fe 3+ | Fe 2+

Anode on left cathode on right

OFB Chapter 13 1811/1/2004

13-4 Concentration Effects and the Nernst Equation

• For Electrochemical Cells which are NOT at standard states (1M or 1 atm)

• From Chapter 11

– Where Q is the reaction quotient• From Chapter 13

Can substitute -n F ∆E° for ∆G° and -n F ∆E for ∆G, and rearranging, becomes the Nernst Equation

OFB Chapter 13 1911/1/2004

If Temp. at 25C and

R = 8.3 JK-1mol-1 and

F = 96,485 C mol-1 and use

J/C = V,

then the equation becomes

Nernst Equation

Nernst Equation

OFB Chapter 13 2011/1/2004

Example 13-6

The cell

Zn | Zn 2+ || MnO4- | Mn2+ | Pt

(the same cell used in Example 3-3) is setup at 298.15 K with the following non-standard concentrations: [H+] = 0.010M (i.e., acidic Ph=2), [MnO4

-] = 0.12M, [Mn2+] = 0.0010M, and [Zn2+] = 0.015M. Calculate the cell voltage.

Anode (oxidation) | cathode (reduction)

This is a Nernst equation problem

1. Balanced the two half reactions

2. Determined overall # electrons transferred

3. Calculate ∆E° for standard conditions (1M)

4. Using the Nernst equation, use ∆E° andsubstitute the non-standard concentrations and solve.

OFB Chapter 13 2111/1/2004


1. Balanced the two half reactions and determined overall # electrons transferred



OFB Chapter 13 2211/1/2004

1. Balanced the two half reactions and determine # electrons transferred

Cathode (Reduction)

MnO4-1 → Mn 2+

Zn | Zn 2+ || MnO4- | Mn2+ | Pt


Anode (Oxidation)

Zn → Zn 2+

To balance electrons flow,

Multiply Reduction X2 and Oxidation X5

n =10

OFB Chapter 13 2311/1/2004

Overall reaction is

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O




Next calculate ∆E° (look-up in tables)

OFB Chapter 13 2411/1/2004





Overall reaction is

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O

Qn

V.∆∆10

log05920EE −°=

1621

4

5222

][][][][

+−

++

=HMnO

ZnMnQ

OFB Chapter 13 2511/1/2004

Qn

V.∆∆10

log05920 EE −= °

1621

4

5222

][][][][

+−

++

=HMnO

ZnMnQ

1621

4

5222

10 ][][][][log

100592025.2E

+−

++

−=HMnO

ZnMnV.V∆

162

5223

10 ]01.0[]12.0[]105.1[]10[log

100592025.2E

−−

−=xV.V∆

OFB Chapter 13 2611/1/2004 voltage) (Cell Volts 2.16 E

)103.5(log10

0592025.2E 18

10

=

−=

∆

xV.V∆

Review Example 13-6

The cell

Zn | Zn 2+ || MnO4- | Mn2+ | Pt

(the same cell used in Example 3-3) is setup at 298.15 K with the following non-standard concentrations: [H+] = 0.010M (i.e., acidic Ph=2), [MnO4-] = 0.12M, [Mn2+] = 0.0010M, and [Zn2+] = 0.015M. Calculate the cell voltage.

Qn

V.∆∆10

log05920EE −°=

OFB Chapter 13 2711/1/2004

Thus far we have used the Nernst equation to calculate the potential difference in a cell.

If given a voltage for a cell, we can calculate an [X] or unknown concentration given the concentrations of the other species

Qn

V.∆∆10

log05920EE −°=

Rearrange the Nernst Equation

OFB Chapter 13 2811/1/2004

]EE[05920

log10

∆∆V.

nQ −°=

Q is the reaction quotient

e.g. for the general reaction,

a A + b B → c C + d D

ba

dc

BADCQ

][][][][

=

OFB Chapter 13 2911/1/2004

13-5 Equilibrium Constants from Electrochemistry

Previously

If a reaction is at equilibrium, after rearrangement this becomes

°

= E

RTnK ∆

Fln

We can now use cell reactions to find Equilibrium constant for Redox reactions, which are difficult to determine directly.

RT

r∆GlnK

°−= ∆Gr° = − n F ∆E°

OFB Chapter 13 3011/1/2004

°∆

= EFln

RTnK

If Temp. at 25C and

R = 8.3 JK-1mol-1 and

F = 96,485 C mol-1 and use

J/C = V,

then the equation becomes

°

= ∆E

V.nK

0592010log

OFB Chapter 13 3111/1/2004

Example 13-8

Calculate the equilibrium constant of the Redox reaction

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O

At 25C using the standard potential difference established in Example 13-3

The cell

Zn | Zn 2+ || MnO4- | Mn2+ | Pt


∆E°= 2.27 volts

N = 10 electrons

OFB Chapter 13 3211/1/2004

°

= E

0592.0log10 ∆

VnK

( )

380

10

10

380 27.2 0592.0

10log

=

=

=

K

VV

K

An enormous equilibrium constant, which means this reaction equilibrium lies to the right.

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O

Permanganate is an extremely strong oxidizing agent and Zinc is an extremely strong reducing agent

OFB Chapter 13 3311/1/2004

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O

Permanganate is an extremely strong oxidizing agent and Zinc is an extremely strong reducing agent

110

][][][][ 380

16214

5222== +−

++

HMnOZnMnK

OFB Chapter 13 3411/1/2004

• Electrolytic Cells can also be used to measure– Acidity constants– Basicity constants– Solubility product constants

AgCl (s) excess → Ag +1 + Cl -1

To determine Ksp, just find the [Ag+1] in an electrolytic cell containing a known amount of [Cl-1]

OFB Chapter 13 3511/1/2004

• Example 13-9A galvanic cell consists of a standard hydrogen half-cell (with platinum electrode) operating as the anode, and a silver half-cell:

Pt|H2 (1 atm)|H+ (1M)||Ag+|AgThe Ag+ (aq) in the cathode compartment is in equilibrium with some solid AgCl and Cl- (aq); the concentration of the Cl- (aq) is 0.00100 M. The measured cell voltage is ∆E=0.398V. Calculate the silver-ion concentration in the cell and the Ksp of silver chloride at 25°C

OFB Chapter 13 3611/1/2004

Pt|H2 (1 atm) | H+ (1M)||Ag+|Ag

Anode (Oxidation)

H2 → 2H + + 2e-

Cathode (reduction)

2 Ag + + 2e- → 2 Ag

][05920

log10 EE

∆∆V.

nQ −°=

Given

0.397 V

From half reactions n=2

Find?Find?

The simplified Nernst equation

OFB Chapter 13 3711/1/2004

Pt|H2 (1 atm)|H+ (1M)||Ag+|Ag

]397.0[05920

2log10 −°= E

∆V.

Q

OFB Chapter 13 3811/1/2004

Overall reaction

2 Ag + + H2 → 2 H + + 2 Ag

7

132

132

22

2

22

106.1][

104][

1

1041][

1]1[

][][][

−+

+

+

+

+

=

==

==

=

xAg

xAg

Q

xAg

Q

PAgAgHQ

H[H+] Given

Ag ppt

Calculated

Answer10

37

106.1

]101][106.1[

]][[

−

−−

−+

=

=

=∴

xK

xxK

ClAgK

AgClsp

AgClsp

AgClsp

Pt|H2 (1 atm)|H+ (1M)||Ag+|Ag

PH2Given

[Cl-] Given

OFB Chapter 13 3911/1/2004

Chapter 13 Summary

∆G = - Q ∆E = -n F ∆E(at constant T,P)

∆G = change in Gibbs EnergyQ = charge∆E = potential differenceN = # electrons transferredF = Faraday constant

][05920

log10 EE

∆∆V.

nQ −°=

∆E° = E° (cathode) - E° (anode)

SuppliesIncreaseReducing AgentPicksDecreaseOxidizing AgentGainDecreaseReductionLossIncreaseOxidation∆Electron∆ OxidizationTerm

OFB Chapter 13 4011/1/2004

13-6 (skip) Batteries and Fuel Cells

OFB Chapter 13 4111/1/2004


OFB Chapter 13 4211/1/2004


Chapter 13 Electrochemistry and Cell...

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