Chapter 13 Electrochemistry and Cell...

OFB Chapter 13 13/23/2004

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


OFB Chapter 13 103/23/2004

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 13 113/23/2004

Standard Reduction potentials E °

Ni2+ + 2e- → Ni (s) -0.257 Volts

Cu2+ + 2e- → Cu (s) +0.345 Volts

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

Cu2+ reduction potential is more positive (+0.34) 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

OFB Chapter 13 123/23/2004

By international convention

2H+ + 2e- → H2 (g)E = 0.0 volts

All reduction potentials can be determined relative to H+

reduction.

OFB Chapter 13 133/23/2004


• 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 13 143/23/2004

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

OFB Chapter 13 153/23/2004

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

OFB Chapter 13 163/23/2004

Oxidizing and Reducing Agents

Supplies ElectronsIncrease

4. Reducing Agent, does the reducing

Picks Up electronsDecrease

3. Oxidizing Agent, does the oxidizing

Gain of ElectronsDecrease2. Reduction

Loss of ElectronsIncrease1. Oxidation

Electron Change

Oxidation Number ChangeTerm

OFB Chapter 13 173/23/2004

Figure 13-5

OFB Chapter 13 183/23/2004

Example 13-5

(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).

(b) Determine which is the stronger reducing agent under these conditions. Refer to 13-5 for necessary data

OFB Chapter 13 193/23/2004

Example 13-5



This is a disproportionation problem, write to get both H3PO3 on the left and then add and calculated ∆E

H3PO3 + 2H+ + 2e-→ H3PO2 + H2O

H3PO3 + H2O → H3PO4 + 2H+ + 2e-

Now Calc. ∆E

OFB Chapter 13 203/23/2004

Example 13-5


Only two half reactions involving H3PO3

H3PO3 + 2H+ + 2e-→ H3PO2 + H2O E= (-0.499) V

H3PO4 + 2H+ + 2e- → H3PO3 + H2O E= (-0.276) V

This is disproportionation, write to get both H3PO3 on the left and then add and calculated ∆E

H3PO3 + 2H+ + 2e-→ H3PO2 + H2O

H3PO3 + H2O → H3PO4 + 2H+ + 2e-

Reduction

Oxidation

OFB Chapter 13 213/23/2004

Why stable?

Recall

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

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

Disproportionation is

2 H3PO3 → H3PO2 + H3PO4

OFB Chapter 13 223/23/2004

Example 13-5


Only two half reactions involving SO2

2SO2 + 2H+ + 4e-→ S2O3 -2 + H2O

HSO4- + 3H+ + 2e- → SO2 + 2H2O

Disproportionation of SO2

Rewrite to get both SO2 on the left, balance electrons and stoichiometry and add

OFB Chapter 13 233/23/2004

Example 13-5


∆E = E (cathode) – E (anode)

= 0.40 – (+ 0.172) = +0.228 ∴unstable

Why unstable? It’s a spontaneous reaction.

RecallFor a Galvanic Cell, a spontaneous reaction takes place.

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

Disproportionation of SO2

4 SO2 + 3 H2O → 2 HSO4- + S2O3

-2 + 4H+

OFB Chapter 13 243/23/2004

Example 13-5


H3PO3 is stable

(non-spontaneous disproportionation)

and

SO2 is unstable

(spontaneous disproportionation)


OFB Chapter 13 253/23/2004

Example 13-5



Only two half reactions involving SO2

2SO2 + 2H+ + 4e-→ S2O3 -2 + H2O E= +0.40 V

HSO4- + 3H+ + 2e- → SO2 + 2H2O E= +0.172 V

Only two half reactions involving H3PO3

H3PO3 + 2H+ + 2e-→ H3PO2 + H2O E= (0.499) V

H3PO4 + 2H+ + 2e- → H3PO3 + H2O E= (0.276) V

Which is the most negative?

OR

OFB Chapter 13 263/23/2004

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 13 273/23/2004

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 13 283/23/2004

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 13 293/23/2004


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



OFB Chapter 13 303/23/2004

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+

OFB Chapter 13 313/23/2004

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 13 323/23/2004





Overall reaction is

2 MnO4-1 + 5 Zn + 16H+

→ 2Mn 2+ + 5 Zn 2+ + 8 H2O

OFB Chapter 13 333/23/2004

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 13 343/23/2004 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 13 353/23/2004

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 13 363/23/2004

]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 13 373/23/2004

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 13 383/23/2004

°∆

= 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 13 393/23/2004

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 13 403/23/2004

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 13 413/23/2004

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 13 423/23/2004

• 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 13 433/23/2004

• 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 13 443/23/2004

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

][05920

log10 EE

∆∆V.

nQ −°=

OFB Chapter 13 453/23/2004


]397.0[05920

2log10 −°= E

∆V.

Q

OFB Chapter 13 463/23/2004

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

[Cl-] Given

Ag ppt

Calculated

Answer10

37

106.1

]101][106.1[

]][[

−

−−

−+

=

=

=∴

xK

xxK

ClAgK

AgClsp

AgClsp

AgClsp


PH2Given

OFB Chapter 13 473/23/2004

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 13 483/23/2004

13-6 (skip) Batteries and Fuel Cells

OFB Chapter 13 493/23/2004


OFB Chapter 13 503/23/2004


OFB Chapter 13 513/23/2004


• Examples /exercises13-2, 13-3, 13-4, 13-5, 13-6, 13-7, 13-8, 13-9

• Problems3, 5, 9, 11, 15, 19, 29, 31, 35, 37, 41, 45

Date post:	09-Feb-2018
Category:	Documents
Upload:	hoangliem
View:	229 times
Download:	6 times

Chapter 13 Electrochemistry and Cell...

Documents