11678_lecture15 2-14-07 voltaic cells

7/30/2019 11678_Lecture15 2-14-07 Voltaic Cells

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A Hydrogen ElectrodeCell potentials can bemeasured

Half-cell potential cannot be

measured

An arbitrary choice sets

the half cell potential of astandard hydrogen

electrode = 0

Eo1/2(H+/H2) = 0 V

Half Reaction is always

written as a reduction

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


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Questions

Anode?

Cathode?

Ox An

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

Red Cat

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

Spontanteous Cell Reaction?

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

Zinc metal dissolves in 1 M H+

Zn/H2 Voltaic Cell

Cell Potential = 0.763 Vunder standard conditions


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Voltage of Voltaic Cells

All voltaic cells are based on a

spontaneous chemical reaction

Therefore, G for the reaction must be

negative This means that the voltage of a voltaic

cell is always positive.

G = -nFe


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Standard Potential of Zn/H2 Cell2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)

Cathode: 2 H+ (aq) + 2e- H2 (g)Anode: Zn (s) Zn2+ (aq) + 2 e-

Eo

(cell) = E1/2o

(H+

/H2) E1/2o

(Zn2+

/Zn)=+0.763V

E0(cell) = 0 E1/2o (Zn2+/Zn) = 0.763 V

E1/2o (Zn2+/Zn) = -0.763 V

Zn2+ (s) + 2e- Zn(s) E1/2o = -0.763 V

Cathode Anode


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Determination of Half-Cell PotentialsAnode and cathode?

H+/H2 is anode and Cu2+/Cu is

cathode

H2 (g) + Cu2+ 2 H+ (aq) + Cu (s)

E0(cell) = E01/2(Cu2+/Cu) E01/2(H

+/H2)

E0(cell) = E01/2(Cu2+/Cu) 0 = 0.340 V

E01/2(Cu2+/Cu) = +0.340 V

Cu/H2 Voltaic CellCell Potential = 0.340 V

under standard conditions


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Cell EMFCell EMF

Standard Reduction PotentialsStandard Reduction Potentials

The larger the difference

between Ered values, thelargerEcell.

In a voltaic cell (spontaneous)

Ered(cathode) is more positivethan Ered(anode).

RecallEcell = Ered(cathode) -

Ered(anode)


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Half-reaction

cells

Eo(cell) = 1.10 V (measured)

E1/2o (Cu2+/Cu) E1/2

o (Zn2+/Zn)

Eo (cell) = 0.340 V (-0.763 V)

Eo (cell) = 1.10 V (Calculated from E01/2)


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Selected StandardElectrode Potentials at 25 oC

EOS

StrongerO

xidizingAge

nts

StrongerReducing

Agents


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Cell EMFCell EMF

Oxidizing and Reducing AgentsOxidizing and Reducing Agents


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Cell Potential Determine the cell potential for a galvanic cellbased on the redox reaction.

Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

Fe+3(aq) + e- Fe+2(aq) E1/20= 0.77 V

Cu+2(aq)+2e- Cu(s) E1/20= 0.34 V

Cu(s) Cu+2(aq)+2e- -E1/20= -0.34 V

2Fe+3(aq) + 2e- 2Fe+2(aq) E1/20= 0.77 V

Cu(s) + 2Fe3+ Cu2+ + 2Fe2+ Ecell0= 0.43V


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Shorthand Cell Notation

eg: Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

solidAqueousAqueoussolid

Anode on the leftCathode on the right

Single line different phases.

Double line porous disk or salt bridge. If all the substances on one side are

aqueous, a platinum electrode is indicated.

Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)Anode CathodeSolution Bridge Solution


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Shorthand Notation Voltaic Cell

Zn Zn2+

(anode)

porous

plate

Cu2+ Zn2+

Cu2+ Cu(cathode)

e-

1.10 V

Start HereEnd Here

Tell the story. What do you see during your trip from the anode to the cathode?

Zn(s) Zn2+ (aq, 1M) Cu2+ (aq, 1M) Cu(s)

1.00 M1.00 M

Trip


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Practice Completely describe the galvanic cell based on

the following half-reactions under standardconditions.

MnO4- + 8 H+ +5e- Mn+2 + 4H2O E=1.51

Fe+3

+3e-

Fe(s) E=0.036V Anode? Cathode?

Cell Potential? Oxidizing agent and reducing agent?

Shorthand cell notation?


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Fe3+ (aq)

MnO4- + 8 H+ +5e- Mn+2 + 4H2O E=1/2 1.51

E0cell = 1.51V 0.036V = 1.47V

Fe+3 +3e- Fe(s)E1/2=0.036V

Fe(s) Fe3+ + 3 e- -E1/2 o = -0.036V

Questions

1. Whats missing in Diagram?

2. Anode?

3. Cathode?

4. Cell Reaction?

5. Direction electron flow?

6. Cell Potential?

7. Shorthand Representation?Fe(s)Fe3+MnO4- (aq H+), Mn2+Pt

Practice3 MnO

4

- + 24 H+ + 5 Fe(s)

3 Mn2+ + 12 H2O + 5 Fe3+

e-e-

e-


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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions

Consider the displacement of silver by nickel:

Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s)

E0 = E0red(Ag+/Ag) E0red(Ni2+/Ni)= (0.80 V) - (-0.28 V)

= 1.08 VThe positive E0 indicates a

spontaneous process.

EMF and FreeEMF and Free--Energy ChangeEnergy Change

G= -nFE


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Spontaneity of Redox ReactionsSpontaneity of Redox Reactions

EMF and FreeEMF and Free--Energy ChangeEnergy Change

G is the change in free-energy, n is the

number of moles of electrons transferred, FisFaradays constant, and Eis the emf of the

cell.

Since n and Fare positive, ifG > 0 then E

11678_lecture15 2-14-07 voltaic cells

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