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S.E. Van Bramerechem_a.mcd4/24/01

Solutions for Electrochemistry Problem Set

Constants:F96484.56coul.mole

1.T273.1525()K.M

mole literR8.31441joule.mole 1.K

1.Equations

E std_cellEcathodeEanodeE cellEstd_cellRT .nF .ln

Canode

C

cathode.1 a. Calculate the cell potential and free energy available for the following electrochemical systems

Ag ( s)| Ag1+ (aq 1.0M) || Cu2+ (aq 1.0M) | Cu(s)

Anode (oxidation) Ag (s) <--> Ag1+ + 1 e-

Cathode (reduction Cu2+ + 2e- <--> Cu(s)

From the table of reduction potentials, we can findE anode0.7996volt.E cathode0.3402volt.And then calculate the standard cell potential E std_cellEcathodeEanodeE std_cell0.4594volt=Next determine the cell potential at the concentrations given

Balance the oxidation and reduction reactions

2 Ag (

s) + Cu2+(aq) <--> 2 Ag1+(aq) + Cu (s)

The number of electrons exchangedn2Calculate Q

Q CAg2 C

CuAnode:Cathode:C

Ag1.0C

Cu1.0Calcuations:

E cellEstd_cellRT.nF .ln CAg2 C Cu.E cell0.4594volt=

S.E. Van Bramerechem_a.mcd4/24/01

1.b This is the same reaction for everything except the concentrations so:

Anode:Cathode:C

Ag0.1C

Cu0.1Calcuations:

E cellEstd_cellRT.nF .ln CAg2 C Cu.E cell0.4298volt=1.c This is the same reaction for everything except the concentrations so:

Anode:Cathode:C

Ag1.0C

Cu0.1Calcuations:

E cellEstd_cellRT .nF .ln CAg2 C Cu.E cell0.489volt=1.d This is the same reaction for everything except the concentrations so:

Anode:Cathode:C

Ag1C

Cu0.01Calcuations:

E cellEstd_cellRT .nF .ln CAg2 C Cu.E cell0.5186volt=1.e This is the same reaction for everything except the concentrations so:

Anode:Cathode:C

Ag0.1C

Cu1.0Calcuations:

E cellEstd_cellRT .nF .ln CAg2 C Cu.E cell0.4002volt=

S.E. Van Bramerechem_a.mcd4/24/01

1 f In this problem the cell is reversed so that:

Cu( s) | Cu2+ (aq 1.0M) || Ag1+ (aq 1.0M) | Ag (s)

Anode (oxidation) Cu (s) <--> Cu2+ + 2e-

Cathode (reduction Ag1+ + 1 e- <--> Ag(s)

From the table of reduction potentials, we can findE anode0.3402volt.E cathode0.7996volt.And then calculate the standard cell potential E std_cellEcathodeEanodeE std_cell0.4594volt =Next determine the cell potential at the concentrations given

Balance the oxidation and reduction reactions

2 Ag

1+(aq) + Cu (s) <--> 2 Ag (s) + Cu2+(aq)

The number of electrons exchangedn2Calculate Q

Q CCuC Ag2

Anode:Cathode:C

Ag1.0C

Cu1.0Calcuations:

E cellEstd_cellRT .nF .ln CCu C Ag2 .E

cell0.4594volt=Notice that in this reaction the cell potential is positive, this electrochemical cell is

spontaneous (the reactions are going the way they want to). So this is the voltage produced by the cell. It is acting like a battery here. In the previous examples, the reactions were all going in the non-spontaneous direction. The voltage was negative, indicating that this is the voltage that must be applied to the system to push it backwards. The previous cells were electrolytic, they were being charged.

S.E. Van Bramerechem_a.mcd4/24/01

2. If the electrochemical cell discussed is used as a battery and begins with electrodes and

150 mL of solution. Identify the limiting reagent and calculate the moles of electrons

exchanged when the reaction goes to completion. We use the balanced chemical equation from 1f, where the reaction was spontaneous. 2 Ag

1+(aq) + Cu (s) <--> 2 Ag (s) + Cu2+(aq)

So, silver ions and copper metal are the reactants.moles

Ag_ion0.250L.()1.0M.().moles

Ag_ion0.25mol

=moles

Cu_solid10gm

.63.546 gm mole .moles

Cu_solid0.1574mol

=Since the balanced equation shows that two moles of silver ions are required for each mole of copper solid, silver is the limiting reagent.moles

Cu_solid_usedmolesAg_ion

2moles

Cu_solid_used0.125mol

=Each silver ion requires one electron to be reduced. So the moles of electrons are the same as the moles of silver.moles electronmolesAg_ionmoles electron0.25mol

S.E. Van Bramerechem_a.mcd4/24/01

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