the end is in site! nernst and electrolysis. electrochemistry
TRANSCRIPT
The End is in Site!
Nernst and Electrolysis
ElectrochemistryElectrochemistry
Galvanic (Electrochemical) CellsGalvanic (Electrochemical) Cells
Spontaneous redox processes have:
A positive cell potential, E0
A negative free energy change, (-G)
Zn - Cu Zn - Cu Galvanic Galvanic CellCell
Zn2+ + 2e- Zn E = -0.76V
Cu2+ + 2e- Cu E = +0.34V
From a From a table of table of reduction reduction potentials:potentials:
Zn - Cu Zn - Cu Galvanic Galvanic CellCell
Cu2+ + 2e- Cu E = +0.34V
The less positive, The less positive, or more negative or more negative reduction reduction potential potential becomes the becomes the oxidation…oxidation…
Zn Zn2+ + 2e- E = +0.76V
Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V
Line Line NotationNotation
Zn(Zn(ss) | Zn) | Zn2+2+((aqaq) || Cu) || Cu2+2+((aqaq) | Cu() | Cu(ss))
An abbreviated An abbreviated representation representation of an of an electrochemical electrochemical cellcell
AnodeAnodesolutionsolution
AnodeAnodematerialmaterial
CathodeCathodesolutionsolution
CathodeCathodematerialmaterial|| ||||||
E0cell
G0 K
E0cell = RT/nF (lnK)
G0 = -nFE0cell
G = -RTlnK
Calculating Calculating GG00 for a Cell for a Cell
)10.1)(48596)(2(0
Coulomb
Joules
emol
coulombsemolG
GG00 = -nFE = -nFE00
nn = moles of electrons in balanced redox equation= moles of electrons in balanced redox equation
FF = = Faraday constant = 96,485 coulombs/mol eFaraday constant = 96,485 coulombs/mol e--
Zn + Cu2+ Zn2+ + Cu EE00 = + 1.10 V
kJJoulesG 2122122670
The Nernst EquationThe Nernst Equation
)ln(0 QnF
RTEE
Standard potentials assume a Standard potentials assume a concentration of 1 M. The Nernst concentration of 1 M. The Nernst equation allows us to calculate potential equation allows us to calculate potential when the two cells are not 1.0 M.when the two cells are not 1.0 M.
RR = 8.31 J/(mol= 8.31 J/(molK) K)
TT = Temperature in K = Temperature in K
nn = moles of electrons in balanced redox equation = moles of electrons in balanced redox equation
FF = Faraday constant = 96,485 coulombs/mol e = Faraday constant = 96,485 coulombs/mol e--
Nernst Equation SimplifiedNernst Equation Simplified
)log(0591.00 Qn
EE
At 25 At 25 C (298 K) the Nernst Equation C (298 K) the Nernst Equation is simplified this way:is simplified this way:
Equilibrium Constants and Cell PotentialEquilibrium Constants and Cell Potential
At equilibriumequilibrium, forward and reverse reactions occur at equal rates, therefore:1.1. The battery is “dead”
2. The cell potential, E, is zero volts
)log(0592.0
0 0 Kn
Evolts
Modifying the Nernst Equation (at 25 C):
Zn + Cu2+ Zn2+ + Cu EE00 = + 1.10 V
Calculating an Equilibrium Constant Calculating an Equilibrium Constant from a Cell Potentialfrom a Cell Potential
)log(2
0591.010.10 Kvolts
)log(0591.0
)2)(10.1(K
)log(2.37 K
372.37 1058.110 xK
Concentration Concentration CellCell
Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.
Both sides have the same
components but at different
concentrations.
??????
Concentration Concentration CellCell
Both sides have the same
components but at different
concentrations.
The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction) Zn Zn2+ (0.10M) + 2e-
(oxidation)
??????
CathodeCathodeAnodeAnode
Zn2+ (1.0M) Zn2+
(0.10M)
Concentration Concentration CellCell
Both sides have the same
components but at different
concentrations.
Spontaneous: e- move from anode to cathode (+Voltage)Use Le Chatelier: Increasing [Zn2+ (1.0M)] forces the reverse to take placeTherefore, Decreasing the Ecell
??????
CathodeCathodeAnodeAnode
Concentration CellConcentration Cell
)log(0591.00 Qn
EE
Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).
Both sides have the same
components but at different
concentrations.
??????
CathodeCathodeAnodeAnode
Zn2+ (1.0M) Zn2+
(0.10M)
ConcentratioConcentration Celln Cell
VoltsE 0.00 )0.1(
)10.0(Q2n
VoltsE 030.0)0.1
10.0log(
2
0591.00.0
Nernst CalculationsNernst CalculationsZn2+ (1.0M) Zn2+
(0.10M)
)log(0591.00 Qn
EE
Electrolytic Electrolytic ProcessesProcesses
A negative cell potential, (--EE00)
A positive free energy change, (++GG)
Electrolytic processes are NOTNOT spontaneous. They have:
Electrolysis of Electrolysis of WaterWater
eHOOH 442 22
OHHeOH 4244 22
In acidic solutionIn acidic solution
Anode rxn:Anode rxn:
Cathode rxn:Cathode rxn:-1.23 V-1.23 V
-0.83 V-0.83 V
-2.06 V-2.06 V222 22 OHOH
Electroplating of Electroplating of SilverSilver
Anode reactionAnode reaction::
Ag Ag Ag Ag++ + e + e--
Electroplating requirementsElectroplating requirements::1. Solution of the plating metal1. Solution of the plating metal
3. Cathode with the object to be plated3. Cathode with the object to be plated2. Anode made of the plating metal2. Anode made of the plating metal
4. Source of current4. Source of current
Cathode reactionCathode reaction::
AgAg++ + e + e-- Ag Ag
A couple of important relationships to remember:
C = coulomb = charge transported by a steady current of one ampere in one second
C = amp x sec
F = C/n then C = nF
If you know the moles of e- generated, then you use stoichiometry to determine moles of the metal that are deposited.
Example #1Assume that 1.50amps of current flow through a solution containing silver ions for 15.0 minutes. The voltage is such that silver is deposited at the cathode. How many grams of silver metal are deposited? Ag+ + e- Ag
Hint: C = amp x sec
C = 1.50 amp x (15min x 60sec/min)
C = 1.35 x 103 C
Hint: C = nF therefore n = C/F
n = 1.35 x 103 C / 9.65 x 104
n = 1.40 x 10-2 mole e- therefore 1.40 x 10-2 mole Ag
= 1.51g Ag are deposited
Example #2One ½ reaction occurring in the lead storage battery is:
Pb + SO42- PbSO4 + 2e-
If the battery delivers 1.50 amps and if its lead electrode contains 454g of Pb, how long can current flow?
Hint: convert to mole Pb
454g = 2.19mol Pb which means 4.38 mole e- are produced
C = nF = 4.38 (9.65 x 104)
C = 4.23 x 105 C
Hint: C = amp x sec
4.23 x 105 = 1.50amp x time
Time = 2.82 x 105sec = 78.3 hours
