determination of electrode potentials
TRANSCRIPT
DETERMINATION OF ELECTRODE POTENTIALS
Electrochemistry
• Electrochemistry deals with the chemical changes produced by electric current and with the production of electricity by chemical reactions
• => all electrochemical reactions involve the transfer of of electron and are therefore redox reactions
• => the sites of oxidation and reduction are separated physically so that oxidation occurs at one location and reduction occurs at the other
Electrochemical cells: 2 types
• 1. Galvanic or Voltaic cells- spontaneous chemical reactions produce electricity and supply it to sn external circuit, provides a useful source of energy
• 2. Electrolytic cells- electrical energy from an external source causes nonspontaneous reactions to occur, no need for a salt bridge!!
Galvanic vs. Electrolytic cellANODE CATHODE
Rxn Ion moving towards it
Sign Rxn Ion moving towards it
Sign
Galvanic Cell oxidation anions _ reduction cations +
Electrolytic cell oxidation anions + reduction cations _
=> the difference between the anode and cathode of galvanic and electrolytic cells is their polarity!!!
Flow of electrons: ANODE → CATHODE (always!!!)
What are we measuring?• Electromotive force (emf) or Cell potential –
measures the tendency of whether a reduction-oxidation reaction will proceed and in what direction; measured by Ecell
• i.e. • Ecell = (+) => reaction proceeds from left to right (Product Favored)
• Ecell = (-) => reaction proceeds from right to left (Reactant Favored)
• • Ecell = 0 => system is at equilibrium• • Ecell = Eo
cell => all species are in their standard concentrations
What are we measuring?• Note: standard electrode potentials are relative
values based on the standard reference hydrogen electrode: H+
(aq) + 2e- → ½H2, which has an assigned half-cell potential of zero
• * Standard states- the standard concentration for 1.) pure substances like pure solid and pure liquid (ex. H2O) is equal to 1, 2.) ionic species (ex. Cu2+) is 1 M, 3.) gases is 1 atm
GALVANIC CELLS - useful energy is produced
• Consider the reaction
• Zn(s) + Cu2+(aq) → Cu(s) + Zn2+
(aq)
• => in a mixed system, there is a direct transfer of electrons from the Zn atom to each Cu2+ ion => work done by the system is not harnessed
• => but if contact between the Cu2+ and Zn is confined to a wire connection, electron transfer is converted to useful electrical work through the wire
• => equivalent to the Gibbs free energy (ΔG) of the reaction system
• => this useful electrical work is harnessed through a galvanic cell. So, how should a galvanic cell be set up?
Galvanic cell
• Cell notation: Zn(s)|Zn2+(1M)||Cu2+(1M)|Cu(s
• Anodic reaction: Zn(s) → Zn2+ + 2 e- Eo = -0.76 V• Cathodic reaction: Cu2+ + 2 e- → Cu(s) Eo = 0.34 V• Overall reaction: Zn(s)+Cu2+
(aq)→Cu(s)+Zn2+(aq) Eo = 1.1 V
A galvanic cell is a device that converts electron transfer into useful electrical work
The Salt Bridge• Maintains electroneutrality
• Provides the contact between the 2 solutions
• => without the salt bridge, no reaction will occur since no ions will replenish the charge imbalance in the half-cells
• Ideal salt for salt bridge: the mobility of the cation and the anion of the salt should be nearly equal (ex: KNO3)
• CATCH: YOU DO NOT NEED A SALT BRIDGE IN AN ELECTROLYTIC CELL. DID WE?
ELECTROLYTIC CELLS - forcing a non-spontaneous reaction to happen at the
expense of energy-no salt bridge is required
Electrolytic cell
• Cell notation: C(graphite)|I2(s)|I-(1M)||OH-(10-7M)|H2(1atm)|C(graphite)
• Anodic reaction: 2 I- → I2 + 2 e-
• Cathodic reaction: 2 H2O + 2 e- → 2 H2 + 2 OH-
• Overall reaction: 2 I- + 2 H2O → 2 H2 + 2 OH-
Zero oxidation states must be placed near the electrode
No need to write water in the cell notation
This process is called electrolysis
Note that graphite was used since it does not react with the solution and it is a good conductor of electricity
Cell Potential, Ecell
• Ecell = Ecathode – Eanode
• At standard conditions,
• Ecell = Eocell = Eo
cathode – Eoanode
• Example: Eocell for Zn(s) + Cu2+ (aq) → Cu(s) + Zn2+ (aq)
Zn(s) → Zn2+ + 2 e- Eo = -0.76 VCu2+ + 2 e- → Cu(s) Eo = 0.34 V
• Eocell = Eo
cathode – Eoanode = 0.34 V – (-0.76 V) = 1.10 V
The Nernst Equation
• used to measure the cell potential for species not in their standard conditions (not 1M, not 1 atm, etc)
• For any temperature:
• At 25oC (298 K):
QlognF
RT303.2EE cello
cell
where R = 8.314 J/mol K, F = 96485 C/mol e-, n = no. of e- transferred, T = temp. in K
Qlogn
0592.0EE cello
cell
For Cu2+ + 2 e- → Cu(s),
]Cu[
1log
n
0592.0EE
2cello
cell
Equilibrium constants from the nernst equation
• At equilibrium, ΔG = 0, Ecell = 0, and Q = Keq thus, becomes
• Solving for Keq, we get:
• Note that Keq can be Ksp, Ka, Kb, Kf, etc.
Qlogn
0592.0EE cello
cell
eqcello Klog
n
0592.0E
0592.0
)E)(n(logantiK cell
o
eq
ΔG and Ecell
• Thus, the reaction is spontaneous only if ΔG is negative and Ecell is positive
• Example: Calculate the ΔGo in J/mol for 3 Sn4+ + 2Cr(s) → 3 Sn2+ + 2 Cr3+
Cathode: 3(Sn4+ + 2e- → Sn2+) Eo = + 0.15 VAnode: 2(Cr(s) → Cr3+ +3e-) Eo = - 0.74 V
Eocell = (+0.15 V) –(-0.74 V) = + 0.89 V
=> The very negative value of ΔG indicates that the reaction is product-favored. This is consistent with the positive value of Ecell
cellnFEG
where F = 96485 C/mol e-, n = no. of e- transferred
mol/J230,515V89.0emol
C485,96
rxnmol
emol6nFEG cell
oo
Note: 1 J = 1 CV
Equilibrium constant?
• Let us now calculate for the Keq of the previous example
• Recall: Eocell = (+0.15 V) –(-0.74 V) = + 0.89 V
• Thus,
• => the very large value of Keq reinforces our previous conclusion that the reaction is product-favored
90eq 10x59.1
0592.0
)89.0)(6(logantiK
EXPERIMENT PROPER
Part A
Cell Notation Anode Cathode
Cu(s)|Cu2+(aq) (0.01 M)|| Cu2+
(aq) (0.1M)|
Cu(s)
Cu(s)|Cu2+(aq) (0.01M) Cu2+
(aq) (0.1M)|Cu(s)
Zn(s)|Zn2+(aq) (0.1M)|| Cu2+
(aq) (0.1M)|Cu(s) Zn(s)|Zn2+(aq) (0.1M) Cu2+
(aq) (0.1M)|Cu(s)
C(graphite)|Fe2+(aq) (0.5M), Fe3+
(aq) (1M)|| Cu2+
(aq) (0.5M)|Cu(s)
C(graphite)|Fe2+(aq) (0.5M),
Fe3+(aq) (0.5M)
Cu2+(aq) (0.5M)|Cu(s)
Note that [Fe2+] and [Fe3+] are both 1 M not 2 M!!!
Theoretical Ecell = Eocell = Eo
cathode - Eoanode since the cells have
standard concentrations (all 1 M)
Concentration cell
• Concentration cells- cells wherein both half cells are composed of the same species but in different ion concentrations
• Eocell in concentration cells is always 0
• For Cu(s)|Cu2+(aq) (1M)|| Cu2+
(aq) (0.1M)|Cu(s)
]solutionedconcentrat[
]solutiondilute[log
n
0592.00Ecell
V030.01
1.0log
2
0592.00Ecell
Concentration cell• As the reaction proceeds, [Cu2+] decreases in the
more concentrated half-cell and increases in the more dilute half-cell until the two concentrations are equal
• At that point, Ecell = 0, and equilibrium has been reached
• => in any concentration cell, the spontaneous reaction is always in the direction that equalizes the concentrations
ELECTROLYSIS OF HALIDESWe did the same thing to KBr and KCl
It may be assumed that the concentration of the halide ion, X-, does not significantly change with the electrolysis of the solution
ELECTROLYSIS
• Next, we connected this C(graphite)| Cl2(g) (1M)|Cl-
(aq) (1M)|| half-cell to a ||Cu2+(aq) (1M)|Cu(s)
half-cell and measured its potential
Cell Notation of Sample Half Cell Half-reaction (reduction) Standard Reduction Potential
Experimental Value
C(graphite)| Cl2(g) |Cl-(aq) (0.1M)||Cu2+
(aq) (0.1M)|Cu(s)
Cl2 + 2 e- → 2 Cl- 0.89 V
C(graphite)| Br2(l),Br-(aq) (0.1M)||Cu2+
(aq) (0.1M)|Cu(s)
Br2 + 2 e- → 2 Br- 0.96 V
C(graphite)| I2(s),I-(aq) (0.1M)||Cu2+
(aq)
(0.1M)|Cu(s)
I2 + 2 e- → 2 I- 0.92 V
Quantitative Aspects of Electrolysis
• Example: Calculate the mass of copper metal produced at the cathode during the passage of 2.50 amperes of current through a solution of copper(II) sulfate for 50.0 minutes.
• Cu2+(aq) + 2e- → Cu(s)
Current X Time
no. of Coulombs
Useful Conversions:
1 J = 1 CV
1 A = 1 C/s
Mass of substance
Mol. of e- passed
Cug47.2emol2
Cug5.63
C485,96
emol1
s
C50.2
min1
s60min0.50Cug