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