17_Electrochemistry

Electrochemistry

Electrochemistry Terminology #1 A process in which an element attains a more positive oxidation state Na(s)  Na+ + eReduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2ClOxidation –

Electrochemistry Terminology #2 An old memory device for oxidation and reduction goes like this…

LEO says GER Lose Electrons = Oxidation Gain Electrons = Reduction

Electrochemistry Terminology #3 Oxidizing agent  The substance that is reduced is the oxidizing agent  Reducing agent  The substance that is oxidized is the reducing agent 

Electrochemistry Terminology #4 



Anode  The electrode where oxidation occurs Cathode  The electrode where reduction occurs

Memory device:

Reduction at the Cathode

Table of Reduction Potentials

Measured against the Standard Hydrogen Electrode

Measuring Standard Electrode Potential

Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E0 A negative free energy change, (-G)

Zn - Cu Galvanic Cell From a table of reduction potentials: Zn2+ + 2e-  Zn Cu2+ + 2e-  Cu

E = -0.76V E = +0.34V

Zn - Cu Galvanic Cell The less positive, or more negative reduction potential becomes the oxidation…

Zn  Zn2+ + 2eCu2+ + 2e-  Cu Zn + Cu2+  Zn2+ + Cu

E = +0.76V E = +0.34V E0 = + 1.10 V

Line Notation An abbreviated representation of an electrochemical cell

Zn(s ) | Zn2+(aq ) || Cu2+(aq ) | Cu(s ) Anode Anode | material solution

||

Cathode solution

|

Cathode material

Calculating G0 for a Cell G0 = -nFE0  n  F

= moles of electrons in balanced redox equation = Faraday constant = 96,485 coulombs/mol e -

Zn + Cu2+  Zn2+ + Cu 0

G   ( 2 mol e 0



)(96 485

 E0

coulombs mol e

G   212267 Joules  



= + 1.10 V

)(1.10

212 kJ 

 Joules Coulomb

)

The Nernst Equation Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. 0

 E  E    R

= 8.31 J/(mol K)

 RT  nF 

ln(Q)

T  =  n

Temperature in K = moles of electrons in balanced redox equation

 F

= Faraday constant = 96,485 coulombs/mol e -

Nernst Equation Simplified At 25 C (298 K) the Nernst Equation is simplified this way: 0

 E  E 



0.0591 n

log(Q )

Equilibrium Constants and Cell Potential At equilibrium, forward and reverse reactions occur at equal rates, therefore: 1. The battery is “dead” 2. The cell potential, E, is zero volts Modifying the Nernst Equation (at 25 C): 0

0 volts  E 



0.0591 n

log( K )

Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu 0 volts 1.10 

(1.10)( 2) 0.0591

 E0

0.0591 2

log( K )

 log( K )

37.2  log( K )

10

37.2

 K   1.58 x 10

= + 1.10 V

37

???

Concentration Cell Both sides have the same components but at different concentrations.

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

???

Concentration Cell Anode

Cathode

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 concentration Zn2+ (1.0M) + 2e-  Zn (reduction) Zn  Zn2+ (0.10M) + 2eZn2+ (1.0M)  Zn2+ (0.10M)

(oxidation)

???

Concentration Concentration Cell Cell

Anode

Cathode

Both sides have the same components but at different concentrations.

Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C). Zn2+ (1.0M)  Zn2+ (0.10M) 0

 E  E 



0.0591 n

log(Q )

Nernst Calculations Zn2+ (1.0M)  Zn2+ (0.10M) 0

 E  E 



0

 E   0.0 Volts

 E  0.0



0.0591 n

log( Q )

n2

0.0591 2

log(

0.10 1 .0

Q

(0.10) (1.0)

)  0.030 Volts

Electrolytic Processes Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0)

A positive free energy change, (+

G)

Electrolysis of Water In acidic solution

Anode rxn: 2 H 2O  O2  4 H   4e   Cathode rxn: 4 H 2O  4e  2 H 2  4OH  

2 H 2O  2 H 2  O2



-1.23 V -0.83 V -2.06 V

Electroplating of Silver Anode reaction: Ag Ag+ + eCathode reaction: Ag+ + eAg Electroplating requirements: 1. Solution of the plating metal 2. Anode made of the plating metal 3. Cathode with the object to be plated 4. Source of current

Solving an Electroplating Problem Q: How many seconds will it take to plate out  5.0 grams of silver from a solution of AgNO 3  using a 20.0 Ampere current? 

Ag+ + e5.0 g

Ag

1 mol Ag 1 mol e-

96 485 C 1 s 107.87 g 1 mol Ag 1 mol e- 20.0 C

= 2.2 x 102 s