Introduction to Electrochemical Cells

Thanks to Eric Edens ‘07

Bollens, Rachel A Pitkin, Julia D

Calhoun, Corinne A Liu, Melinda D

Clute-Reinig, Nicholas A Takahashi, Alexis D

Crawford, Catherine A Tarhan, Leyla D

Cyr, Christina B Valdez, Stephanie E

Evarkiou-Kaku, Anatolia B Vega, Alejandra E

Ferguson, Luke B Vega, Marvin E

Fernald, Samuel B Wang, Kevin E

Folse, Katherine C Washington, Shannon F

Forrester, William C Winninghoff, Hayley F

Ho, Vanessa C Zahedi, Leila F

Kuo, Linus C    

Experimental Goals

• Reinforce topics from last semester– Provide experience working with electrochemical cells– Investigate dependence of cell potential on

concentration• Nernst Equation

– Using cell potential to determine equilibrium constants

• Introduce related material– Concentration Cells– Ionic Strength

Two Types of Electrochemical Cells

Information on Procedure• Working with a partner

• Check out kit from stockroom

• Using small volumes for measurements– Take only what you need, a few mLs– Avoid making cells that form a precipitate,

such as silver with halides

• If precipitates form, combine half-cell of interest with different half-cell.– Why?

Part A: Potential Measurements• Use filter crucible and beaker to construct

electrochemical cells– Red lead is Cathode– Black lead is Anode– Make sure that your multimeters are set to

measure VDC

• All half-cells will be expressed vs Ag+|Ag half-reaction in the end

• Need to make measurements that allow this comparison

Example

Suppose you have data for the following two reactions:

2Ag+ + Cu 2Ag + Cu2+ ΔE1 = 0.46V

Zn2+ + Cu Zn + Cu2+ ΔE2 = -1.10V

You can then calculate ΔE for

2Ag+ + Zn 2Ag + Zn2+ ΔE3 = ?Under standard conditions 1atm pressure, each metal 1.0M

Example

2Ag+ + Cu 2Ag + Cu2+ ΔE1 = 0.46V

Zn + Cu2+ Zn2+ + Cu - ΔE2 = 1.10V

2Ag+ + Zn 2Ag + Zn2+ ΔE3 = ΔE1+ (- ΔE2)

ΔE3 = 0.46 + 1.10 = 1.56V

Part A - Important Notes

• For Fe3+/Fe2+ half-reaction, you will need to combine equal volumes of the two solutions immediately prior to making your measurements

• All measurements should be made three times to ensure that you have obtained the correct values

• I2/I- half-reaction– What is really going on?

Remember the Nernst Eq?

Qn

RT

QRTnn

nGnG

QRTGG

rr

rr

ln

ln

ln

FEE

EFEF

EF EF

Demonstrates cell voltage (ΔE) depends on concentration (Q)

Part B: Concentration Dependence

• Exploring the Nernst Equation with real cells

• Since E°cell=E°cathode-E°anode=0.00V if both Cu+2

Experimentally, Cu/Cu concentration cell is hard to measure, so we will vary the copper concentration (crucible) and measure it against a zinc half-cell (beaker).

Part B: Ionic Strength

• When salt concentrations are large enough, we begin to see deviations from Nernstian behavior

• Maintain the same ionic strength for all measurements

• Make dilutions with stock, water, & KNO3

Calculating Ionic Strength

ii czI 2

21

03.

solution nitrate copper(II) 0.01M For the

3.

2.0)1(1.0)2(2

1

solution nitrate copper(II)stock For the

)(2

)(2

32

I

I

MMINOCu

The difference in ionic strengths is .3-.03 =.27

How much KNO3 needs to be added to your dilutions?

All dilutions made in 100ml volumetric flasks (in kits).

How much 0.4M KNO3 is required to make up for the .27 difference?

M1V1=M2V2

0.4M*V1=0.27M*100ml

V1=67.5 ml

Each time a dilution is made, the ionic strength is maintained.

Parts C - Equilibrium Constant

• Using concentration cells, determine the value of Q.

• Use Q to determine the unknown concentration

Part C: Ksp for AgCl• Starting with ~50 ml of 0.1M KCl in a beaker, add AgNO3

dropwise until a precipitate forms.

• Place inside this half-cell, a crucible containing 0.1M AgNO3

• The resulting cell is:

Ag|Ag+(unknown concentration)|| Ag+(0.1 M)|Ag

Use E = E°- (RT/nF)ln([Ag+unknown]/[.1M Ag+])

then calculate Ksp = [Ag+unknown]x[Cl-]

Safety and Waste Disposal

• Two separate waste bottles in the lab– One for silver nitrate– One for all other solutions (halides)

• Silver nitrate stains hands, so be careful!

Before Starting

• Find a partner

• Check out kit from stockroom

• Have fun!

ALSO, make sure to read your experiment for next week for the round robin…

Part D Formation Constant for [Cu(NH3)4]2+

• The overall reaction is:

• Use the same type of arrangement as in Part C, replacing Ag with Cu in the cathode half-cell

243(aq)3

2(aq) NHCuNHCu 4

Table of Changes Using Moles**Cu2+ NH3 Cu(NH3)4

2+

Initial 0.03L*0.1M 0.003L*6M 0.000

Change -X -4X +X

Equilibrium

moles

[Cu2+]unk*.033L (0.018-4X) X

** Assumes 30ml Cu(NO3)2 solution and 3 ml NH3

For equilibrium concentrations divide by 0.033L

M

CueQ unk

ERT

nFcell

1.0

][ 2

Helpful hints

QK

nFRT

Qn

RTK

nF

RT

QnRT

KnF

RT

nKRTso

KRTnGr

ln

lnln

ln

ln

ln:

ln

F

FEE

E

EF

EF:Recall

1?Q whenhappens What