TO CATCH LOTS OF FISH, YOU MUST FIRST GO TO THE WATER.

-ANON-

GALVANIC (VOLTAIC) CELLS

THE ENERGY RELEASED IN A REDOX REACTION CAN BE USED TO PERFORM ELECTRICAL WORK.

Cu+2 (aq) + Zn (s) Zn+2 (aq) + Cu (s)

IF WE PLACE A Zn STRIP IN A SOLUTION OF Cu+2, THE Cu+2 WILL BE REDUCED TO Cu, AND THE Zn METAL WILL BE OXIDIZED TO Zn+2.

THE TWO HALF REACTIONS ARE

Cu+2 + 2e- Cu REDUCTION

Zn Zn+2 + 2e- OXIDATION

WE CAN SEPARATE THESE TWO HALF REACTIONS INTO SEPARATE COMPARTMENTS.

THE ENERGY DRIVING THE REACTION IS THE DIFFERENCE IN POTENTIAL ENERGY. IN THIS CASE, WE REFER TO THIS AS THE ELECTROMOTIVE FORCE OR EMF. WE MEASURE THIS IN VOLTS.

THIS FORCE DRIVES ELECTRONS FROM THE ANODE TO THE CATHODE IN THE EXTERNAL CIRCUIT.

FOR A VOLTAIC CELL TO WORK, THE TWO HALF-CELLS MUST REMAIN ELECTRICALLY NEUTRAL.

THAT IS WHERE THE SALT BRIDGE COMES IN. IT ALLOWS IONS TO MIGRATE INTO THE CELL COMPARTMENTS TO MAINTAIN ELECTRICAL NEUTRALITY.

A POROUS PLUG OR BARRIER WOULD WORK EQUALLY AS WELL – SOMETHING THAT WOULD ALLOW ION MIGRATION, BUT WOULD PREVENT THE SOLUTIONS FROM MIXING.

ANIONS WOULD MIGRATE TOWARD THE ANODE COMPARTMENT.

CATIONS WOULD MIGRATE TOWARD THE CATHODE COMPARTMENT.

ELECTRONS IN THE EXTERNAL CIRCUIT WOULD MIGRATE FROM THE ANODE TO THE CATHODE.

AGAIN, YOU CAN ONLY USE THE ENERGY FROM A REDOX REACTION AS ELECTRICAL ENERGY IF YOU CAN SEPARATE THE TWO HALF REACTIONS.

THE EMF OF A CELL DEPENDS ON THE SPECIFIC REACTIONS THAT OCCUR AT THE ANODE AND CATHODE.

IT ALSO DEPENDS ON THE CONCENTRATIONS.

FOR REFERENCE PURPOSES, WE MEASURE CELL VOLTAGES UNDER STANDARD CONDITIONS. THESE ARE:

CONCENTRATIONS OF IONS IN SOLUTION = 1 MTEMPERATURE = 25o CGAS PRESSURE = 1 atm

Zn (s) + Cu+2 (aq, 1M) Cu (s) + Zn+2 (aq, 1M) @ 25o C

Eocell = +1.10 V

CELL POTENTIAL IS ALWAYS THE DIFFERENCE IN POTENTIAL BETWEEN TWO HALF-CELL POTENTIALS. WE CAN’T MEASURE THE POTENTIAL OF A SINGLE ELECTRODE.

BY CONVENTION, WE TALK ABOUT STANDARD REDUCTION POTENTIAL, SUCH THAT

Eocell = Eo

red (CATHODE) – Eored (ANODE)

NOW, IF WE PICK A HALF CELL REACTION AND SET ITS STANDARD REDUCTION POTENTIAL EQUAL TO ZERO, WE CAN MEASURE EVERY OTHER HALF-CELL POTENTIAL RELATIVE TO THIS ONE STANDARD AND COME UP WITH A LIST OF STANDARD REDUCTION POTENTIALS.

THIS WILL ALLOW US TO CALCULATE CELL POTENTIALS USING DIFFERENT ELECTRODES.

REDUCTION WILL OCCUR AT THE CATHODE.

OXIDATION WILL OCCUR AT THE ANODE.

THE HALF-CELL THAT WE PICK IS THE HYDROGEN ELECTRODE.

THIS IS REFERED TO AS THE STANDARD HYDROGEN ELECTRODE OR SHE.

2H+ (aq, 1M) + 2e- H2 (g, 1 atm) Eored = 0 V

THIS GIVES US A REFERENCE THAT WE CAN USE TO MEASURE OTHER HALF-CELL POTENTIALS RELATIVE TO.

THESE STARNDARD REDUCTION POTENTIALS THAT COULD BE OBTAINED FROM A SERIES OF SUCH MEASUREMENTS COULD BE USED FOR CALCULATING CELL VOLTAGES.

METAL/METAL ION EoRED

Mg/Mg+2 -2.37 v

Zn/Zn+2 -0.76

Cu/Cu+2 +0.35

Ag/Ag+ +0.80

THESE ARE ALL RELATIVE TO THE HYDROGEN ELECTRODE AT 0 volt.

YOU CAN THINK OF THESE HALF-CELL REDUCTION POTENTIALS AS A MEASURE OF THEIR TENDENCY TO GAIN ELECTRONS.

A GOOD REFERENCE SOURCE FOR ADDITONAL READING ON THE TOPIC IS GIVEN AT:http://www.chemguide.co.uk/physical/redoxeqia/introduction.html

LET’S GO BACK TO OUR TABLE. SUPPOSE WE HAVE A CELL THAT INVOLVES Mg/Mg+2 AND Ag/Ag+.

THE HALF-CELL REACTIONS ARE:

Mg+2 (aq) + 2e- Mg (s) Eored = -2.37 volt

Ag+ (aq) + e- Ag (s) Eored = +0.80 volt

IF WE MAKE A CELL USING THESE METALS, SILVER HAS THE MOST POSITIVE REDUCTION POTENTIAL, SO IT WILL BE THE CATHODE, AND MAGNESIUM WILL BE THE ANODE.

Eocell = Eo

red (CATHODE) – Eored (ANODE) = Eo

red (Ag) – Eo

red (Mg)

Eocell = +0.80 – (-2.37) = +3.17 volt

THE CELL REACTION WOULD BE:

Mg (s) + 2Ag+ (aq) Mg+2 (aq) + 2Ag (s)

THE GREATER THE POSITIVE VALUE FOR Eored , THE

GREATER THE DRIVING FORCE FOR REDUCTION.

EXAMPLE: A VOLTAIC CELL IS BASED ON THE FOLLOWING TWO STANDARD HALF-REACTIONS:

Cd+2 (aq) + 2e- Cd (s) Eored = -0.403 volt

Sn+2 (aq) + 2e- Sn (s) Eored = -0.136 volt

WHAT IS THE CELL POTENTIAL, AND WHAT REACTIONS OCCUR AT THE CATHODE AND ANODE?

THE STANDARD REDUCTION POTENTIAL FOR Sn/Sn+2 IS LESS NEGATIVE (MORE POSITIVE) THAN FOR Cd/Cd+2, SO IT WILL BE THE CATHODE.

CATHODE Sn+2 (aq, 1M) + 2e- Sn (s) Eored = -

0.136 volt

ANODE Cd (s) Cd+2 (aq, 1M) + 2e- Eored = -

0.403 volt

CELL Sn+2 (aq, 1M) + Cd (s) Sn (s) + Cd+2 (aq, 1M)

Eocell = -0.136 – (-403) = 0.267 volt

EXAMPLE: A VOLTAIC CELL IS BASED ON THE FOLLOWING HALF REACTIONS:

ANODE: In+ (aq) In+3 (aq) + 2e-

CATHODE: Br2 (l) +2e- 2Br- (aq) Eored = +1.065

volt

IF Eocell = +1.46 volt, CALCULATE Eo

red FOR In+/In+3

Eocell = Eo

red (Br2/Br-) – Eored (In+/In+3)

1.46 = 1.065 – Eored (In+/In+3)

Eored (In+/In+3) = -1.46 + 1.065 = -0.395 volt = -0.40

volt

STANDARD REDUCTION POTENTIAL TABLES CAN ALSO BE USED TO PREDICT DIRECTIONS OF OXIDATION REDUCTION REACTIONS.

BATTERIES

A BATTERY IS SIMPLY A SELF-CONTAINED ELECTROCHEMICAL POWER SOURCE THAT CONSISTS OF ONE OR MORE VOLTAIC CELLS.

FOR EXAMPLE, THE TYPICAL FLASHLIGHT BATTERY IS AN ALKALINE BATTERY THAT PRODUCES +1.55 VOLT AT ROOM TEMPERATURE.

CATHODE: 2MnO2 + 2H2O + 2e- 2MnO(OH) + 2OH-

ANODE: Zn + OH- Zn(OH)2

IF YOU WANTED A 9V BATTERY, YOU WOULD CONSTRUCT ONE BY CONNECTING 6 +1.5V BATTERIES.

YOU CAN CONSTRUCT BATTERY PACKS TO GIVE YOU WHATEVER VOLTAGE YOU NEED BY HOOKING INDIVIDUAL CELLS IN SERIES, OR YOU CAN HOOK THEM IN PARALLEL TO GIVE YOU POWER OVER A LONGER PERIOD OF TIME.

THE CONVENTIONAL AUTOMOBILE BATTERIES ARE LEAD-ACID BATTERIES.

CATHODE: PbO2 + HSO4- + 3H+ + 2e- PbSO4 +

2H2O

ANODE: Pb + HSO4- PbSO4 + H+ + 2e-

Eocell = +2.041 volt

THE REACTANTS, Pb AND PbO2 ARE SOLIDS, SO THERE IS NO NEED TO SEPARATE THE CELL INTO ANODE AND CATHODE COMPARTMENTS – JUST KEEP THEM FROM TOUCHING.

FIBER GLASS SPACERS CAN BE USED.

IF YOU WANTED 12 VOLTS, YOU WOULD HAVE A BATTERY CONTAINING 6 CELLS.

THIS BATTERY CAN BE RECHARGED BY CONNECTING IT TO AN EXTERNAL POWER SOURCE AND REVERSING THE DIRECTION OF THE CURRENT.

THE PbSO4 FORMED DURING DISCHARGE ADHERES TO THE ELECTRODES. DURING CHARGING, THE PbSO4 AT ONE ELECTRODE IS CONVERTED INTO Pb AND INTO PbO2 AT THE OTHER.