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Basic Concepts of Electrochemical Cells. Electrifying!. Anode. Cathode. CHEMICAL CHANGE ---> ELECTRIC CURRENT. With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”. Zn is oxidized and is the reducing agent Zn(s) ---> Zn 2+ (aq) + 2e-

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Basic Concepts of Electrochemical Cells

Electrifying!

Anode

Cathode


CHEMICAL CHANGE --->ELECTRIC CURRENT

With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”

  • Zn is oxidized and is the reducing agent Zn(s) ---> Zn2+(aq) + 2e-

  • Cu2+ is reduced and is the oxidizing agentCu2+(aq) + 2e- ---> Cu(s)


Electrons are transferred from Zn to Cu2+, but there is no useful electric current.

CHEMICAL CHANGE --->ELECTRIC CURRENT

Oxidation: Zn(s) ---> Zn2+(aq) + 2e-

Reduction: Cu2+(aq) + 2e- ---> Cu(s)

--------------------------------------------------------

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)


CHEMICAL CHANGE --->ELECTRIC CURRENT

  • To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.

This is accomplished in a GALVANIC or VOLTAIC cell.

A group of such cells is called a battery.


Zn --> Zn2+ + 2e-

Cu2+ + 2e- --> Cu

Oxidation

Anode

Negative

Reduction

Cathode

Positive

•Electrons travel thru external wire.

  • Salt bridge allows anions and cations to move between electrode compartments.

<--Anions

Cations-->


The Cu|Cu2+ and Ag|Ag+ Cell

Electrons move from anode to cathode in the wire.

Anions & cations move thru the salt bridge.


Anode, site of oxidation,

negative

Cathode, site of reduction, positive


1.10 V

1.0 M

1.0 M

CELL POTENTIAL, E

  • Electrons are “driven” from anode to cathode by an electromotive force or emf.

  • For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M.

Zn and Zn2+,

anode

Cu and Cu2+,

cathode


CELL POTENTIAL, E

  • For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M.

  • This is the STANDARD CELL POTENTIAL, Eo

  • —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

  • This means pure solids or in solution at a concentration of 1M!!!!


Calculating Cell Voltage

  • Balanced half-reactions can be added together to get overall, balanced equation.

Zn(s) ---> Zn2+(aq) + 2e-

Cu2+(aq) + 2e- ---> Cu(s)

--------------------------------------------

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

  • If we know Eo for each half-reaction, we could get Eo for net reaction.

  • Lets revisit my haiku!


Oxidation Haiku!

  • Lost an electron

  • But now feeling positive

  • Oxidized is cool!

  • What is that? You want a reduction Haiku?


Reduction Haiku!!!

  • Gained some electrons

  • Gave me a negative mood!

  • Now I can say Ger!

  • Thank you… Enjoy the buffet… Don’t eat the chemicals or furniture kids!


CELL POTENTIALS, Eo

Can’t measure 1/2 reaction Eo directly. Therefore, measure it relative to a STANDARD HYDROGEN CELL, SHE.

2 H+(aq, 1 M) + 2e- <----> H2(g, 1 atm)

Eo = 0.0 V


Negative electrode

Positive electrode

Zn/Zn2+ half-cell hooked to a SHE.

Eo for the cell = +0.76 V

Supplier of electrons

Acceptor of electrons

Zn --> Zn2+ + 2e-

Oxidation

Anode

2 H+ + 2e- --> H2

Reduction

Cathode


Reduction of H+ by Zn

Figure 20.10


Overall reaction is reduction of H+ by Zn metal.

Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = +0.76 V

Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V

Zn is a (better) (poorer) reducing agent than H2.


Cu/Cu2+ and H2/H+ Cell

Eo = +0.34 V

Positive

Negative

Acceptor of electrons

Supplier of electrons

Cu2+ + 2e- --> Cu

Reduction

Cathode

H2 --> 2 H+ + 2e-

Oxidation

Anode


Cu/Cu2+ and H2/H+ Cell

Overall reaction is reduction of Cu2+ by H2 gas.

Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq)

Measured Eo = +0.34 V

Therefore, Eo for Cu2+ + 2e- ---> Cu is

+0.34 V


+

Zn/Cu Electrochemical Cell

Zn(s) ---> Zn2+(aq) + 2e-Eo = +0.76 V

Cu2+(aq) + 2e- ---> Cu(s)Eo = +0.34 V

---------------------------------------------------------------

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

Eo (calc’d) = +1.10 V

Anode, negative, source of electrons

Cathode, positive, sink for electrons


Yes It is finally time for a DEMO!!

  • Do you feel like bridging that salt?


oxidizing

o

ability of ion

E

(V)

2+

Cu

+ 2e- Cu

+0.34

+

2 H

+ 2e- H

0.00

2+

Zn

+ 2e- Zn

-0.76

reducing ability

of element

TABLE OF STANDARD REDUCTION POTENTIALS

2


Potential Ladder for Reduction Half-Reactions

Figure 20.11


Table 21.1 Page 970


2+

Cu

+ 2e- Cu

+0.34

+

2 H

+ 2e- H

0.00

2

2+

Zn

+ 2e- Zn

-0.76

Standard Redox Potentials, Eo

Any substance on the right will reduce any substance higher than it on the left.

Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner (anode) and oxidizing agent at northwest corner (cathode).


Standard Redox Potentials, Eo

Any substance on the right will reduce any substance higher than it on the left.

  • Zn can reduce H+ and Cu2+.

  • H2 can reduce Cu2+ but not Zn2+

  • Cu cannot reduce H+ or Zn2+.


Using Standard Potentials, EoTable 20.1

  • In which direction do the following reactions go?

  • Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)

  • 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)

  • What is Eonet for the overall reaction?


Standard Redox Potentials, Eo

E˚net = “distance” from “top” half-reaction (cathode) to “bottom” half-reaction (anode)

E˚net = E˚cathode - E˚anode

Eonet for Cu/Ag+ reaction = +0.46 V


Eo for a Voltaic Cell

Cd --> Cd2+ + 2e-

or

Cd2+ + 2e- --> Cd

Fe --> Fe2+ + 2e-

or

Fe2+ + 2e- --> Fe

All ingredients are present. Which way does reaction proceed?


Eo for a Voltaic Cell

From the table, you see

•Fe is a better reducing agent than Cd

•Cd2+ is a better oxidizing agent than Fe2+

Overall reaction

Fe + Cd2+ ---> Cd + Fe2+

Eo = E˚cathode - E˚anode

= (-0.40 V) - (-0.44 V)

= +0.04 V


More About Calculating Cell Voltage

2 H2O + 2e- ---> H2 + 2 OH- Cathode

2 I----> I2 + 2e- Anode

-------------------------------------------------

2 I- + 2 H2O --> I2 + 2 OH- + H2

Assume I- ion can reduce water.

Assuming reaction occurs as written,

E˚net = E˚cathode - E˚anode

= (-0.828 V) - (+0.535 V) = -1.363 V

Minus E˚ means rxn. occurs in opposite direction


If you have reached this far, you need a break!


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