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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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chemical change electric current
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)
chemical change electric current1

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 current2
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.

slide5

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 cu 2 and ag ag cell
The Cu|Cu2+ and Ag|Ag+ Cell

Electrons move from anode to cathode in the wire.

Anions & cations move thru the salt bridge.

slide7

Anode, site of oxidation,

negative

Cathode, site of reduction, positive

cell potential e

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 e1
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
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
Oxidation Haiku!
  • Lost an electron
  • But now feeling positive
  • Oxidized is cool!
  • What is that? You want a reduction Haiku?
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 e o
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

slide14

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

slide16

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 cu 2 and h 2 h cell
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 cu 2 and h 2 h cell1
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/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
Yes It is finally time for a DEMO!!
  • Do you feel like bridging that salt?
table of standard reduction potentials

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

standard redox potentials e o

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 e o1
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 e o table 20 1
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 e o2
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

e o for a voltaic cell
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?

e o for a voltaic cell1
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
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

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