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Electrochemistry. To obtain a useful current, we separate the oxidation and reduction half-reactions so that electron transfer occurs thru an external.

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Presentation on theme: "Electrochemistry. To obtain a useful current, we separate the oxidation and reduction half-reactions so that electron transfer occurs thru an external."— Presentation transcript:

1 Electrochemistry

2 To obtain a useful current, we separate the oxidation and reduction half-reactions so that electron transfer occurs thru an external wire.To obtain a useful current, we separate the oxidation and reduction half-reactions so that electron transfer occurs thru an external wire. ElectrochemistryElectrochemistry This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a battery. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

3 Galvanic Cells 19.2 spontaneous redox reaction Anode= oxidation Cathode= reduction - +

4 Galvanic Cells 19.2 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Standard emf (E 0 ) cell E 0 = E cathode - E anode cell 00

5 Standard Electrode Potentials 19.3 Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced In a cell, the substance with the greater E 0 value is at the cathode (where reduction takes place)

6 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd 2+ will be reduced (at the cathode) Cr(s) will be oxidized (at the anode) E 0 = -0.40 - (-0.74) cell E 0 = 0.34 V cell 19.3 E 0 = E cathode - E anode cell 00


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