Harnessing the changes in oxidation and reduction

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Presentation transcript:

Harnessing the changes in oxidation and reduction Electrochemistry Harnessing the changes in oxidation and reduction

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

Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction

Standard Reduction Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2e- + 2H+ (1 M) H2 (1 atm) E0 = 0 V Standard hydrogen electrode (SHE)

E0 is for the reaction as written The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

Remember: Anode: electrode in the half-cell where oxidation takes place Metal electrode atoms are oxidized and become aqueous ions Anions must flow from the salt bridge into this half cell to balance out the addition of new metal ions

Remember: Cathode: electrode in the half-cell where reduction takes place Metal ions become neutral atoms Cations must flow from the salt bridge into this half cell to balance loss of metal ions

Remember: Electrons flow through the wire from the anode towards the cathode Anions flow from the salt bridge into the anode half-cell Cations flow from the salt bridge into the cathode half-cell

Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

Standard Reduction Potentials E0 = 0.76 V cell Standard emf (E0 ) cell E0 = Ecathode - Eanode cell Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) E0 = EH /H - EZn /Zn cell + 2+ 2 0.76 V = 0 - EZn /Zn 2+ EZn /Zn = -0.76 V 2+ Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

Standard Reduction Potentials E0 = 0.34 V cell E0 = Ecathode - Eanode cell Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 2+ ECu /Cu = 0.34 V 2+ Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) 2H+ (1 M) + 2e- Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)