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Reactions of Metals. Reactions of Metals with H 2 O The metal is the anode and will be oxidized. 2H 2 O + 2e-  2OH - + H 2 E° = -0.83 V Mg  Mg 2+ +

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Presentation on theme: "Reactions of Metals. Reactions of Metals with H 2 O The metal is the anode and will be oxidized. 2H 2 O + 2e-  2OH - + H 2 E° = -0.83 V Mg  Mg 2+ +"— Presentation transcript:

1 Reactions of Metals

2 Reactions of Metals with H 2 O The metal is the anode and will be oxidized. 2H 2 O + 2e-  2OH - + H 2 E° = -0.83 V Mg  Mg 2+ + 2e- E° = +2.37 V The water and substances in it will be the cathode. Need to consider: i) neutral water ii) oxygenated water iii) acidic solution O 2 + 2H 2 O + 4e-  4OH - E° = +0.40 V 2H + + 2e-  H 2 E° = +0.00 V

3 Metals below the cathodic water ½ rxn will be spontaneous That’s why: Not all metals react with neutral water eg. ones that will:ones that won’t: Not all metals react with oxygenated water eg. ones that will:ones that won’t: Not all metals react with acidic water eg. ones that will:ones that won’t: Cr, Mn, V, Al, Ti, Mg, Na, Ca, Ba, K, Li Zn, Cr, Fe, Cd, Co, Ni, Sn, Pb, Cu, Hg, Ag, Au Cu, Hg, Ag, Au Cu, Fe, Pb, Sn, Ni, Co, Cd, Cr, Zn, Mn,V, Al, Ti, Mg, Na, Ca, Ba, K, Li Cu, Hg, Ag, Au Fe, Pb, Sn, Ni, Co, Cd, Cr, Zn, Mn, V, Al, Ti, Mg, Na, Ca, Ba, K, Li

4 Metals with Metals Alloys are created to change a metal’s properties As most metals undergo oxidation – rust, corrosion, patina – other metals are added to stop the oxidation. eg. Fe or Cu need protection. The protecting metal, called a sacrificial anode, is below on the table relative to the metal in question. It will undergo the oxidation and save the metal. eg. Which metals will protect Zn? eg. Which metals will not protect Zn? Cr, Mn, V, Al, Ti, Mg, Na, Ca, Ba, K, Li Cr, Fe, Cd, Co, Ni, Sn, Pb, Cu, Hg, Ag, Au

5 eg.a)Pick a metal that will protect Zn and determine the cell potential if the cathode is oxygenated water. an: Mg  Mg 2+ + 2e- E° = +2.37 V cat: O 2 + 2H 2 O + 4e-  4OH - E° = +0.40 V 2 2 4 2x E cell ° is positive  spontaneous 2Mg + O 2 + 2H 2 O  2Mg 2+ + 4OH - E cell ° = +2.77 V

6 an: Zn  Zn 2+ + 2e- E° = +0.76 V cat: O 2 + 2H 2 O + 4e-  4OH - E° = +0.40 V 22 4 2x E cell ° is positive  spontaneous b)Do the same for the case where Zn is not protected. 2Zn + O 2 + 2H 2 O  2Zn 2+ + 4OH - E cell ° = +1.16 V


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