Reactions of Metals

Reactions of Metals with H2O The metal is the anode and will be oxidized. Mg  Mg2+ + 2e- E° = +2.37 V The water and substances in it will be the cathode. Need to consider: i) neutral water – WOA of the 3 2H2O + 2e-  2OH- + H2 E° = -0.83 V ii) oxygenated water - requires a very long rxn time, or an accelerant, as O2 is not very water soluble O2 + 2H2O + 4e-  4OH- E° = +0.40 V iii) acidic solution - as with acid rain 2H+ + 2e-  H2 E° = +0.00 V

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: Cr, Mn, V, Al, Ti, Be, Mg, Ce, Na, Ca, Ba, Cs, Ra, K, Li Zn, Cr, Ga, Fe, Cd, Tl, Co, Ni, Sn, Pb, Cu, Hg, Ag, Au Not all metals react with oxygenated water eg. ones that will: ones that won’t: All the above, plus Cu, Fe, Pb, Sn, Ni, Co, Tl, Cd, Ga, Cr, Zn Cu, Hg, Ag, Au Not all metals react with acidic water eg. ones that will: ones that won’t: All the above except Cu Cu, Hg, Ag, Au

Metals with Metals eg. Which metals will not protect Fe? 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 Fe? All the metals below Fe in the Standard Reduction Table eg. Which metals will not protect Fe? All the metals above Fe in the Table

eg. a) Will Cr protect Fe? Determine the anode and determine the cell potential if the cathode is oxygenated water. Cr  Cr2+ + 2e- E° = +0.91 V SRA Cr  Cr3+ + 3e- E° = +0.74 V Fe  Fe2+ + 2e- E° = +0.45 V Fe  Fe3+ + 3e- E° = +0.04 V 4 2x an: Cr  Cr2+ + 2e- E° = +0.91 V 2 2 cat: O2 + 2H2O + 4e-  4OH- E° = +0.40 V 2Cr + O2 + 2H2O  2Cr2+ + 4OH- Ecell° = +1.31 V Ecell° is positive  spontaneous and Fe is protected

eg. b) Do the same for Ag and Fe. Ag  Ag1+ + e- E° = -0.80 V Fe  Fe2+ + 2e- E° = +0.45 V SRA Fe  Fe3+ + 3e- E° = +0.04 V 4 2x an: Fe  Fe2+ + 2e- E° = +0.45 V 2 2 cat: O2 + 2H2O + 4e-  4OH- E° = +0.40 V 2Fe + O2 + 2H2O  2Fe2+ + 4OH- Ecell° = +0.85 V  spontaneous and Fe is NOT protected