Given the information below, calculate E o cell for the following reactions in t

Question

Given the information below, calculate Eocellfor the following reactions in terms of x
a) 2VO2+ (aq) + Br2(aq) +6H2O(l) --> 2VO2+(aq) +2Br-(aq) + 4H3O+(aq) b) 2VO2+(aq) + 2Br-(aq)+ 4H3O+(aq)--> 2VO2+ (aq) + Br2(aq) +6H2O(l)
Given Info: Cathode Reaction Eo (V) VO2+ + 2H + e- -->VO2+ + H2O x H2O2 + 2H3O++ 2e- --> 4H2O 1.78 Br2 + 2e- --> 2Br-        1.07

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a) 2VO2+ (aq) + Br2(aq) +6H2O(l) --> 2VO2+(aq) +2Br-(aq) + 4H3O+(aq) b) 2VO2+(aq) + 2Br-(aq)+ 4H3O+(aq)--> 2VO2+ (aq) + Br2(aq) +6H2O(l)
Given Info: Cathode Reaction Eo (V) VO2+ + 2H + e- -->VO2+ + H2O x H2O2 + 2H3O++ 2e- --> 4H2O 1.78 Br2 + 2e- --> 2Br-        1.07

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Cathode Reaction Eo (V) VO2+ + 2H + e- -->VO2+ + H2O x H2O2 + 2H3O++ 2e- --> 4H2O 1.78

Explanation / Answer

Eocell = Eooxidation +Eoreduction a) oxidation:      2VO2+ +2H2O ---> 2VO2+ + 4H + 2e- E = -2x reduction:      Br2 +2e- --> 2Br-       E=1.07 ---------------------------------------------------------------------- Eocell = 1.07-2x b) oxidation:      2Br- ---->Br2 + 2e-       E=-1.07 reduction:     2VO2+ + 4H + 2e- -->2VO2+ + 2H2O       E = 2x ----------------------------------------------------------------------------- Eocell = 2x -1.07 P.S. Don't get confused with the number of waters (H2O)and hydroniums (H3O+). There is basically thefollowing unseen equation in solution: 4H2O ----->4H2O

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