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2H 2 O → O 2 + 4H + + 4e − Oxidation (generation of dioxygen) 4H + + 4e − → 2H 2 Reduction (generation of dihydrogen) 2H 2 O → 2H 2 + O 2 Total Reaction Of the two half reactions, the oxidation step is the most demanding because it requires the coupling of 4 electron and proton transfers and the formation of an oxygen-oxygen bond.
The 4e − pathway reaction is the cathode reaction in fuel cell especially in proton-exchange membrane fuel cells, alkaline fuel cell and solid oxide fuel cell. While the 2e − pathway reaction is often the side reaction of 4e- pathway or can be used in synthesis of H 2 O 2 .
O 2 + 4H + + 4e − → 2H 2 O +0.82 In classical electrochemistry, E° for O 2 = +1.23 V with respect to the standard hydrogen electrode (SHE). At pH = 7, E red = 1.23 – 0.059 V × 7 = +0.82 V: P680 + + e − → P680 ~ +1.0 Half-reaction independent of pH as no H + is involved in the reaction
Thus, a reduction half reaction can be written for the O2 as it gains 4 electrons: O 2 ( g ) + 4 e − 2 O 2 − {\displaystyle {\ce {O2_{(g)}{}+ 4e- -> 2O^2-}}} The overall reaction is the sum of both half reactions:
Reduction half reaction: O 2 + 4e − → 2 O 2−; Iron (Fe) has been oxidized because the oxidation number increased. Iron is the reducing agent because it gave electrons to the oxygen (O 2). Oxygen (O 2) has been reduced because the oxidation number has decreased and is the oxidizing agent because it took electrons from iron (Fe).
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4 + H + + e −: ⇌ HMnO − 4: 0.9 1 Po Po 4+ + 2 e −: ⇌ Po 2+ 0.9 2 [8] Hg 2 Hg 2+ + 2 e −: ⇌ Hg 2+ 2: 0.91 2 [13] Pd Pd 2+ + 2 e −: ⇌ Pd(s) 0.915 2 [15] Au [AuCl 4] − + 3 e −: ⇌ Au(s) + 4 Cl −: 0.93 3 N: NO − 3 + 3 H + + 2 e −: ⇌: HNO 2 (aq) 0.94: 2 [6]: 789 Mn MnO 2 (s) + 4 H + + e −: ⇌ Mn 3+ + 2 H 2 O: 0.95 1 ...
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