redox chemistry simple question
http://s13.postimg.org/nr3nnh3lz/screenshot_25.png
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
Original post by bubakazouba
http://s13.postimg.org/nr3nnh3lz/screenshot_25.png
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
The cell invovles both manganate(VII) and hydrogen peroxide. The cell EMF is the difference between the electrode potentials = +0.83 V
Original post by bubakazouba
http://s13.postimg.org/nr3nnh3lz/screenshot_25.png
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
I don't get it why isn't the answer just -0.68V, as it is the reverse reaction of the reduction of O2 to H2O2 which is the oxidation of H2O2 to O2 which is what the question wants?
Original post by charco
The cell invovles both manganate(VII) and hydrogen peroxide. The cell EMF is the difference between the electrode potentials = +0.83 V
This. ^^
Original post by charco
The cell invovles both manganate(VII) and hydrogen peroxide. The cell EMF is the difference between the electrode potentials = +0.83 V
ok I got it, many thanks dude
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