Respuesta :
Answer:
(a) Two electrochemical half reactions for ethanol [tex](C_{2}H_{5}OH)[/tex] is written below
[tex]C_{2}H_{5} + 60[/tex] ⇒ [tex]2CO_{2} + 3H_{2}O + 12e^{-} (anode)[/tex]
[tex]3O_{2} + 12e^{-}[/tex] ⇒ [tex]60^{=}[/tex] [tex](Cathode)[/tex]
(b) The direct oxidation of the fuel will occur in a solid oxide fuel cell but we have to compete with other sets of chemical electrochemical reaction such as water-gas-shift reaction
[tex]CO + H_{2}O[/tex] ⇆ [tex]CO_{2} + H_{2}(WGSR)[/tex]
(c) The electrochemical half reaction that could convert ethanol directly into a fuel cell that conducts hydrogen ions is shown below
[tex]C_{2}H_{5}OH + 3H_{2}O[/tex] ⇒ [tex]2CO_{2} + 12H^{+} + 12e^{-} (anode)[/tex]
[tex]3O_{2} + 12H^{+} + 12e^{-} - 6H_{2}O (Cathode)[/tex]
[tex]C_{2} H_{5} OH + 3O_{2}[/tex] ⇒ [tex]3H_{2}O + 2CO_{2}[/tex]
(d) The half reactions that would be required are shown below
[tex]C_{2}H_{5} OH + 12OH^{-}[/tex] ⇒ [tex]CO_{2} + 9H_{2}O + 12e^{-} (anode)[/tex]
[tex]3O_{2} + 6H_{2}O + 12e^{-}[/tex] ⇒ [tex]12OH^{-} (Cathode)[/tex]
[tex]C_{2}H_{5}OH + 3O_{2}[/tex] ⇒ [tex]3H_{2}O + 2CO_{2}[/tex]