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I need to compare solid working with

[itex]

\frac{1}{2}O_{2}+H_{2}O+2e^{-} \rightarrow 2OH^{-}

[/itex]

and

[itex]H_{2}+2OH^{-} \rightarrow 2H_{2}O+2e^{-}[/itex]

and liquid fuel cells working with

[itex]

H_{2}\rightarrow 2H^{+}+2e^{-}

[/itex]

and

[itex]

\frac{1}{2}O_{2}+2H^{+}+2e^{-} \rightarrow H_{2}O

[/itex]

My idea was to compare the required electrical energy (which I know by [itex] dt U I [/itex] and the chemical energy given by [itex] Q \cdot n_{H_{2}O}=Q\cdot n_{H_{2}O}=Q\cdot\nu_{gas}\cdot n_{O_{2}}=Q\cdot\nu_{O_{2}}\cdot\frac{PV_{O_{2}}}{n_{O_{2}}RT}

[/itex]

where [itex] \nu _{O_{2}} [itex] is a coefficient that is equal to 1/2 in the case of solid fluel cells. My question is simple: what is [itex] \nu _{O_{2}} [itex] equal to in the case of liquid cell since there is a double equation and that we don't know which portion of OH^{-} of the reductive equation is used in the oxydative equation.

Any help would be helpful, thanks