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This problem is composed of three parts: (a) For the reaction 1/2 O_2 +2H+ +2e− ⇌...

This problem is composed of three parts:
(a) For the reaction
1/2 O_2 +2H+ +2e− ⇌ H_2O

the standard electrode potential is + 1.23 V. Under standard-state conditions, if the electrode potential is reduced to 1.0 V, will this bias the reaction in the forward or reverse direction?  

(b) For the reaction
H_2 ⇌ 2H+ +2e−

the standard electrode potential is 0.0 V. Under standard-state conditions, if the electrode potential is increased to 0.10V, will this bias the reaction in the forward or reverse direction?

(c) Considering your answers to parts (a) and (b), in an H2–O2 fuelcell, if we increase the overall rate of the fuel cell reaction,
H_2 + 1/2 O_2 ⇌ H_2O
which is made up of the half reactions
H_2 ⇌ 2H+ +2e−
1/2 O_2 +2H+ +2e− ⇌ H_2O
what happens to the potential difference (voltage output) for the reaction?

Engineering   Chemical-Engineering   
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Answer #1

a) If the electrode potential is reduced to 1 V from 1.23 V ,the reaction is forward biased as the reactants will convert more into products , such that the potential moves to 1.23 V.

b) If the electrode is increased to 0.1V, then the reaction occurs in reverse direction as the products reunite to bring back the 0.0V.

c)When we increase the overall rate of fuel cell reaction, the potential difference for the reaction reamains constant (if temperature is not changed).

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