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?