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Calculate the standard electrode potential (E°) for the reaction below at 298K using the standard reduction potentials given:2H⁺(aq) + 2e⁻ → H₂(g)          E°= 0.00VFe³⁺(aq) + e⁻ → Fe²⁺(aq)        E°=+0.77V 2H⁺(aq) + Fe³⁺(aq) → H₂(g) + Fe²⁺(aq)

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ago by (510 points)
To calculate the standard electrode potential  E  for the overall reaction, we need to determine which half-reaction is being reduced and which is being oxidized. Then, we can use the standard reduction potentials given to find the E for the overall reaction.In the overall reaction:2H aq  + Fe aq   H g  + Fe aq We can see that:- Hydrogen ions  H  are gaining electrons to form hydrogen gas  H , so this half-reaction is being reduced.- Iron III  ions  Fe  are losing an electron to form iron II  ions  Fe , so this half-reaction is being oxidized.Now, we can use the standard reduction potentials given for each half-reaction:Reduction half-reaction: 2H aq  + 2e  H g           E = 0.00VOxidation half-reaction: Fe aq  + e  Fe aq         E = +0.77VSince the reduction half-reaction is already given, we don't need to change it. However, we need to reverse the oxidation half-reaction to represent the actual process occurring in the overall reaction:Reversed oxidation half-reaction: Fe aq   Fe aq  + eWhen we reverse the half-reaction, we also need to change the sign of the standard reduction potential:Reversed oxidation half-reaction E: -0.77VNow, we can add the standard reduction potentials of the two half-reactions to find the E for the overall reaction:E overall  = E reduction  + E reversed oxidation E overall  = 0.00V +  -0.77V E overall  = -0.77VSo, the standard electrode potential  E  for the overall reaction at 298K is -0.77V.

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