Is this correct?

Question

2H2O2 (aq)à 2H2O (l) +O2 (g) +O2 (g)          standard electrode potential: 0.55 V

The equation and standard cell potential for the decompositions of H2O2 (aq) in acidic solution at 25 degrees Celsius is given above. The reduction half-reactions for the process are listed below.

O2 (g) + 4H+ (aq) +4e- à 2H2O (l)  Electrode potential= 1.23 V

O2(g) + 2H+ (aq) +2e- à H2O2 (aq)   Electrode potential= ?

O2 (g) +2H+ (aq)+ 2e- à H2O2 (aq)

What is the standard reduction potential for the half reaction represented above?

Options include

A)-1.78 V

B) +0.68 V

C) -0.68 V

D) +1.78 V

Answer 1

B) +0.68 V

The relationship that decides this is E°cell = E°(cathode) - E°(anode), where both numbers are read off the table as reduction potentials. Cell potential is intensive, so you never multiply a potential by a coefficient - that is the single most common error on this problem type.

Setting it up. In the decomposition of hydrogen peroxide the two half-reactions you were given are the pieces. The reduction that runs is O2 + 4 H+ + 4 e- -> 2 H2O, E° = +1.23 V, and the oxidation is the reverse of O2 + 2 H+ + 2 e- -> H2O2, which is the potential you are solving for. So 0.55 = 1.23 - x, and x = +0.68 V.

Why the other three choices exist. C) -0.68 V is the right magnitude with the sign flipped - put it back in and you get 1.23 - (-0.68) = 1.91 V, not 0.55 V. A) and D) both come from adding instead of subtracting: 1.23 + 0.55 = 1.78. D is not a random distractor, though. +1.78 V is very close to the real standard potential for the other peroxide couple, H2O2 + 2 H+ + 2 e- -> 2 H2O. It is a genuine table value for a half-reaction this question did not ask about, which is exactly what makes it tempting.

Check it against a real table. The tabulated values are about +0.70 V for O2/H2O2 and about +1.76 V for H2O2/H2O. Your answer of +0.68 V lands on the first one, which tells you the setup was right and not just arithmetically lucky. It also says something real about peroxide: it sits between O2 and H2O on the potential ladder, so it can be pushed either direction - which is why it decomposes on its own.

One thing to flag in the problem itself: the overall equation is printed as 2 H2O2 -> 2 H2O + O2 + O2, with O2 written twice. The balanced reaction is 2 H2O2 -> 2 H2O + O2. Mark that on your copy so it does not trip you up later.