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

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

Short answer: B) +0.68 V.

Start by naming what is happening. In 2 H2O2 goes to 2 H2O + O2, the peroxide oxygen sits at oxidation state -1 and ends up in two places at once: down at -2 in water and up at 0 in O2. That is a disproportionation, so hydrogen peroxide is its own oxidizing agent and its own reducing agent. It is also why both half reactions you were handed are written around O2 rather than around two different elements.

The one rule that does all the work here: a standard potential is intensive. You may multiply a half reaction by whatever coefficient you need to make electrons cancel, but you never multiply its E value. Potentials get subtracted, never scaled.

Given:  O2 + 4 H+ + 4 e-  ->  2 H2O          E = 1.23 V
Given:  O2 + 2 H+ + 2 e-  ->  H2O2           E = x  (what we want)
Given:  2 H2O2  ->  2 H2O + O2               E(cell) = 0.55 V

Now build the cell out of the two given couples. The 1.23 V couple runs forward as written, so it is the cathode. The unknown couple runs backward, because peroxide is being consumed to make O2, so it is the anode. Double the anode half so the four electrons cancel.

cathode:      O2 + 4 H+ + 4 e-  ->  2 H2O
anode  :  2 H2O2  ->  2 O2 + 4 H+ + 4 e-
net    :  O2 + 2 H2O2  ->  2 H2O + 2 O2
cancel :      2 H2O2  ->  2 H2O + O2      (the target reaction)

Notice that the anode half got doubled and x did not. That is the rule from two paragraphs up, and it is the single place most people lose this problem.

E(cell) = E(cathode) - E(anode)
0.55 V  = 1.23 V - x
x       = 1.23 - 0.55 = 0.68 V

So O2 + 2 H+ + 2 e- goes to H2O2 has E = +0.68 V, which is choice B. The accepted literature value for that couple is +0.695 V, so the number the problem is steering you toward is the real one.

Because the two given couples carry different electron counts, it is worth checking the result on a quantity that actually adds. Free energies add; potentials do not.

dG = -n F E
dG(1.23 V couple, n = 4) = -4(96485)(1.23) = -474.7 kJ
dG(0.68 V couple, n = 2) = -2(96485)(0.68) = -131.2 kJ
dG(H2O2 + 2 H+ + 2 e- -> 2 H2O) = -474.7 - (-131.2) = -343.5 kJ
E for that couple = 343500 / (2 x 96485) = 1.78 V

That 1.78 V is the reduction potential of the other peroxide couple, H2O2 + 2 H+ + 2 e- going to 2 H2O, and it deserves a second look, because it is also answer choice D. Its literature value is 1.776 V. Adding 1.23 + 0.55 instead of subtracting produces exactly 1.78, so D is not a random wrong number, it is the right answer to a question you were not asked.

The four choices sort cleanly by mistake:

B) +0.68 V is correct. D) +1.78 V means you solved for the wrong half reaction, by adding instead of subtracting. C) -0.68 V is the right magnitude with the subtraction reversed, that is, E(anode) minus E(cathode). A) -1.78 V is both errors stacked.

One more habit worth building: run a direction check before any arithmetic. Hydrogen peroxide decomposing is spontaneous, so E(cell) must be positive, and the given 0.55 V agrees. For 1.23 minus x to come out positive and equal to 0.55, x has to be positive and smaller than 1.23. Both negative choices are dead before you touch a calculator.

Finally, the thing this problem is quietly teaching. Peroxide is thermodynamically unstable at every concentration: the decomposition runs at 0.55 V, which is a free energy near -212 kJ for the equation as written. The only reason a bottle survives on a shelf is that the reaction is kinetically slow. Drop in a speck of MnO2, or a piece of liver with catalase in it, or just leave it in sunlight, and it goes off in front of you. Exams love to ask why a strongly favorable reaction is not observed, and this is the cleanest example of the answer: thermodynamics says whether, kinetics says when.