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 you need

E°_cell = E°_cathode − E°_anode

Both values are looked up as reductions, then subtracted. Do not flip the anode's sign and add — that is the same operation done twice, and it is the most common way to lose this question.

Substituting

You are given E°_cell = 0.55 V for the decomposition, and the O₂/H₂O couple at 1.23 V. Solving for the unknown couple:

0.55 = 1.23 − E°_anode

E°_anode = 1.23 − 0.55 = +0.68 V

Two checks that the answer is right

1. Sign. The unknown couple must be less positive than 1.23 V, or E°_cell would come out negative and the decomposition would not be spontaneous. That rules out D (+1.78 V) immediately, and both negative options are far too low.
2. It matches the real value. The accepted standard potential for O₂ + 2H⁺ + 2e⁻ → H₂O₂ is about +0.69 V. Landing on 0.68 from the problem's data is a good sign you did it correctly.

Why the distractors were chosen

Option D, +1.78 V, is not random — it is the genuine potential for the other peroxide couple, H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O. The two negative options are those same numbers with flipped signs, which is exactly what you would get by mishandling the anode term.

The chemistry underneath

This decomposition is a disproportionation: hydrogen peroxide is simultaneously oxidised and reduced. Oxygen in H₂O₂ is in the unusual −1 oxidation state, sitting between 0 (in O₂) and −2 (in H₂O), so it can go either way.

Half of it is oxidised to O₂, half reduced to H₂O — and because both processes are favourable, hydrogen peroxide is thermodynamically unstable with respect to its own decomposition. That is why the bottle in your medicine cabinet is brown glass and slowly loses strength: it is decomposing on its own, just slowly, until a catalyst such as the catalase in blood makes it visibly fizz.