I'm not sure if I got this wrong because I interpreted the cell incorrectly. When they have an arrow pointing in to the tube for $\ce{O2}$, does this imply that $\ce{O2}$ is a reactant? If so then B is definitely wrong, which was my answer.

Is there any other reason why B is wrong? I wasn't sure if it was right or wrong when I did this question.


1 Answer 1


I think the arrow for $\ce{O2}$ just means there's oxygen gas in that part of the apparatus.

The reason B is wrong is that conversion of hydrogen peroxide to oxygen gas and water is the spontaneous reaction at these conditions. Assuming SLC implies 1 atm of pure oxygen, then everything is under standard conditions so we don't need to bring Nernst into it. The half reactions are as follows with the corresponding potentials.

$$ \ce{H2O2 + 2H+ + 2e- -> 2H2O} (\Delta E^{\circ} = +1.78) $$ $$ \ce{H2O2 -> O2 + 2H+ + 2e-} (\Delta E^{\circ} = -0.7) $$

so the net reaction is $$ \ce{2H2O2 -> O2 + 2H2O} (\Delta E^{\circ} = +1.08) $$

Since $\Delta E^{\circ} $ is positive, this reaction is spontaneous.

The fact that this reaction doesn't happen means that there's a high kinetic barrier. I think that's what A is getting at, though I have a little bit of an issue with how it's worded. Regardless it's still better than B.


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