Question

Hydrogen peroxide (H2O2) is a possible product of the reduction of oxygen in acidic solution:

02(g)+2H+(aq) + 2 e-→ H2O2(,) H2O2(,) + 2 H+(aq) + 2 e-→ 2 H2O(aq) 02(g)+ 4 H+(aq)+4 e-→ 2 H20(,) E3 =0.70 V It can then be further reduced to water E2。 = 1.78 V The direct reduction of O2 to water has the following half-cell potential: E1 = 1.23 V (a) Compute ΔE° for the disproportionation of H2O2 in acidic solution. (b) Is H202 stable with respect to disproportionation in acidic solution?

Answer 1

Solution.

The disproportionation of H₂O₂ in acidic solution proceeds as follows:

$2H_2O_2=2H_2O+O_2;$

It can be obtained combining the opposite first equation and the direct second equation. Therefore, the cell potential is E = 1.78 - 0.70 = 1.08 V.

As the standard cell potential is positive, the reaction of hydrogen peroxide disproportionation is favorable under the acidic conditions.