Question

Consider the reaction between ozone and a metal cation, m2+, to form the metal oxide, mo2, and dioxygen: o3 + m2+(aq) + h2o(l) →o2(g) + mo2(s) + 2 h+ for which eocell = 0.44. given that eored of ozone is 2.07 v, calculate eored of mo2.

Answer 1

O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V

Answer 2

Answer:

1.63 V

Explanation:

Let's consider the following redox reaction.

O₃(g) + M²⁺(aq) + H₂O(l) → O₂(g) + MO₂(s) + 2 H⁺(aq)

We can identify both half-reactions.

Reduction (cathode): O₃(g) + 2 H⁺(aq) + 2 e⁻ → O₂(g) + H₂O(l)  E°red = 2.07 V

Oxidation (anode): M²⁺(aq) + 2 H₂O(l)  → MO₂(s) + 4 H⁺(aq) + 2 e⁻ E°red = ?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

0.44 V = 2.07 V - E°red, an

E°red, an = 1.63 V