Question

Be sure to answer all parts. In acidic solution, O3 and Mn2+ ions react spontaneously: O3(g) + Mn2+(aq) + H2O(l) → O2(g) + MnO2(s) + 2H+(aq) E o cell = 0.84 V

Write the balanced half-reactions and calculate E o manganese when given E o ozone = 2.07 V.

Include states of matter in your answers.

Reduction half-reaction:

Oxidation half-reaction:

E o manganese=

Answer 1

Reaction : (acidic solution)

O₃(g) + Mn²⁺(aq) + H₂O(l) ==> O₂(g) + MnO₂(s) + 2H⁺(aq) Eo Cell = 0.84 V ...i

Half reactions : (as reduction reactions):   Acidic conditions

1. O₃(g) + 2H⁺(aq) + 2e ==> O₂(g) + H₂O(l)       Eo _(O3/O2) = + 2.07 V

2. MnO₂(s) + 4H⁺(aq) + 2e ==> Mn²⁺(aq) + 2 H₂O(l)

so,

reduction Half reaction : O₃(g) + 2H⁺(aq) + 2e ==> O₂(g) + H₂O(l)    Eo _(O3/O2) = + 2.07 V        ....ii

oxidation Half reaction : Mn²⁺(aq) + 2 H₂O(l)==>   MnO₂(s) + 4H⁺(aq) + 2e   Eo= ?

From (i) -(ii) :

    O₃(g) + Mn²⁺(aq) + H₂O(l) ==> O₂(g) + MnO₂(s) + 2H⁺(aq) Eo Cell = 0.84 V

-        O₃(g) + 2H⁺(aq) + 2e ==> O₂(g) + H₂O(l)    Eo _(O3/O2) = + 2.07 V

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             Mn²⁺(aq) + 2 H₂O(l)==>   MnO₂(s) + 4H⁺(aq) + 2e     Eo = - 1.23 V

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or in form of Reduction half reaction :

MnO₂(s) + 4H⁺(aq) + 2e ==> Mn²⁺(aq) + 2 H₂O(l)         Eo _(Mn2+/MnO2) = + 1.23 V