Given below are half cell reactions:
$\begin{array}{l}{\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{E}}_{{\mathrm{Mn}}^0}^2/{\mathrm{MnO}}_4^{-}=-1.510\mathrm{V}\\ \frac{1}{2}{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\mathrm{O}\\ {{\mathrm{E}}^{\circ }}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}=+1.223\mathrm{V}\end{array}$
Will the permanganate ion, ${\mathrm{MnO}}_4^{-}$ liberate O₂ from water in the presence of an acid?

To determine if the permanganate ion (MnO₄^-) can liberate oxygen (O₂) from water in acidic conditions, we need to examine the cell potential of the proposed reaction.

Step 1: Identify the Anode and Cathode Reactions

The given half-reactions are:

• MnO4- + 8H+ + 5e- → Mn2+ + 4H2O 
  EMnO4-/Mn2+0 = +1.510 V (cathode reaction)
• (1/2) O2 + 2H+ + 2e- → H2O 
  EO2/H2O0 = +1.223 V (anode reaction)

Step 2: Calculate the Cell Potential (E_cell⁰)

The standard cell potential (E_cell⁰) for a reaction can be calculated using the formula:

E_cell⁰ = E_cathode⁰ - E_anode⁰

Substituting the values:

Ecell0 = EMnO4-/Mn2+0 - EO2/H2O0 
Ecell0 = +1.510 V - (+1.223 V) 
Ecell0 = +0.287 V

Step 3: Analyze the Result

The positive E_cell⁰ value (+0.287 V) indicates that the reaction is thermodynamically favorable. Therefore, MnO₄^- will indeed oxidize H₂O, causing the liberation of O₂ gas in acidic conditions.

Conclusion:

Yes, the permanganate ion (MnO₄^-) can liberate O₂ from water in the presence of an acid, as the positive cell potential confirms the feasibility of the reaction.