Given below are half cell reactions: MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O, E^ (MnO4^-/Mn^2+) = +1.510 V 1/2 O2 + 2H+ + 2e^- -> H2O, E^ (O2/H2O) = +1.223 V Will the permanganate ion, MnO4^-, liberate O2 from water in the presence of an acid?

Correct answer: (A) (a) Yes, because E^ cell = +0.287 V. Answer: (a) Yes, because E^ cell = +0.287 V. Solution: Here MnO4^- is reduced (cathode) and water is oxidised to O2 (anode). E^ cell = E^ cathode - E^ anode = (+1.510) - (+1.223) = +0.287 V. Since E^ cell > 0, the reaction is feasible, so MnO4^- can liberate O2 from water in acidic medium.

NEET 2022 · ChemistryOxidantsPrevious Year Question

Given below are half cell reactions: $MnO X_{4} X^{-} + 8 H X^{+} + 5 e X^{-} Mn X^{2 +} + 4 H X_{2} O$ , $E_{(MnO X_{4} X^{-} / Mn X^{2 +})}^{\circ} = + 1.510$ V $\frac{1}{2} O X_{2} + 2 H X^{+} + 2 e X^{-} H X_{2} O$ , $E_{(O X_{2} / H X_{2} O)}^{\circ} = + 1.223$ V Will the permanganate ion, $MnO X_{4} X^{-}$ , liberate $O X_{2}$ from water in the presence of an acid?

Answer: (A) (a) Yes, because $E_{ce l l}^{\circ} = + 0.287$ V. \textbf{Answer:} (a) Yes, because $E_{ce l l}^{\circ} = + 0.287$ V.

A.(a) Yes, because $E_{ce l l}^{\circ} = + 0.287$ V✓
B.(b) No, because $E_{ce l l}^{\circ} = - 0.287$ V
C.(c) Yes, because $E_{ce l l}^{\circ} = + 2.733$ V
D.(d) No, because $E_{ce l l}^{\circ} = - 2.733$ V

Correct Answer

(A) (a) Yes, because $E_{ce l l}^{\circ} = + 0.287$ V

Solution & Explanation

\textbf{Answer:} (a) Yes, because $E_{ce l l}^{\circ} = + 0.287$ V. \textbf{Solution:} Here $MnO X_{4} X^{-}$ is reduced (cathode) and water is oxidised to $O X_{2}$ (anode). $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = (+ 1.510) - (+ 1.223) = + 0.287$ V. Since $E_{ce l l}^{\circ} > 0$ , the reaction is feasible, so $MnO X_{4} X^{-}$ can liberate $O X_{2}$ from water in acidic medium.

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Given below are half cell reactions: MnOX4​X−+8HX++5eX−​MnX2++4HX2​O, E(MnOX4​X−/MnX2+)∘​=+1.510 V 21​OX2​+2HX++2eX−​HX2​O, E(OX2​/HX2​O)∘​=+1.223 V Will the permanganate ion, MnOX4​X−, liberate OX2​ from water in the presence of an acid?

Solution & Explanation

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