Given below are the 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) 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 = E^ MnO4^-/Mn^2+ - E^ O2/H2O E^ cell = 1.510 - 1.223 = +0.287 V Since E^ cell > 0, _r G^° = -nFE^ cell < 0, so the reaction is spontaneous and MnO4^- liberates O2 from water.

NEET 2022 · ChemistryGalvanic & EMFPrevious Year Question

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

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

Correct Answer

(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). $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{MnO X_{4} X^{-} / Mn X^{2 +}}^{\circ} - E_{O X_{2} / H X_{2} O}^{\circ}$ $E_{ce l l}^{\circ} = 1.510 - 1.223 = + 0.287 V$ Since $E_{ce l l}^{\circ} > 0$ , $Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ} < 0$ , so the reaction is spontaneous and $MnO X_{4} X^{-}$ liberates $O X_{2}$ from water.

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Given below are the half-cell reactions: MnOX4​X−+8HX++5eX−​MnX2++4HX2​O, EMnOX4​X−/MnX2+∘​=−1.510 V 21​OX2​+2HX++2eX−​HX2​O, EOX2​/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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