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: For O2 to be liberated, MnO4^- must act as the oxidant (reduced to Mn^2+) and water as the reductant (oxidised to O2). E^ cell = E^ cathode - E^ anode = E^ (MnO4^-/Mn^2+) - E^ (O2/H2O). Using the standard reduction potentials, E^ (MnO4^-/Mn^2+) = +1.510 V and E^ (O2/H2O) = +1.223 V (the value -1.510 V is quoted for the reverse, oxidation, direction). Thus E^ cell = 1.510 - 1.223 = +0.287 V. Since E^ cell is positive, G^° < 0 and the reaction is spontaneous, so MnO4^- will liberate O2. Hence (a).

NEET 2022 · ChemistryPrevious 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 \textbf{Solution:} For $O X_{2}$ to be liberated, $MnO X_{4} X^{-}$ must act as the oxidant (reduced to $Mn X^{2 +}$ ) and water as the reductant (oxidised to $O X_{2}$ ).

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:} For $O X_{2}$ to be liberated, $MnO X_{4} X^{-}$ must act as the oxidant (reduced to $Mn X^{2 +}$ ) and water as the reductant (oxidised to $O X_{2}$ ). $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{(MnO X_{4} X^{-} / Mn X^{2 +})}^{\circ} - E_{(O X_{2} / H X_{2} O)}^{\circ}$ . Using the standard reduction potentials, $E_{(MnO X_{4} X^{-} / Mn X^{2 +})}^{\circ} = + 1.510$ V and $E_{(O X_{2} / H X_{2} O)}^{\circ} = + 1.223$ V (the value $- 1.510$ V is quoted for the reverse, oxidation, direction). Thus $E_{ce l l}^{\circ} = 1.510 - 1.223 = + 0.287$ V. Since $E_{ce l l}^{\circ}$ is positive, $Δ G^{\circ} < 0$ and the reaction is spontaneous, so $MnO X_{4} X^{-}$ will liberate $O X_{2}$ . Hence (a).

🎯

43,000+ questions in हिंदी & English · one-tap toggle

Practice NEET 2022 Chemistry — free, with instant solutions

43,000+ NEET questions solved step-by-step, chapter-wise, in the MedicNEET app.

Open →📚Practice all NEET Chemistry PYQs chapter-wise →💡Don't get it? Learn the Redox Equilibrium concepts →

← Chemistry Q14 Chemistry Q16 →

More NEET PYQ solutions

ChemIdentify the suitable reagent for the following conversion: C6H5COOCH3 -> C6H5CHO (methyl benzoate to benzaldehyde)ChemThe correct order of decreasing acidity of the following aliphatic acids is ChemThe major product of the following reaction is: 4-oxo-4-phenylbutanenitrile C6H5-CO-CH2-CH2-CN is treated with (i) CH3MgBr (excess) and then (ii) H3O+.ChemOut of the following complex compounds, which compound will have the minimum conductance in solution?ChemWhich of the following are paramagnetic? A. [NiCl4]^2- B. Ni(CO)4 C. [Ni(CN)4]^2- D. [Ni(H2O)6]^2+ E. Ni(PPh3)4 ChemThe correct order of the wavelength of light absorbed by the following complexes is: A. [Co(NH3)6]^3+ B. [Co(CN)6]^3- C. [Cu(H2O)4]^2+ D. [Ti(H2O)6]^3+ChemThe correct order of decreasing basic strength of the given amines is: (N-ethylethanamine = (C2H5)2NH; ethanamine = C2H5NH2; N-methylaniline = C6H5NH(CH3); benzenamine/aniline = C6H5NH2)ChemPredict the major product P in the following sequence of reactions. The starting material is 1-methylcyclopent-1-ene: (i) HBr, benzoyl peroxide; (ii) KCN; (iii) Na(Hg)/C2H5OH P (major). The structures of the options (A)-(D) are shown in the image below:

More: All NEET 2022 Chemistry PYQs · NEET Chemistry PYQs · All NEET PYQs · Previous year papers

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

More NEET PYQ solutions