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.

🎯

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

Practice NEET 2022 Chemistry — free, with instant solutions

The app builds a daily goal from your weak topics and brings each question back just before you forget it — so what you solve today turns into marks on exam day.

Open →📚Practice all NEET Chemistry PYQs chapter-wise →💡Don't get it? Learn the D And F Block Elements concepts →

← Chemistry Q19 Chemistry Q21 →

More NEET PYQ solutions

ChemIn the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from ChemWhich of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu^2+ + Cu (b) 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O (c) 2KMnO4 -> K2MnO4 + MnO2 + O2 (d) 2MnO4^- + 3Mn^2+ + 2H2O -> 5MnO2 + 4H+ Select the correct option from the following:ChemWhen neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is ChemWhich one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?ChemThe correct order of decreasing acidity of the following aliphatic acids is:ChemWhich one of the following reactions does NOT belong to Lassaigne's test?ChemHow many products (including stereoisomers) are expected from the monochlorination of 2-methylbutane, (CH3)2CH-CH2-CH3?ChemThe total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

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