Given below are half-cell reactions: Will the permanganate ion, liberate from water in the presence of an acid? [2022]

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Q.
Given below are half-cell reactions:

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {Mn}^{2 +} + 4 H_{2} O;$ $E_{{Mn}^{2 +} / {MnO}_{4}^{-}}^{\circ} = - 1.510$ $V$

$\frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} ⟶ H_{2} O;$ $E_{O_{2} / H_{2} O}^{\circ} = + 1.223$ $V$

Will the permanganate ion, ${MnO}_{4}^{-}$ liberate $O_{2}$ from water in the presence of an acid [2022]

1 Yes, because $E_{cell}^{\circ} = + 0.287$ $V$

2 No, because $E_{cell}^{\circ} = - 0.287$ $V$

3 Yes, because $E_{cell}^{\circ} = + 2.733$ $V$

4 No, because $E_{cell}^{\circ} = - 2.733$ $V$

Ans.
(1)

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = + 1.510 - 1.223 = + 0.287$ $V$

As $E_{cell}^{\circ}$ is positive, hence the reaction is feasible.

Solution

1 Yes, because $E_{cell}^{\circ} = + 0.287$ $V$

2 No, because $E_{cell}^{\circ} = - 0.287$ $V$

3 Yes, because $E_{cell}^{\circ} = + 2.733$ $V$

4 No, because $E_{cell}^{\circ} = - 2.733$ $V$

Ans.
(1)

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = + 1.510 - 1.223 = + 0.287$ $V$

As $E_{cell}^{\circ}$ is positive, hence the reaction is feasible.

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Solution

Q. Given below are half-cell reactions: MnO4-+8H++5e-⟶Mn2++4H2O ; EMn2+/MnO4-∘=-1.510 V 12O2+2H++2e-⟶H2O ; EO2/H2O∘=+1.223 V Will the permanganate ion, MnO4- liberate O2 from water in the presence of an acid [2022]

1 Yes, because Ecell∘=+0.287 V

2 No, because Ecell∘=-0.287 V

3 Yes, because Ecell∘=+2.733 V

4 No, because Ecell∘=-2.733 V

Ans. (1) Ecell∘=Ecathode∘-Eanode∘ =+1.510-1.223=+0.287 V As Ecell∘ is positive, hence the reaction is feasible.

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1 Yes, because $E_{cell}^{\circ} = + 0.287$ $V$

2 No, because $E_{cell}^{\circ} = - 0.287$ $V$

3 Yes, because $E_{cell}^{\circ} = + 2.733$ $V$

4 No, because $E_{cell}^{\circ} = - 2.733$ $V$

Ans.
(1)

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = + 1.510 - 1.223 = + 0.287$ $V$

As $E_{cell}^{\circ}$ is positive, hence the reaction is feasible.