Verified Solution • 2025–26 Session moderate

QNEET-2022-Ch-413: NEET 2022 Chemistry - Class 12 NEET-2022-Ch-413 Solution: Given below are half cell reactions: \(MnO_4^- + 8H^+ + 5e^-…

Given below are half cell reactions: MnO4−+8H++5e−→Mn2++4H2OEMn2+/MnO4−∘=−1.510 VMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O E_{Mn^{2+}/MnO_4^-}^\circ = -1.510\ \text{V} 12O2+2H++2e−→H2OEO2/H2O∘=+1.223 V\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O E_{O_2/H_2O}^\circ = +1.223\ \text{V} Will the permanganate ion MnO4−MnO_4^- liberate O2O_2 from water in the presence of an acid? Yes because Ecell∘=+0.287 VE_{cell}^\circ = +0.287\ \text{V} No because Ecell∘=−0.287 VE_{cell}^\circ = -0.287\ \text{V} Yes because Ecell∘=+2.733 VE_{cell}^\circ = +2.733\ \text{V} No because Ecell∘=−2.733 VE_{cell}^\circ = -2.733\ \text{V}

Multiple Choice Options

  • A Yes, because EcelloE_{cell}^o E ce ll o _ = + 0.287 V
  • B No, because EcelloE_{cell}^o E ce ll o _ = −- _ 0.287 V
  • C Yes, because EcelloE_{cell}^o E ce ll o _ = + 2.733 V
  • D No, because EcelloE_{cell}^o E ce ll o _ = −- _ 2.733 V

Step-by-Step Verified Solution

_ M n O 4 _ + 8 H + + 5 e _ _ M n 2 + + 4 H 2 O ..... (i) E M n O 4 _ / M n 2 + o = _ E M n 2 + / M n O 4 _ o = 1.51 V _ H 2 O _ 1 2 O 2 + 2 H + + 2 e _ ..... (ii) E O 2 / H 2 O o = 1.223 V Using 2 _ (i) + 5 _ (ii), net cell reactions is 2 M n O 4 _ + 6 H + _ 2 M n 2 + + 5 2 O 2 + 3 H 2 O E c e l l o = E C o _ E A o = E M n O 4 _ / M n 2 + o _ E O 2 / H 2 O o = 1.51 _ 1.223 = 0.287 V Since E c e l l o > 0 , therefore net cell reaction is spontaneous and so M n O 4 _ liberate O 2 from H 2 O in presence of an acid.

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