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NEET · Chemistry · STD 12 - 2. Electrochemistry

Given below are half cell reactions: \(MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O\), \(E^{o} _{ Mn ^{2+} / MnO _{4}^{-}}=-1.510 \,V\) \(\frac{1}{2} O _{2}+2 H ^{+}+2 e ^{-} \rightarrow H _{2} O\), \(E _{ O _{2} / H _{2} O }^{o}=+1.223 \,V\) Will the permanganate ion, \(MnO _{4}^{-}\)liberate \(O _{2}\) from water in the presence of an acid ?

  1. A No, because \(E _{\text {cell }}^{\circ}=-0.287\, V\)
  2. B Yes, because \(E _{\text {cell }}^{\circ}=+2.733\, V\)
  3. C No, because \(E_{\text {cell }}^{\circ}=-2.733 \,V\)
  4. D Yes, because \(E _{\text {cell }}^{\circ}=+0.287\, V\)
Verified Solution

Answer & Solution

Correct Answer

(D) Yes, because \(E _{\text {cell }}^{\circ}=+0.287\, V\)

Step-by-step Solution

Detailed explanation

Cathode : \(\underset{ E _{ RP }^{\circ}=1.510 V }{2 MnO _{4}^{-}}+16 H ^{+}+10 e^{-} \rightarrow 2 Mn ^{+2}+8 H _{2} O\)