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 ?
- A No, because \(E _{\text {cell }}^{\circ}=-0.287\, V\)
- B Yes, because \(E _{\text {cell }}^{\circ}=+2.733\, V\)
- C No, because \(E_{\text {cell }}^{\circ}=-2.733 \,V\)
- D Yes, because \(E _{\text {cell }}^{\circ}=+0.287\, V\)
Answer & Solution
Correct Answer
(D) Yes, because \(E _{\text {cell }}^{\circ}=+0.287\, V\)
Step-by-step Solution
Detailed explanation
Cathode : \(\underset{ E _{ RP }^{\circ}=1.510 V }{2 MnO _{4}^{-}}+16 H ^{+}+10 e^{-} \rightarrow 2 Mn ^{+2}+8 H _{2} O\)
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