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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

दिया गया है \(Co ^{3+}+ e ^{-} \rightarrow Co ^{2+} ; E ^{\circ}=+1.81 \,V\) \(Pb ^{4+}+2 e ^{-} \rightarrow Pb ^{2+} ; E ^{\circ}=+1.67 \,V\) \(Ce ^{4+}+ e ^{-} \rightarrow Ce ^{3+} ; E ^{\circ}=+1.61 \,V\) \(Bi ^{3+}+3 e ^{-} \rightarrow Bi ; E ^{\circ}=+0.20 \,V\) स्पीशीज की उपचायक सामर्थ्य इस क्रम में बढ़ेगी

  1. A \(C{e^{4 + }} < P{b^{4 + }} < B{i^{3 + }} < C{o^{3 + }}\)
  2. B \(C{o^{3 + }} < P{b^{4 + }} < C{e^{4 + }} < B{i^{3 + }}\)
  3. C \(B{i^{3 + }} < C{e^{4 + }} < P{b^{4 + }} < C{o^{3 + }}\)
  4. D \(C{o^{3 + }} < C{e^{4 + }} < B{i^{3 + }} < P{b^{4 + }}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(B{i^{3 + }} < C{e^{4 + }} < P{b^{4 + }} < C{o^{3 + }}\)

Step-by-step Solution

Detailed explanation

Lower the standard reduction potential, more the ability to get reduced higher the oxidizing power
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