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Question: Answered & Verified by Expert
Consider the following relations for emf of a electrochemical cell
(A) Emf of cell = (oxidation potential of anode) - (reduction potential of cathode)
(B) Emf of cell = (oxidation potential of anode) + (reduction potential of cathode)
(C) Emf of cell = (reduction potential of anode) + (reduction potential of cathode)
(D) Emf of cell = (oxidation potential of anode)-(Oxidation potential of cathode)
Which of the above relations are correct?
ChemistryElectrochemistryNEETNEET 2010 (Mains)
Options:
  • A (C) and (A)
  • B (A) and (B)
  • C (C) and (D)
  • D (B) and (D)
Solution:
2782 Upvotes Verified Answer
The correct answer is: (B) and (D)
$\begin{aligned}
& \mathrm{E}_{\text {cell }}=\underset{\text { cathode }}{D}-\underset{\text { (red) }}{\mathrm{E}_{\text {anode }}^{\circ}} \\
& \text {or } E_{\text {cell }}=\underset{\text { cred) }}{\mathrm{\mathrm{E}}_{\text {cathode }}^{\circ}}+\underset{\text { (oxide) }}{E_{\text {Anode }}^{\circ}} \\
& \text {or } \mathrm{E}_{\text {cell }}=\underset{\text { (oxide) }}{\mathrm{E}} \underset{\text { anode }}{\mathrm{E}}-\mathrm{E}_{\text {cathode }}^{\circ} \text { (oxide) }
\end{aligned}$

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