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Electrochemistry question

2010 · Q55
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Electrochemistry question

2010 · Q55

NEETChemistryElectrochemistryMCQ+4 / −1
Consider the following relations for emf of an electrochemical cell
(i)   EMF of cell = (Oxidation potential of anode) −-− (Reduction potential of cathode)
(ii)  EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) −-− (Oxidation potential of cathode)

Which of the above relations are correct?
  1. A
    (iii) and (i)
  2. B
    (i) and (ii)
  3. C
    (iii) and (iv)
  4. D
    (ii) and (iv)
View written solutionFree

Correct answer: D

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode – Oxidation potential of cathode.

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