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)

Question

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?NEET 2010 (Mains)

Options:

Answer:

D

Solution:

$E_{cell} = cathode D - (red) E_{anode}^{\circ} or E_{cell} = cred) E_{cathode}^{\circ} + (oxide) E_{Anode}^{\circ} or E_{cell} = (oxide) E anode E - E_{cathode}^{\circ} (oxide)$

TestHub · Accepted Answer

​ E cell ​ = cathode D ​ − (red) E anode ∘ ​ ​ or E cell ​ = cred) E cathode ∘ ​ ​ + (oxide) E Anode ∘ ​ ​ or E cell ​ = (oxide) E ​ anode E ​ − E cathode ∘ ​ (oxide) ​

Chemistry - Electrochemistry Question with Solution | TestHub

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