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Electrochemistry

Question
CBSEENCH12011289

Consider the following relations for emf of an electrochemical cell

A) Emf of cell = (oxidation potential of anode) - (reduction potential of cathode)

B) Emf of cell = (oxidation potential of anode) - ( reducation potential of cathode)

C) Emf of cell = (oxidation potential of anode)- (reduction potential of cathode)

D) Emf of cell = (oxidation potential of anode) - (oxidation potential of anode)-(oxidation potential  of cathode)

  • (C) and (A)

  • (A) and (B)

  • (C) and (D)

  • (B) and (D)

Solution

D.

(B) and (D)

straight E subscript cell space equals space straight E subscript cathode space left parenthesis red right parenthesis end subscript superscript straight o minus straight E subscript anode space left parenthesis red right parenthesis end subscript superscript straight o
straight E subscript cell space equals space straight E subscript cathode space left parenthesis red right parenthesis end subscript superscript 0 space plus space straight E subscript Anode space left parenthesis oxid right parenthesis end subscript superscript 0
straight E subscript cell space equals space straight E subscript cathode space left parenthesis oxid right parenthesis end subscript superscript 0 space plus space straight E subscript Anode space left parenthesis oxid right parenthesis end subscript superscript 0