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a
emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
b
emf ofcell = (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)
answer is B, D.
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