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ID: 512734 • Letter: D

Question

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Explanation / Answer

from data table:

Eo(Cu+/Cu(s)) = 0.520

Eo(Ag+/Ag(s)) = 0.800

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ag+/Ag(s))

anode is (Cu+/Cu(s))

The chemical reaction taking place is

Ag+ + Cu(s) --> Ag(s) + Cu+

Eocell = Eocathode - Eoanode

= (0.800) - (0.520)

= 0.280 V