The emf of a galvanic cell, with electrode potentials of Zn=+0.76V , is:
+0.34V
+0.76V
−1.1V
+1.1V
A
+0.34V
B
+0.76V
C
+1.1V
D
−1.1V
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Solution
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Left electrode : Zn(s)→Zn2++2e−. Right electrode : Cu2+(aq)+2e−→Cu(s)
The overall reaction of the cell is the sum of above two reactions:
Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s)
Emf of the cell,Eocell=Eo(cathode)−Eo(anode)
Emf of the cell =0.34V−(−0.76)V=+1.1V
Hence, option B is correct.
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