Redox Reactions And Electrochemistry Ques 216
Question: The e.m.f. of a cell whose half cells are given below is $ M{g^{2+}}+2{e^{-}}\to Mg(s)\ E{}^\circ =-2.37\ V $ $ C{u^{2+}}+2{e^{-}}\to Cu(s)\ E{}^\circ =+0.34\ V $ [Pb.CET 2001]
Options:
A) + 1.36 V
B) + 2.71 V
C) + 2.17 V
D) - 3.01 V
Show Answer
Answer:
Correct Answer: B
Solution:
$ E_{Cu}^{0}>E_{Mg}^{0} $ hence Cu acts as cathode and Mg acts as anode.
$ E_{cell}^{0}=E_{Cu}^{0}-E_{Mg}^{0} $ $ =(0.34)-(-2.37)=+2.71\ V $ .
