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 $ .



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