Redox Reactions And Electrochemistry Ques 216

Question: The e.m.f. of a cell whose half cells are given below is Mg2++2eMg(s) E=2.37 V Cu2++2eCu(s) E=+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:

ECu0>EMg0 hence Cu acts as cathode and Mg acts as anode.

Ecell0=ECu0EMg0 =(0.34)(2.37)=+2.71 V .



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