Redox Reactions And Electrochemistry Ques 212

Question: For the electrochemical cell, $ M|{M^{+}}||{X^{-}}|X, $ $ E{}^\circ ({M^{+}}|M) $ $ =0.44\ V $ $ E{}^\circ (X|{X^{-}})=0.33\ V $ From this data, one can deduce that [Pb.CET 2004]

Options:

A) $ E{{{}^\circ }_{cell}}=-0.77,V $

B) $ {M^{+}}+{X^{-}}\to M+X $ is the spontaneous reaction

C) $ M+X\to {M^{+}}+{X^{-}} $ is the spontaneous reaction

D) $ E{{{}^\circ }_{cell}}=.77\ V $

Show Answer

Answer:

Correct Answer: B

Solution:

For $ {M^{+}}+{X^{-}}\to M+X $
$ E_{cell}^{0}=E_{Cathode}^{0}+E_{Anode}^{0} $

$ =0.44-0.33=+0.11\ V $

Since $ E_{cell}^{0}=(+)\ 0.11\ V $ is positive hence this reaction should be spontaneous.



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक