Redox Reactions And Electrochemistry Question 32
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 reverse of a 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 MX $ $ 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.
BETA
