Redox Reactions And Electrochemistry Question 770
Question: For the electrochemical cell, $ M|{M^{+}}||{X^{-}}|X, $ $ E^{o}({M^{+}}/M) $ = 0.44 V and $ E^{o}(X/{X^{-}}) $ = 0.33 V. From this data one can deduce that [IIT-JEE (Screening) 2000]
Options:
A) $ M,+,X,\to {M^{+}}+{X^{-}} $ is the spontaneous reaction
B) $ {M^{+}}+{X^{-}}\to M+X $ is the spontaneous reaction
C) $ E_{cell} $ = 0.77 V
D) $ E_{cell} $ = - 0.77 V
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Answer:
Correct Answer: B
Solution:
For the given cell $ M|{M^{+}}||{X^{-}}|X $ , the cell reaction is derived as follows: RHS: reduction $ X+{e^{-}}\to {X^{-}} $ -..(i) LHS: Oxidation $ M\to {M^{+}}+{e^{-}} $ -..(ii) Add (i) and (ii) $ M+X\to {M^{+}}+{X^{-}} $ The cell potential = $ -0.11,V $ Since $ E_{cell}= $ - ve, the cell reaction derived above is not spontaneous. In fact, the reverse reaction will occur spontaneously.
BETA
