Q. For the electrochemical cell, $M\ |\ M^{+}\ ||\ X^{-}\ |\ X,\ E^{{^\circ}}\left( M^{+}\ |\ M \right) = 0.44\ V\$and$\ E^{{^\circ}}\left( X\ |\ X^{-} \right) = 0.33\ V.$ From this data one can deduce that
A) $E_{cell} = 0.77\ V$
B) $- 0.77\ V$
C) $M^{+} + \ X^{-}\ \rightarrow M + X$ is the spontaneous reaction
D) $M + X \rightarrow M^{+} + \ X^{-}$ is the spontaneous reaction
Explanation
(c)
For the given cell, $M\left| M^{+} \right||\ X^{-}|\ X,$ the cell reaction is derived as follows :
RHS :Reduction$\ :X + e^{-} \rightarrow X^{-}\$ …(i)
LHS :Oxidation$\ :M \rightarrow M^{+} + e^{-}$ …(ii)
Add Eqs. (i) an (ii)
$M + X \rightarrow M^{+} + X^{-}$
The cell potential $= \ - \ 0.11$ V
Since, $E_{cell} = - ve,$ the cell reaction derived above is non-spontaneous. In fact, the reverse reaction will occur spontaneously
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.