Q. For the electrochemical cell, $M\ \left| \ M^{+}\ ||X^{-}\ \right|X\ ,\$ $E^{{^\circ}}\ (M^{+}|\ M)\ = \ 0.44\ V,\ E^{{^\circ}}\ = \ (X\ |\ X^{-})\ = \ 0.33\ V$ . From this datAone can deduce that
A) $E_{cell}^{{^\circ}}\ = \ - 0.77\ V$
B) $M^{+}\ + \ X^{-}\ \rightarrow \ M\ + \ X$ is the spontaneous reaction
C) $M\ + \ X\ \ \ \rightarrow \ M^{+}\ + \ X^{-}$ is the spontaneous reaction
D) $E_{cell}^{{^\circ}}\ = \ 0.77\ V$
Explanation
(b)
$E_{cell}^{{^\circ}}$ for the reaction $M^{+}$ + $X^{-}$ → $M\ + \ X\$ is given as follows
= $E_{cathode}^{{^\circ}}$ + $E_{anode}^{{^\circ}}$
= 0.44 V + 0.33 V
= + 0.11 V
Since, $E_{cell}^{{^\circ}}$ is positive for the above reaction .
Hence, this reaction is spontaneous.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.