For the electrochemical cell, $M | M^{+} ∥ X^{-} | X, E_{M^{+} / M}^{o} = 0.44 V a n d E_{X / X^{-}}^{o} = 0.33 V$ . From this data we can deduce that :

M + X ⟶ M^{+} + X^{-}

is the spontaneous reaction

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M^{+} + X^{-} ⟶ M + X

is the spontaneous reaction

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E_{c e l l} = 0.77 V

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E_{c e l l} = - 0.77 V

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Solution

The correct option is A $M^{+} + X^{-} ⟶ M + X$ is the spontaneous reaction
For option A, $E_{o} = 0.33 - 0.44 = - 0.11$
Since, $E_{o}$ is negative, $Δ G$ is positive hence, the reaction is non-spontaneous.

For option B, $E_{o} = - 0.33 + 0.44 = 0.11$
Since, $E_{o}$ is +ve, $Δ G$ is negative hence, the reaction is spontaneous.
So, option B is correct.

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Similar questions

For the electrochemical cell, M|M⁺| | X |X^-, Pt. E^o(M⁺/M) = 0.44 V and E^o(X/X^-) = 0.33 V. From this data one can deduce that

M | M^{+} | | X | X^{-}

E_{M^{+} / M}^{0} = 0.44 V

E_{X / X^{-}}^{0} = 0.33 V

M | M^{+} | | X^{-} | X, E_{M^{+} / M}^{\circ} = 0.44 V

E_{X / X}^{\circ} = 0.33 V

M / M^{+} | | X^{-} / X, E_{M^{+} / M}^{\circ} = 0.44

E_{X / X^{-}}^{\circ} = 0.33 V

M + X \to M^{+} + X^{-}

M | M^{+} ∥ X^{-} | X

E_{(M^{+} | M)}^{o} = 0.44 V

E_{(X | X^{-})}^{o} = 0.33 V

0.5 a m p

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