For the electrochemical cell,M|M^{+}||X^{-}|X | vedclass

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Question

(B) The cell reaction is $M + X 	o M^{+} + X^{-}$.For this cell,the anode reaction is $M 	o M^{+} + e^{-}$ $(E^{\circ}_{ox} = -E^{\circ}(M^{+}|M) =

Vedclass · Accepted Answer

$M^{+} + X^{-} 	o M + X$ is the spontaneous reaction

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(B) The cell reaction is $M + X 	o M^{+} + X^{-}$.For this cell,the anode reaction is $M 	o M^{+} + e^{-}$ $(E^{\circ}_{ox} = -E^{\circ}(M^{+}|M) = -0.44 \; V)$ and the cathode reaction is $X + e^{-} 	o X^{-}$ $(E^{\circ}_{red} = E^{\circ}(X|X^{-}) = 0.33 \; V)$.$E^{\circ}_{cell} = E^{\circ}_{red} + E^{\circ}_{ox} = 0.33 \; V + (-0.44 \; V) = -0.11 \; V$.Since $E^{\circ}_{cell}$ is negative,the forward reaction $M + X 	o M^{+} + X^{-}$ is non-spontaneous.Conversely,the reverse reaction $M^{+} + X^{-} 	o M + X$ has $E^{\circ}_{cell} = +0.11 \; V$,which is positive,making it spontaneous.

For the electrochemical cell,$M|M^{+}||X^{-}|X,$ given $E^{\circ}(M^{+}|M) = 0.44 \; V$ and $E^{\circ}(X|X^{-}) = 0.33 \; V$. From this data,one can deduce that:

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