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

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Question

(B) The cell reaction for $M|M^{+}||X^{-}|X$ is:Anode (Oxidation): $M 	o M^{+} + e^{-}$,${E^{o}}_{ox} = -0.44 \ V$Cathode (Reduction): $X + e^{-} 	o

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

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(B) The cell reaction for $M|M^{+}||X^{-}|X$ is:Anode (Oxidation): $M 	o M^{+} + e^{-}$,${E^{o}}_{ox} = -0.44 \ V$Cathode (Reduction): $X + e^{-} 	o X^{-}$,${E^{o}}_{red} = 0.33 \ V$Cell reaction: $M + X 	o M^{+} + X^{-}$${E^{o}}_{cell} = {E^{o}}_{cathode} - {E^{o}}_{anode} = 0.33 \ V - 0.44 \ V = -0.11 \ V$Since ${E^{o}}_{cell}$ is negative,the forward reaction is non-spontaneous.Therefore,the reverse reaction,$M^{+} + X^{-} 	o M + X$,is spontaneous.

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:

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