For the electrochemical cell M mid M^{2+} par | vedclass

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Question

(C) The cell reaction is $M + X ightarrow M^{2+} + X^{2-}$.Standard cell potential $E^{\circ}_{	ext{cell}} = E^{\circ}_{	ext{cathode}} - E^{\circ}

Vedclass · Accepted Answer

$M^{2+} + X^{2-} \rightarrow M + X$ is a spontaneous reaction

Vedclass · Answer

(C) The cell reaction is $M + X ightarrow M^{2+} + X^{2-}$.Standard cell potential $E^{\circ}_{	ext{cell}} = E^{\circ}_{	ext{cathode}} - E^{\circ}_{	ext{anode}}$.Here, the cathode is $X/X^{2-}$ and the anode is $M/M^{2+}$.Given $E^{\circ}_{(X/X^{2-})} = 0.34 \ V$ (reduction potential of cathode).Given $E^{\circ}_{(M^{2+}/M)} = 0.46 \ V$ (reduction potential of anode).$E^{\circ}_{	ext{cell}} = E^{\circ}_{	ext{cathode}} - E^{\circ}_{	ext{anode}} = 0.34 \ V - 0.46 \ V = -0.12 \ V$.Since $E^{\circ}_{	ext{cell}} Therefore, the reverse reaction $M^{2+} + X^{2-} ightarrow M + X$ is spontaneous.

For the electrochemical cell $M \mid M^{2+} \parallel X \mid X^{2-}$, if $E^{\circ}_{(M^{2+} / M)} = 0.46 \ V$ and $E^{\circ}_{(X / X^{2-})} = 0.34 \ V$, which of the following is correct?

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