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

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Question

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

Vedclass · Accepted Answer

$E_{	ext{cell}} = 0.80 \ V$

Vedclass · Answer

(D) The cell reaction is $M + X^{2-} ightarrow M^{2+} + X$.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 $E^{\circ}_{(M^{2+}/M)} = 0.46 \ V$,so $E^{\circ}_{(M/M^{2+})} = -0.46 \ V$.Given $E^{\circ}_{(X/X^{2-})} = 0.34 \ V$.$E^{\circ}_{	ext{cell}} = 0.34 \ V - (-0.46 \ V) = 0.80 \ V$.Since $E^{\circ}_{	ext{cell}} > 0$,the reaction $M + X^{2-} ightarrow M^{2+} + X$ is spontaneous.However,checking the options,if we consider the reverse reaction $M^{2+} + X^{2-} ightarrow M + X$,it would be non-spontaneous. Given the options provided,$E^{\circ}_{	ext{cell}} = 0.80 \ V$ is the correct value.

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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