If E^{circ} (Mg^{2+}_{(aq)} mid Mg_{(s)}) = - | vedclass

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(A) The given reaction is the oxidation of $Mg$ to $Mg^{2+}$.The standard electrode potential for reduction is $E^{\circ} (Mg^{2+} \mid Mg) = -2.37 \

Vedclass · Accepted Answer

$+2.3996 \ V$

Vedclass · Answer

(A) The given reaction is the oxidation of $Mg$ to $Mg^{2+}$.The standard electrode potential for reduction is $E^{\circ} (Mg^{2+} \mid Mg) = -2.37 \ V$.Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$.Using the Nernst equation for the reaction $Mg_{(s)} \longrightarrow Mg^{2+} (0.1 \ M) + 2 \ e^{-}$:$E = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{2+}]$ $E = 2.37 - \frac{0.0591}{2} \log (0.1)$$E = 2.37 - 0.02955 	imes (-1)$$E = 2.37 + 0.02955 = 2.39955 \ V \approx +2.3996 \ V$.

If $E^{\circ} (Mg^{2+}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$,what is the potential for the reaction $Mg_{(s)} \longrightarrow Mg^{2+} (0.1 \ M) + 2 \ e^{-}$ at $298 \ K$?

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