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By aerial oxidation $\ce{Cu+}$ can easily oxidise to $\ce{Cu^2+}$. Which of the following should be added to $\ce{Cu+}$ to prevent it from oxidation: $\ce{F-}$, $\ce{Mg^2+}$, $\ce{Mg}$ or $\ce{F2}$? $$ \begin{align} E(\ce{Cu^2+/Cu+}) &= \pu{0.15 V}\\ E(\ce{F2/F-}) &= \pu{2.87 V}\\ E(\ce{Mg^2+/Mg}) &= \pu{-2.37 V} \end{align} $$

Why it isn't $\ce{Mg^2+}$?

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    $\begingroup$ What would Mg2+ do to prevent oxidation of Cu+? $\endgroup$ – Ivan Neretin Dec 28 '17 at 7:11
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    $\begingroup$ Teacher asking this question should get fired. One of the given would indeed prevent Cu+ from getting oxidised, but that's not the only thing it would do. $\endgroup$ – Karl Dec 28 '17 at 10:52
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We need a better reducing agent (that itself oxidises) than $\ce {Cu^+}$. So choose the ones which have greater reduction potential than $\ce {Cu^+}$. It isn't $\ce {Mg^2+}$ as it cannot oxidise any further hence it has a very low(negative) reduction potential, so it cannot stop $\ce {Cu^+}$ from oxidising.

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In one sense none of the options really works. Metallic magnesium certainly prevents the oxidation of $\ce{Cu^+}$ to $\ce{Cu^{2+}}$, but also reduces $\ce{Cu^+}$ to elemental copper. To stabilize $\ce{Cu^+}$ for real use a soft-base complexing agent like thiourea.

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