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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+? $\endgroup$ Dec 28, 2017 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, 2017 at 10:52
  • $\begingroup$ Ditch all the choices and go with thiourea. It works beautifully, try adding it to a copper(II) sulfate solution (color disappears) or better yet acopper(II) chloride solution (color disappears and a hite precipitate of copper(I) chloride appears which rediscover with more thiourea). $\endgroup$ Oct 16, 2022 at 16:27

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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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