Preventing oxidation of copper(I) 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 $\ce{F2}$? 
$$
\begin{align}
E(\ce{Cu^2+/Cu+}) &amp;= \pu{0.15 V}\\
E(\ce{F2/F-}) &amp;= \pu{2.87 V}\\
E(\ce{Mg^2+/Mg}) &amp;= \pu{-2.37 V}
\end{align}
$$

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