What will happen when a silver ring is placed in an acidic solution?

Assertion : $Cu^{2+}$ ions get reduced more easily than $H^{+}$ ions.
Reason : Standard electrode potential of copper is $0.34 \ V$.

Given the standard electrode potentials $E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$,calculate the standard cell potential $E^{\circ}_{cell}$ for the reaction: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ (in $V$)

If the half-cell reaction $A + e^- \to A^-$ has a large negative reduction potential,it follows that

If $Cu^{+} + e^- \to Cu$ ; $E^o = X_1$ and $Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2$,then the value of $E^o$ for $Cu^{2+} + e^- \to Cu^{+}$ will be:

The cell reaction of a cell is $Mg_{(s)} + Cu^{2+}_{(aq)} \to Cu_{(s)} + Mg^{2+}_{(aq)}$. If the standard reduction potentials of $Mg$ and $Cu$ are $-2.37 \ V$ and $+0.34 \ V$ respectively,the $EMF$ of the cell is ................. $V$.