$Cu^{+} + e^- \to Cu$ ; $E^o = X_1 \ V$
$Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2 \ V$
Then for $Cu^{2+} + e^- \to Cu^{+}$ ; $E^o$ will be ?

For the given reactions:
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)

Consider the following ${E^0}$ values:
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$ and $Fe^{2+}/Fe$ are $-0.76 \ V$,$-0.23 \ V$ and $-0.44 \ V$ respectively.
The reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ will be spontaneous when:

Will $Fe_{(s)}$ be oxidised to $Fe^{2+}$ by the reaction with $1 \ M$ $HCl$ ($E^o$ for $Fe/Fe^{2+} = +0.44 \ V$)?

Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)