When oxidation and reduction take place in a cell,its electromotive force $(EMF)$ 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)

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

For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

For the reaction $Cr_2O_7^{2-} + I^- \to I_2 + Cr^{3+}$,if $E_{cell}^o = 0.79 \ V$ and $E_{Cr_2O_7^{2-}}^o = 1.33 \ V$,then $E_{I_2}^o = $ ............. $V$.

What does the negative sign in the expression $E_{Zn^{2+}|Zn}^o = -0.76 \ V$ mean?