Iron displaces copper from its salt solution,because

$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.

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

On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

$A$ $1.0 \ M$ solution with respect to each of the metal halides $AX_3, BX_2, CX_3$ and $DX_2$ is electrolysed using platinum electrodes. If
$E^o_{A^{3+}/A} = 1.50 \ V, \quad E^o_{B^{2+}/B} = 0.3 \ V,$
$E^o_{C^{3+}/C} = -0.74 \ V, \quad E^o_{D^{2+}/D} = -2.37 \ V.$
The correct sequence in which the various metals are deposited at the cathode is

The electrode potentials are given as follows:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is: