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 = $ ............. $V$.

Review the $SRP$ (at $25\,\text{°C}$) data in acidic medium:
$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$
$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$
where $x < y$,point out the wrong statement.

Given,the standard potentials $E_{(Cu^{2+}/Cu)}^{\circ}$ and $E_{(Cu^{+}/Cu)}^{\circ}$ as $0.340 \ V$ and $0.522 \ V$ respectively,the value of $E_{(Cu^{2+}/Cu^{+})}^{\circ}$ is $.... \ V$

The $E^o$ values of half-cells are given below. Which combination of two half-cells will form a cell with the maximum potential?
$(i)$ $A + e^- \rightarrow A^- ; E^o = -0.24 \ V$
$(ii)$ $B^- + e^- \rightarrow B^{2-} ; E^o = +1.25 \ V$
$(iii)$ $C^- + 2e^- \rightarrow C^{3-} ; E^o = -1.25 \ V$
$(iv)$ $D + 2e^- \rightarrow D^{2-} ; E^o = +0.68 \ V$

The two half-cell reactions of an electrochemical cell are given as: $Ag^{+} + e^{-} \rightarrow Ag$; $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$. The value of cell $EMF$ will be: (in $V$)

Determine true $(T)$ and false $(F)$ for the following statements:
$(i)$ Always $E_{cell}^o = E_{cell}$
$(ii)$ In standard conditions,$E_{cell}^o = E_{cell}$
$(iii)$ $E_{cell}^o = E_{cell}$ is reached when the cell attains equilibrium.
$(iv)$ $E_{cell}^o = -\frac{\Delta_r G^o}{nF}$