The $E^o$ for half cells $Fe/Fe^{2+}$ and $Cu/Cu^{2+}$ are $-0.44 \, V$ and $+0.32 \, V$ respectively. Then

The $E^{\circ}$ values of half-cells are given below. Which combination of two half-cells will result in a cell with the maximum potential?
$(i) \, A^{3-} \rightarrow A^{2-} + e^{-}; E^{\circ} = 1.5 \, V$
$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ} = 0.5 \, V$
$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ} = 0.5 \, V$
$(iv) \, D \rightarrow D^{2+} + 2e^{-}; E^{\circ} = -1.15 \, V$

Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

$MnO_4^- (aq) + 8H^+ (aq) + 5e^- \to Mn^{2+} (aq) + 4H_2O (l)$; $E_1^o = 1.51 \ V$
$MnO_2 (s) + 4H^+ (aq) + 2e^- \to Mn^{2+} (aq) + 2H_2O (l)$; $E_2^o = 1.21 \ V$
$MnO_4^- (aq) + 4H^+ (aq) + 3e^- \to MnO_2 (s) + 2H_2O (l)$; $E_3^o = ?$
Value of $E_3^o$ will be ............ $V$

When does a cell reaction occur spontaneously?

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: