Given: $E^{0}_{Mn^{+7} \mid Mn^{+2}} = 1.5 \ V$ and $E^{0}_{Mn^{+4} \mid Mn^{+2}} = 1.2 \ V$,then $E^{0}_{Mn^{+7} \mid Mn^{+4}}$ is (in $V$)

The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

Identify which of the following cell reactions is spontaneous under standard state conditions.

Electrode potentials of five elements $A, B, C, D$ and $E$ are respectively $-1.36 \ V, -0.32 \ V, 0 \ V, -1.26 \ V$ and $-0.42 \ V$. The reactivity order of these elements is:

$A$ solution contains $Fe^{2+}$,$Fe^{3+}$ and $I^{-}$ ions. This solution was treated with iodine at $35^{\circ}C$. $E^{\circ}$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and $E^{\circ}$ for $I_2/2I^{-}$ is $+0.536 \ V$. The favourable redox reaction is

The oxidation potentials of the following half-cell reactions are given:
$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$
What will be the $EMF$ of the cell whose cell reaction is:
$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$