For the half-reactions $Zn \rightarrow Zn^{2+} + 2e^{-}$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$,the standard oxidation potentials are given as $E^{0}_{Oxi} = +0.76 \ V$ and $E^{0}_{Oxi} = +0.41 \ V$ respectively. Calculate the cell potential for the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ in $V$.

Calculate $E_{cell}^{\circ}$ for the following cell: $Zn_{(s)} | Zn^{2+}_{(1 \ M)} || Pb^{2+}_{(1 \ M)} | Pb_{(s)}$ given that $E^{\circ}_{Zn^{2+}/Zn} = -0.763 \ V$ and $E^{\circ}_{Pb^{2+}/Pb} = -0.126 \ V$. (in $V$)

For the cell reaction,$2 Al_{(s)} + 3 Cu^{2+}_{(aq)} \rightarrow 2 Al^{3+}_{(aq)} + 3 Cu_{(s)}$. If $\Delta G^{\circ} = -1158 \ kJ$,what is $E^{\circ}_{cell}$ (in $V$)?

In a cell,the following reactions take place:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)

The standard electrode potentials $(E^o)$ for $OCl^{-}/Cl^{-}$ and $\frac{1}{2}Cl_2/Cl^{-}$ are $0.94 \ V$ and $+1.36 \ V$ respectively,the $E^o$ value for $OCl^{-}/\frac{1}{2}Cl_2$ will be ........... $V$.

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: