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:

Given that the standard potentials $(E^{\circ})$ of $Cu^{2+}/Cu$ and $Cu^{+}/Cu$ are $0.34 \ V$ and $0.522 \ V$ respectively,the $E^{\circ}$ of $Cu^{2+}/Cu^{+}$ is (in $V$)

Standard reduction electrode potentials of three metals $A$,$B$,and $C$ are respectively $+0.5 \ V$,$-3.0 \ V$,and $-1.2 \ V$. The reducing powers of these metals are

If $E^o_{A^{+2}/A} = -0.30 \ V$ and $E^o_{A^{+3}/A^{+2}} = 0.40 \ V$,the standard $EMF$ of the reaction: $A + 2A^{+3} \to 3A^{+2}$ will be ............ $V$.

Calculate the standard cell potential (in $V$) of the cell in which the following reaction takes place:
$Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \to Fe^{3+}_{(aq)} + Ag_{(s)}$
Given that:
$E^o_{Ag^{+}/Ag} = x \ V$
$E^o_{Fe^{2+}/Fe} = y \ V$
$E^o_{Fe^{3+}/Fe} = z \ V$

The standard electrode potential for the half-cell reactions are
$Zn^{2+} + 2e^{-} \longrightarrow Zn ; E^{\circ} = -0.76 \ V$
$Fe^{2+} + 2e^{-} \longrightarrow Fe ; E^{\circ} = -0.44 \ V$
The $emf$ of the cell reaction,
$Fe^{2+} + Zn \longrightarrow Zn^{2+} + Fe$ is