The standard electrode potentials $\left( E_{M^{+}/M}^\circ \right)$ of four metals $A, B, C$ and $D$ are $-1.2 \ V, 0.6 \ V, 0.85 \ V$ and $-0.76 \ V$,respectively. The sequence of deposition of metals on applying potential is

For the given reactions:
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)

$K,$ $Ca,$ and $Li$ metals may be arranged in the increasing order of their standard electrode potentials as

When oxidation and reduction take place in a cell,its electromotive force $(EMF)$ will be

Consider the following reduction processes:
$Al^{3+} + 3e^{-} \rightarrow Al_{(s)}, E^{\circ} = -1.66 \ V$
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}, E^{\circ} = +0.77 \ V$
$Co^{3+} + e^{-} \rightarrow Co^{2+}, E^{\circ} = +1.81 \ V$
$Cr^{3+} + 3e^{-} \rightarrow Cr_{(s)}, E^{\circ} = -0.74 \ V$
The tendency to act as a reducing agent decreases in the order:

Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)