The standard reduction electrode potentials of four elements are:
$A = -0.250 \ V$,$B = -0.136 \ V$,$C = -0.126 \ V$,$D = -0.402 \ V$
The element that displaces $A$ from its compounds is:

The reaction $Zn^{2+} + 2e^{-} \to Zn$ has a standard electrode potential of $-0.76 \ V$. This means:

Assume the cell reaction,$A_{(s)} + B_{(aq)}^{+2} \rightarrow A_{(aq)}^{+2} + B_{(s)}$. If $\Delta G^{\circ} = -386 \ kJ$ at $298 \ K$,what is $E_{\text{cell}}^{\circ}$ (in $V$)? (Assume $n = 2$)

The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

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)

Calculate the $emf$ of the cell $Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^+_{(aq)} | Ag_{(s)}$. Given: $E^0_{Cu^{2+}/Cu} = +0.34 \ V$,$E^0_{Ag^+/Ag} = +0.80 \ V$.