The reduction potential ($E^{\circ}$,in $V$) of $MnO_4^-{_{\text{(aq)}}} / Mn_{\text{(s)}}$ is. . . . . $[Given : E^{\circ}_{(MnO_4^{-(aq)} / MnO_{2(s)})} = 1.68 \ V ; E^{\circ}_{(MnO_{2(s)} / Mn^{2+}_{(aq)})} = 1.21 \ V ; E^{\circ}_{(Mn^{2+(aq)} / Mn_{(s)})} = -1.03 \ V]$

What is the procedure to determine the cell potential?

The standard reduction potential for $Fe^{2+}/Fe$ and $Sn^{2+}/Sn$ electrodes are $-0.44 \ V$ and $-0.14 \ V$ respectively. For the given cell reaction $Fe^{2+} + Sn \to Fe + Sn^{2+}$,the standard $EMF$ is ...............$V$.

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

For the reduction of $NO_3^{-}$ ion in an aqueous solution,$E^0$ is $+0.96 \ V$. Values of $E^0$ for some metal ions are given below:
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V \ \ \ \ E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe \ \ \ \ E^0 = -0.04 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au \ \ \ \ E^0 = +1.40 \ V$
$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg \ \ \ \ E^0 = +0.86 \ V$
The pair$(s)$ of metals that is(are) oxidized by $NO_3^{-}$ in aqueous solution is(are):
$(A) V$ and $Hg$
$(B) Hg$ and $Fe$
$(C) Fe$ and $Au$
$(D) Fe$ and $V$

Which among the following reactions will take place spontaneously?