If the half-cell reaction $A + e^- \to A^-$ has a large negative reduction potential,it follows that

The $E^{\ominus}$ for $Cl_{2}/Cl^{-}$ is $+1.36 \ V,$ for $I_{2}/I^{-}$ is $+0.53 \ V,$ for $Ag^{+}/Ag$ is $+0.79 \ V,$ $Na^{+}/Na$ is $-2.71 \ V$ and for $Li^{+}/Li$ is $-3.04 \ V.$ Arrange the following ionic species in decreasing order of reducing strength: $I^{-}, Ag, Cl^{-}, Li, Na$

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 $V$

For the reaction $M^{+n}_{(aq)} + ne^{-} \rightarrow M_{(s)}$,if the standard reduction potentials of elements $M_1, M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively,what is the order of their reducing strength?

Consider the following ${E^0}$ values:
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.

Can the absolute electrode potential of an electrode be measured?