Hydrogen gas is not liberated when the following metal is added to dil $HCl$.

Based on the data given below: $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$. The strongest reducing agent is:

$Cr_2O_7^{2-} + I^{-} \to I_2 + Cr^{3+}$
$E^0_{cell} = 0.79 \ V$
$E^0_{Cr_2O_7^{2-}} = 1.33 \ V$,$E^0_{I_2}$ is ........... $V$

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 potentials of the half-reactions are given below:
$F_{2(g)} + 2e^- \rightarrow 2F^-_{(aq)}$; $E^o = +2.85 \ V$
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$; $E^o = +1.36 \ V$
$Br_{2(l)} + 2e^- \rightarrow 2Br^-_{(aq)}$; $E^o = +1.06 \ V$
$I_{2(s)} + 2e^- \rightarrow 2I^-_{(aq)}$; $E^o = +0.53 \ V$
The strongest oxidising and reducing agents respectively are:

The tendency of an electrode to lose electrons is known as