Given the standard electrode potentials: $E^o_{Cr^{3+}/Cr} = -0.74 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$,$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,and $E^o_{Cl_2/Cl^-} = 1.36 \ V$. Based on this information,which is the strongest oxidizing agent?

Assertion : $Cu^{2+}$ ions get reduced more easily than $H^{+}$ ions.
Reason : Standard electrode potential of copper is $0.34 \ V$.

The standard reduction potentials at $298 \ K$ for the following half reactions are given against each:
$Zn^{2+}(aq.) + 2e^- \rightleftharpoons Zn_{(s)}$; $E^\circ = -0.762 \ V$
$Cr^{3+}(aq.) + 3e^- \rightleftharpoons Cr_{(s)}$; $E^\circ = -0.740 \ V$
$2H^{+}(aq.) + 2e^- \rightleftharpoons H_{2(g)}$; $E^\circ = 0.00 \ V$
$Fe^{3+}(aq.) + e^- \rightleftharpoons Fe^{2+}(aq.)$; $E^\circ = 0.770 \ V$
Which is the strongest reducing agent?

If $E^0_{Fe^{2+} / Fe} = -0.441 \ V$ and $E^0_{Fe^{3+} / Fe^{2+}} = 0.771 \ V$,the standard emf of the cell reaction $Fe_{(s)} + 2 Fe^{3+}_{(aq)} \longrightarrow 3 Fe^{2+}_{(aq)}$ is

Consider the following relations for $EMF$ of an electrochemical cell:
$(i)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Reduction potential of cathode)
$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode)
$(iii)$ $EMF$ of cell = (Reduction potential of anode) $+$ (Reduction potential of cathode)
$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode)
Which of the above relations are correct?

The standard electrode potential $(E^o)$ of $Cu^{2+}/Cu$ is $+0.34 \, V$,while that of $Zn^{2+}/Zn$ is $-0.76 \, V$. Explain the reason for this difference.