The two half-cell reactions of an electrochemical cell are given as: $Ag^{+} + e^{-} \rightarrow Ag$; $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$. The value of cell $EMF$ will be: (in $V$)

Given the standard reduction potentials at $25\,^oC$ for the following half-reactions:
$Zn^{2+}_{(aq)} + 2e^- \rightleftharpoons Zn_{(s)}, E^o = -0.76\,V$
$Cr^{3+}_{(aq)} + 3e^- \rightleftharpoons Cr_{(s)}, E^o = -0.74\,V$
$2H^+_{(aq)} + 2e^- \rightleftharpoons H_{2(g)}, E^o = 0.00\,V$
$Fe^{3+}_{(aq)} + e^- \rightleftharpoons Fe^{2+}_{(aq)}, E^o = +0.77\,V$
Which of the following is the strongest reducing agent?

The values of $E^0$ for metals $A$,$B$,and $C$ are $0.34 \ V$,$-0.80 \ V$,and $-0.46 \ V$ respectively. State the correct order for their ability to act as reducing agents.

Standard oxidation potentials $(SOP)$ of four metals $P, Q, R$ and $S$ are $+2.87 \ V, +3.05 \ V, -0.80 \ V$ and $+0.25 \ V$ respectively. The reducing power of these metals is:

Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be: