$$O.N.$$  of $$C$$ changes form 0 to + 4 by oxidation.
Hence $$HN{O_3}$$  is oxidising agent.

Calculating the oxidation state of nitrogen in given molecules;
Oxidation state of $$N$$ in $$N{H_3}$$  is
$$x + 3 \times \left( { + 1} \right) = 0\,\,{\text{or}}\,\,x = - 3$$
Oxidation state on $$N$$ in $$HN{O_3}$$  is
$$1 + x + 3 \times \left( { - 2} \right) = 0\,\,{\text{or}}\,\,x = + 5$$
Oxidation state of $$N$$ in $$Na{N_3}$$  is
$$ + 1 + 3x = 0\,\,{\text{or}}\,x = - \frac{1}{3}$$
Oxidation state of $$N$$ in $$M{g_3}{N_2}$$  is
$$3 \times 2 + 2x = 0\,\,{\text{or}}\,\,x = - 3$$
Thus 3 molecules ( i.e. $$N{H_3},Na{N_3}$$   and $$M{g_3}{N_2}$$  have nitrogen in negative oxidation states.

$$\eqalign{ & {\text{In}}\,\,SO_3^{2 - } \cr & x + 3\left( { - 2} \right) = - 2;x = + 4 \cr & {\text{In}}\,\,{S_2}O_4^{2 - } \cr & 2x + 4\left( { - 2} \right) = - 2 \cr & 2x - 8 = - 2 \cr & 2x = 6;\,x = + 3 \cr & {\text{In}}\,\,{S_2}O_6^{2 - } \cr & 2x + 6\left( { - 2} \right) = - 2 \cr & 2x = 10;\,x = + 5 \cr & {\text{hence the correct order is}} \cr & {S_2}O_4^{2 - } < SO_3^{2 - } < {S_2}O_6^{2 - } \cr} $$

Standard electrode potential i.e. reduction potential of $$A$$ is minimum $$(–3.05V)$$   i.e. its oxidation potential is maximum which implies $$'A'$$  is most reactive chemically.

The balanced equation is
$${K_2}C{r_2}{O_7} + 4KCl + 6{H_2}S{O_4} \to $$       $$2Cr{O_2}C{l_2} + 6{K_2}S{O_4} + 3{H_2}O$$

$$\mathop {3{N_2}{H_4}}\limits^{ - 2} + \mathop {2BrO_3^ - }\limits^{ + 5} \to $$     $$3\mathop {{N_2}}\limits^0 + \mathop {2B{r^ - }}\limits^{ - 1} + 6{H_2}O$$

Oxidation number of oxygen is -2 in most of its compounds. The exceptions are -1 in peroxides and $$ - \frac{1}{2}$$  in superoxides, +2 in $$O{F_2}$$  and +1 in $${O_2}{F_2}.$$

For the reaction, $$2F{e^{3 + }} + 2{I^ - } \to 2F{e^{2 + }} + {I_2}$$
$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ - E_{\frac{{{I_2}}}{{{I^ - }}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.77 - \left( {0.54} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = + 0.23\,V \cr} $$
Here, $$E_{{\text{cell}}}^ \circ $$  is $$ + ve$$  so, reaction is feasible.
For the reaction, $$Cu + 2A{g^ + } \to C{u^{2 + }} + 2Ag$$
$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{\frac{{A{g^ + }}}{{Ag}}}^ \circ - E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.80 - \left( {0.34} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = + 0.46\,V \cr} $$
Here, $$E_{{\text{cell}}}^ \circ $$  is $$ + ve$$  so, the reaction is feasible.
For the reaction, $$2F{e^{3 + }} + Cu \to 2F{e^{2 + }} + C{u^{2 + }}$$
$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ - E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.77 - \left( {0.34} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = + 0.43\,V \cr} $$
Here, $$E_{{\text{cell}}}^ \circ $$  is $$ + ve$$  so, the reaction is feasible.
For the reaction, $$Ag + F{e^{3 + }} \to A{g^ + } + F{e^{2 + }}$$
$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ - E_{\frac{{A{g^ + }}}{{Ag}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.77 - \left( {0.80} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - 0.03\,V \cr} $$
Here, $$E_{{\text{cell}}}^ \circ $$  is negative so, the reaction is not feasible.

$$\eqalign{ & C{l_2} + O{H^ - } \to C{l^ - } + ClO_4^ - + {H_2}O \cr & \mathop {C{l_2}}\limits^0 \to \mathop {C{l^ - }}\limits^{ - 1} \,{\text{(Reduction)}} \cr & \mathop {C{l_2}}\limits^0 \to \mathop {ClO_4^ - }\limits^{ + 7} \,{\text{(Oxidation)}} \cr} $$