$${H_2}{S_2}{O_8};$$  \[H-O-\underset{\begin{smallmatrix} \parallel \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{S}}}\,-O-O-\underset{\begin{smallmatrix} \parallel \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{S}}}\,-O-H\]
Two oxygen atoms are involved in peroxide linkage, hence oxidation number will be -1 (each).
$${H_2}{S_2}{O_8}:2\mathop {\left( { + 1} \right)}\limits_H + 2\mathop {\left( x \right)}\limits_S + 2\mathop {\left( { - 1} \right)}\limits_{\left( {O - O} \right)} + 6\mathop {\left( { - 2} \right)}\limits_O $$        $$ = 0 \Rightarrow x = + 6$$

Skeletal equation is $$M{n^{3 + }} \to M{n^{2 + }} + Mn{O_2} + {H^ + }$$
Oxidation half equation : $$M{n^{3 + }} \to \mathop {Mn{O_2}}\limits^{ + 4} $$
or $$M{n^{3 + }} + 2{H_2}O \to $$     $$Mn{O_2} + 4{H^ + } + {e^ - }...\left( {\text{i}} \right)$$
Reduction half equation : $$M{n^{3 + }} \to M{n^{2 + }}$$
or $$M{n^{3 + }} + {e^ - } \to M{n^{2 + }}...\left( {{\text{ii}}} \right)$$
Adding eq. (i) and (ii) the balanced equation is $$2M{n^{3 + }} + 2{H_2}O \to $$    $$Mn{O_2} + M{n^{2 + }} + 4{H^ + }$$

$$\eqalign{ & Cr{O_2}\,C{l_2} \cr & {\text{Let }}O.{\text{ No}}{\text{. of }}Cr = x \cr & \therefore \,\,x + 2\left( { - 2} \right) + 2\left( { - 1} \right) = 0 \cr & x - 4 - 2 = 0 \cr & \therefore \,\,x = + 6 \cr} $$

On balancing the given reaction, we find
$$3N{a_2}HAs{O_3} + NaBr{O_3} + 6HCl$$       $$ \to 6NaCl + 3{H_3}As{O_4} + NaBr$$

$$\eqalign{ & {H_3}P{O_2}: + 3 + x - 4 = 0 \Rightarrow x = + 1 \cr & {H_3}P{O_4}: + 3 + x - 8 = 0 \Rightarrow x = + 5 \cr & P{H_3}:x + 3 = 0 \Rightarrow x = - 3 \cr & HP{O_3}: + 1 + x - 6 = 0 \Rightarrow x = + 5 \cr} $$

No explanation is given for this question. Let's discuss the answer together.

Based on the values of standard reduction potentials the sequence is followed.