Standard reduction potentials of the half-reactions are given below: 

$F_{2\left(g\right)}$ $+$ $2e^{-}\to 2{F^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+2.85$ $V$ $$

$C{l}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{C{l}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.36$ $V$ $$

$B{r}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{B{r}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.06$ $V$ $$

$I_{2\left(g\right)}$ $+$ $e^{-}\to 2{I^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+0.53$ $V$ $$

The strongest oxidizing and reducing agents, respectively, are: 

1. $B{r}_2$ and $C{l}^{-}$

2. $C{l}_2$ and $B{r}^{-}$

3. $C{l}_2$ and $I_2$

4. $F_2$ and $l^{-}$

A solution contains Fe²⁺, Fe³⁺ and I^– ions. This solution was treated with iodine at 35^°C. E° for Fe³⁺/Fe²⁺ is +0.77 V and E° for I₂/2I^– = 0.536 V.
The favourable redox reaction is:

1. Fe²⁺ will be oxidized to Fe³⁺.

2. I₂ will be reduced to I^–.

3. There will be no redox reaction.

4. I^– will be oxidized to I₂.

The compound that contains zero oxidation state of Fe is:

Best description of the behavior of bromine in the reaction given below is:

${\mathrm{H}}_2\mathrm{O}+{\mathrm{Br}}_2\to \mathrm{HOBr}+\mathrm{HBr}$

1. Both oxidized and reduced

2. Oxidized only

3. Reduced only

4. Proton acceptor only

If the oxidation numbers of A, B, and C are + 2, +5, and –2 respectively, then the possible formula of the compound is:

A non-feasible reaction among the following is:

1. $2$ $\mathrm{KI}+{\mathrm{Br}}_2\to 2\mathrm{KBr}+{\mathrm{I}}_2$

2. $2$ $\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$

3. $2$ $\mathrm{KBr}+{\mathrm{Cl}}_2\to 2\mathrm{KCl}+{\mathrm{Br}}_2$

4. $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{F}}_2\to 4\mathrm{HF}+{\mathrm{O}}_2$

Standard electrode potentials are:

$F{e}^{+2}/Fe$ ,$E°$ $=$ $-0.\mathrm{44 }$

$\mathrm{ }F{e}^{+3}/F{e}^{+2} ,E°$ $=0.77$

Choose the correct observation if $F{e}^{+2},$ $F{e}^{+3}$, and Fe  block are kept together:

1. $F{e}^{+3}$ increases 

2. $F{e}^{+3}$ decreases 

3. $\frac{F{e}^{+2}}{F{e}^{+3}}$ remains unchanged 

4. $F{e}^{+2}$ decreases

The oxidation states(O.S.) of sulphur in the anions ${{\mathrm{SO}}_3}^{2-},$ ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}$ $\mathrm{and}$ ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$ follow the order:

1. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

2. ${{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

3. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{{\mathrm{SO}}_3}^{2-}$

4. ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}$

From the following, identify the reaction having the top position in the EMF series (standard reduction potential) according to their electrode potential at 298 K.

1. Mg²⁺+ 2e^–$\to$Mg_(s)
2. Fe²⁺ + 2e^–$\to$ Fe_(s)
3. Au³⁺+ 3e^–$\to$Au_(s)
4. K⁺+ le ^–$\to$K_(s)