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)

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 incorrect oxidation number of the underlined atom in the following species is:

Among the following hydrohalic compounds, the best reductant is -

$1. \mathrm{HCl}$
$2. \mathrm{HBr}$
$3. \mathrm{HI}$
$4. \mathrm{HF}$

The compound AgF₂ (unstable) acts as a/ an:

1. Oxidising agent.

2. Reducing agent.

3. Both oxidising and reducing agent.

4. Neither oxidising and reducing agent.

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: