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

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)

Fluorine reacts with ice as per the following reaction

H₂O(s) + F₂(g) → HF(g) + HOF(g)

This reaction is a redox reaction because-

The oxidation number of sulphur and nitrogen in H₂SO₅ and NO₃^- are respectively-

The formulas for the following compounds are: 

(a) Mercury(II) chloride  and (b) Thallium(I) sulphate

1. HgCl₂,  Tl₂SO₄

2. Hg₂Cl₂, Tl₂SO₄

3. HgCl₂,  TlSO₄

4. HgCl₂,  Tl₃SO₄

The correct statement(s) about the given reaction is -

$\stackrel{}{\mathrm{X}}{\mathrm{eO}}_{6\left(\mathrm{aq}\right)}^{4-}+2{{\mathrm{F}}^{-1}}_{\left(\mathrm{aq}\right)}+6{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}\to$
$\stackrel{}{\mathrm{X}}{\mathrm{eO}}_{3\left(\mathrm{g}\right)}+{\stackrel{}{\mathrm{F}}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

1. ${\mathrm{XeO}}_6^{4-}$ oxidises F^-

2. The oxidation number of F increases from -1  to  zero

3. ${\mathrm{XeO}}_6^{4-}$ is a stronger oxidizing agent that ${\mathrm{F}}^{-}$

4. All of the above.

The oxidising agent and reducing agent in the given reaction are

$5{\mathrm{P}}_{4\left(\mathrm{s}\right)}+12{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+12{{\mathrm{HO}}^{-}}_{\left(\mathrm{aq}\right)}\to$
$8{\mathrm{PH}}_{3\left(\mathrm{g}\right)}+12{{\mathrm{HPO}}_2^{-}}_{\left(\mathrm{aq}\right)}$

1. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{P}}_4$

2. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{H}}_2\mathrm{O}$

3. Oxidising agent = ${\mathrm{H}}_2\mathrm{O}$; Reducing agent = ${\mathrm{P}}_4$

4. None of the above

The correct statement about the given reaction is-

(CN)_2(g) + 2OH^-_(aq) $\mathbf{\to }$CN^-_(aq) + CNO^-_(aq) + H₂O_(l)

The ${\mathrm{Mn}}^{3+}$ ion is unstable in solution and undergoes disproportionation reaction to give ${\mathrm{Mn}}^{2+},$ ${\mathrm{MnO}}_2$ $\mathrm{and}$ ${\mathrm{H}}^{+}$ ion. The balanced ionic equation for the reaction is-