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

1.	Cu2O is -1	2.	Cl\(O_{3}^{-}\) is +5
3.	K2Cr2O7 is +6	4.	HAuCl4  is +3

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 $l^{-}$

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

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 correct statement(s) about the given reaction is -

\(\small\mathrm{{X} eO_{6 (aq)}^{4 -} + 2 F^{- 1}_{(aq)} + 6 H^{+}_{(aq )} \rightarrow}~\mathrm{{X} eO_{3 (g)} + {F}_{2 (g)} + 3 H_{2} O_{(l)}}\)${}_{}$

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 balanced equation for the reaction between chlorine and sulphur dioxide in water is:

Consider the given data:

${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{{\mathrm{I}}^{-}/{\mathrm{I}}_2}=$ $-0.54$
${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=0.80;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$
${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$
${\mathrm{E}}_{\mathrm{Ag}/{\mathrm{Ag}}^{+}}=-0.80;$ ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77$

Using the electrode potential values given above, identify the reaction which is not feasible:

Given below are two statements: