Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements is: 

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.

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 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 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-

Which element exhibits both positive and negative oxidation states?
1. Cs
2. Ne
3. I
4. F

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