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:

The set of metals that can show disproportionation reaction is:
1. Cu, Na, Li
2. Mg, F, Ne
3. P, Cl, S
4. Mn, Cu, Ga

The maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen 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:

Arrange the metals Al, Cu, Fe, Mg, and Zn in the correct decreasing order based on their ability to displace each other from the solution of their salts:
1. Al> Zn > Fe > Cu > Mg

2. Zn > Fe > Cu > Mg > Al

3. Mg > Al > Zn > Fe > Cu

4. Fe > Mg > Zn > Al > Cu

The correct statement about the electrolysis of an aqueous solution of ${\mathrm{AgNO}}_3$ with Ag electrode is:

The reactions of electrolysis for a dilute solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with platinum electrode are:
1. H⁺ ions reduce at cathode; H₂O oxidised at anode
2. H⁺ ions reduce at cathode; SO₄^2-​ oxidised at anode
3. H₂O reduces at cathode; H₂O oxidised at anode
4. H₂O reduces at cathode; H⁺ ions​​​ oxidised at anode

The correct statement about electrolysis of an aqueous solution of ${\mathrm{CuCl}}_2$ with Pt electrode is-

The oxidizing agent and reducing agent in the given reaction are :

$3{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{l}\right)}+4{\mathrm{ClO}}_{3\left(\mathrm{aq}\right)}^{-}\to$
$6{\mathrm{NO}}_{\left(\mathrm{g}\right)}+4{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

1. Oxidising agent = ${\mathrm{N}}_2{\mathrm{H}}_4$; Reducing agent = ${\mathrm{ClO}}_3^{-}$

2. Oxidising agent = ${\mathrm{ClO}}_3^{-}$; Reducing agent = ${\mathrm{N}}_2{\mathrm{H}}_4$

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

4. Oxidising agent = ${\mathrm{ClO}}_3^{-}$ ; Reducing agent = ${\mathrm{ClO}}_3^{-}$