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:

1.	9 g	2.	15 g
3.	12 g	4.	11g

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 = ${\mathrm{ClO}}_3^{-}$

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

${}_{}$${\mathrm{Cl}}_2{\mathrm{O}}_{7\left(\mathrm{g}\right)}+4{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\to$
$2{\mathrm{ClO}}_{2\left(\mathrm{aq}\right)}^{-}+4{\mathrm{O}}_{2\left(\mathrm{g}\right)}+5{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$${}_{}$

1. Oxidizing agent = H₂O₂; Reducing agent = ${\mathrm{Cl}}_2{\mathrm{O}}_7$

2. Oxidizing agent = ${\mathrm{Cl}}_2{\mathrm{O}}_7$; Reducing agent = ${\mathrm{H}}_2{\mathrm{O}}_2$

3. Oxidizing agent = H₂O₂; Reducing agent = H₂O₂

4. None of the above

Given a galvanic cell with the following reaction

${\mathrm{Zn}}_{\left(\mathrm{s}\right) } + 2{{\mathrm{Ag}}^{+}}_{\left(\mathrm{aq}\right)} \to {{\mathrm{Zn}}^{+2}}_{\left(\mathrm{aq}\right)} + {\mathrm{Ag}}_{\left(\mathrm{s}\right)}$

The negatively charged electrode will be:

1. Zn

2. Ag

3. Both

4. None of  the above

The oxidation states of the central atom in the given species are, respectively:

${\mathrm{H}}_4{\mathrm{P}}_2{\mathrm{O}}_7$ $$ $\mathrm{and}$ ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7$