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:

1.	Fe3+(aq) and I- aq)
2.	Ag+(aq) and Cu(s)
3.	Fe3+(aq) and Cu(s)
4.	Ag(s) and Fe3+(aq)

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. ${\mathrm{H}}^{+} \mathrm{ion} \mathrm{reduce} \mathrm{at} \mathrm{cathode}; {\mathrm{H}}_2\mathrm{O} \mathrm{oxidised} \mathrm{at} \mathrm{anode}$

2. ${\mathrm{H}}^{+} \mathrm{ion} \mathrm{reduce} \mathrm{at} \mathrm{cathode}; {\mathrm{SO}}_4^{2-} \mathrm{ion} \mathrm{oxidised} \mathrm{at} \mathrm{anode}$

3. ${\mathrm{H}}_2\mathrm{O} \mathrm{ion} \mathrm{reduce} \mathrm{at} \mathrm{cathode}; {\mathrm{H}}_2\mathrm{O} \mathrm{oxidised} \mathrm{at} \mathrm{anode}$

4. ${\mathrm{H}}_2\mathrm{O} \mathrm{ion} \mathrm{reduce} \mathrm{at} \mathrm{cathode}; {\mathrm{H}}^{+} \mathrm{ion} \mathrm{oxidised} \mathrm{at} \mathrm{anode}$

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:

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

KI₃, H₂S₄O₆

The oxidation numbers of iodine and sulphur in the above compounds are, respectively:

1. \(\frac{1}{3}\) ; 4
2. 2.5 ; \(\frac{1}{3}\)
3. \(-\frac{1}{3}\) ; 2.5
4. 2.5 ; 3