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. 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:

${\mathrm{Zn}}_{\left(\mathrm{s}\right)<mspace\; width="1em"></mspace>}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2{{\mathrm{Ag}}^{+}}_{\left(\mathrm{aq}\right)}<mspace\; width="1em"></mspace>\to {{\mathrm{Zn}}^{+2}}_{\left(\mathrm{aq}\right)}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{2Ag}}_{\left(\mathrm{s}\right)}<mspace\; width="1em"></mspace>$

Which electrode will be negatively charged?

1. Zn

2. Ag

3. Both Zn and Ag

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