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-

1.	\(\small{2 \mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}}\)
2.	Mn3+(aq) + H2O(l) → MnO2(s) + 2Mn2+(aq) + 4H+(aq)
3.	\(\small{5 \mathrm{Mn}^{3+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{MnO}_{2(\mathrm{s})}+3 \mathrm{Mn}^{2+}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})}\)
4.	\(\small{2 \mathrm{Mn}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{MnO}_{2(\mathrm{s})}+2 \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}^{+}{ }_{(\mathrm{aq})}}\)

Which element exhibits both positive and negative oxidation states?

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

The set of metals that can show disproportionation reaction is -

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:

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

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

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