Hydrogen resembles halogens as it has similar tendency to gain a single electron to attain stable noble gas configuration.

\[2Al+2NaOH+2{{H}_{2}}O\xrightarrow{\Delta }\]      \[2NaAl{{O}_{2}}+3{{H}_{2}}\]
\[Sn+2NaOH+{{H}_{2}}O\xrightarrow{\Delta }\]      \[N{{a}_{2}}Sn{{O}_{3}}+2{{H}_{2}}\]
\[Zn+2NaOH\xrightarrow{\Delta }\]    \[N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}}\]

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$$S{O_2}$$  changes to $${H_2}S{O_4}$$  ( $$O.N.$$  changes from $$+4$$  to $$+6$$  oxidation )
$$2KI \to {I_2}$$   ( $$O.S.$$  changes from $$–1$$  to $$0$$ oxidation )
$$PbS \to PbS{O_4}$$   ( $$O.S.$$  changes from $$–2$$  to $$+6$$  oxidation )
$$A{g_2}O \to 2Ag$$   ( $$O.S.$$  changes from $$+1$$  to $$0$$ reduction )

Very pure hydrogen can be prepared by the action of water on sodium hydride.
$$NaH + {H_2}O \to \mathop {NaOH + {H_2}}\limits_{\left( {{\text{very}}\,{\text{pure}}\,{\text{Hydrogen}}} \right)} $$

$$Ba{O_2} \cdot 8{H_2}{O_{\left( s \right)}} + {H_2}S{O_{4\left( {aq} \right)}} \to $$       $$BaS{O_{4\left( s \right)}} + {H_2}{O_{2\left( {aq} \right)}} + 8{H_2}{O_{\left( l \right)}}$$

$${H_2}{O_2}$$  decomposes slowly on exposure to light.
$$2{H_2}{O_2}_{\left( l \right)} \to 2{H_2}{O_{\left( l \right)}} + {O_{2\left( g \right)}}$$
In the presence of metal surfaces or traces of alkali ( present in glass containers ), the above reaction is catalysed. It is, therefore, stored in wax-lined glass or plastic vessels in dark. Urea can be added as a stabiliser. It is kept away from dust because dust can induce explosive decomposition of the compound.

Potash alum is used for purifying water.

$${H_2}O$$  acts as a Bronsted acid and gives a proton to react with a base.
$$\mathop {{H_2}O}\limits_{{\text{acid}}} + \mathop {N{H_3}}\limits_{{\text{base}}} \rightleftharpoons O{H^ - } + NH_4^ + $$
In $$\mathop {{H_2}O}\limits_{{\text{base}}} + \mathop {{H_2}S}\limits_{{\text{acid}}} \to {H_3}{O^ + } + H{S^ - },{H_2}O$$        acts as a base with $${H_2}S.$$

$$C{a^{2 + }} + 2RH \to {R_2}Ca + 2{H^ + }$$
$${10^6}\,mL\,\,{\text{of}}\,\,{H_2}O\,\,{\text{has}}\,\,{\text{77}}{\text{.5}}\,{\text{g}}\,C{a^{2 + }}$$       $$ = \frac{{77.5}}{{40}}\,mole\,\,C{a^{2 + }}$$
$$ \Rightarrow 2 \times \frac{{77.5}}{{40}}\,mole\,{H^ + } = $$      $$3.875\,mole\,{H^ + }$$
$$\therefore {10^3}\,mL\,\,{\text{of}}\,\,{H_2}O\,\,{\text{has}}\,\,{H^ + } = $$       $$3.875 \times {10^{ - 3}}\,M$$
$$\therefore pH = - \log \left[ {{H^ + }} \right] = 2.41$$