Hydrogen is the first element in the perodic table and is also the highest element known. Molecular hydrogen is best prepared by the electrolysis of water containing a small amount of acid or alkali. It is also a constituent of water gas (CO+H_(2)) known as syn gas. Hydrogen is releatively inert at room temperatrure due very high bond dissociatin enthalpy (430.0kJ mol^(-1)) of H-H bond. (i) Why does hydrogen act as a powerful reducing agent? (ii) How does hydrogen cause of unsaturatedhydrocarbons? (iii) Hydrogen- oxygen fuel are used for generating electrical energy. How are those cells better than the conventional fuels?

Hydrogen is the first element in the perodic table and is also the hig

1.	Hydrogen is the first element in the perodic table and is also the highest element known. Molecular hydrogen is best prepared by the electrolysis of water containing a small amount of acid or alkali. It is also a constituent of water gas (CO+H_(2)) known as syn gas. Hydrogen is releatively inert at room temperatrure due very high bond dissociatin enthalpy (430.0kJ mol^(-1)) of H-H bond. (i) Why does hydrogen act as a powerful reducing agent? (ii) How does hydrogen cause of unsaturatedhydrocarbons? (iii) Hydrogen- oxygen fuel are used for generating electrical energy. How are those cells better than the conventional fuels?
Answer» Solution :(i) Hydrogen atom(H) can part the only with only electron which it has and therefore. Acts as a reducing agent. (ii) Molecular hydrogen adds on the compounds having multiple bonds `(C-=C and C=C "bonds")` and converts them into SATURATED hydrocarbons. `underset("Ethyne")(HC)-=CH+2H_(2) overset(Ni//573K) to underset("Ethyne")(CH_(3))-CH_(3)` (III) The followingelectrode REACTIONS take place in hydrogen oxygen fuel cell At anode:`2H_(2)(g)+4OH^(-)(aq) to 4H_(2)O(l)+4e^(-)` At cathode: `O_(2)(g)+2H_(2)O(l)+4e^(-) to OH^(-)(aq)` `2H_(2)(g)+O_(2)(g) to 2H_(2)O(l)` Step III Calcination of volume strength of `H_(2)O_(2)` `5 cm^(2) of H_(2)O_(2) "solution gives"=22.4cm^(3) "of oxygen at N.T.P."` `1cm^(3) "of" H_(2)O_(2) "solution will give"=(22.4)/(5)=4.48cm^(3) "of oxygen at N.T.P."` `"Volume strength of" H_(2)O_(2)=4.48`