Hydrogen can be prepared by the action of dil. `H_(2)SO_(4)` on

Options

A. copper
B. iron
C. lead
D. mercury

Answer

B

Explanation

`Cu(E_(Cu^(2+)//Cu)^(@)=+0.34 V)` and `Hg(E_(Hg^(2+)//Hg)^(@)=+0.80 V)`
lie above `H_(2)` in the electrochemical series and hence do not react with dil. `H_(2)SO_(4)` to evolve `H_(2)` gas . Although `Pb(E_(Pb^(2+)//Pb)^(@)=-0.13V)` lies below `H_(2)` in the electrochemical series, it reacts with dil. `H_(2)SO_(4)` in the begining to evolves `H_(2)` gas but `PbSO_(4)` produced during the reaction forms a
`Pb+ H_(2)SO_(4)("dil.") to underset("In soluble")(PbSO_(4))(s) + H_(2)(g)` protective layer around the unreacted Pb and the reaction stops. Only iron `(E_(Fe^(2+)//Fe)^(@)==0.44V)` reacts readily with dil. `H_(2)SO_(4)` to evolve `H_(2)` gas.