Predict reaction of 1N sulphuric acid with following metals : (i) copper (ii) lead (iii) iron Given, `E_(Cu^(2+)|Cu )^(0)`= 0.34volt , `E_(Pb^(2+)|Pb)^(0)` = -0.13 volt, `E_(Fe^(2+)|Fe)^(0)` = -0.44 volt

Answer

(i)No (ii) Yes (iii) Yes

Explanation

If the metal is to react with dil `H_(2)SO_(4) i.e( H^(+)` ions) to produce `H_(2)` gas the metal should have a lowere electrode potential than that of standard hydrogen electrode i.e 0.0 V
(i) since `E_(Cu^(2+),Cu)^(@)=0.34V` is higher than `E_(H^(+),H_(2)^(@)=0.0V` therefore Cu will not react with `1NH_(2)SO_(4)` to produce `H_(2)` gas
(ii)Sinc `E_(pb^(2+),pb)^(@)=-0.13V` and `E_(H^(+),H_(2)^(@)=0.0` therefore lead will react with `1N H_(2)SO_(4)` to produce `H_(2)` gas