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

Predict reaction of 1 N sulhuric acid with the following metals : (i)

1.	Predict reaction of 1 N sulhuric acid with the following metals : (i) copper (ii) lead (iii) iron Given E_(Cu^(2+),Cu)^(@)=+0.34 "volt" ,E_(Pb^(2+),Pb)^(@)=-0.13 "volt" ,E_(Fe^(2+),Fe)^(@)=-0.44 volt
Answer» Solution :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