Adding powdered Pb and Fe to a solution containing 1.0 M each of `Pb^(2+)` and `Fe^(2+)` ions would result into the formation of:

Adding powdered Pb and Fe to a solution containing 1.0 M each of Pb^(+2) and Fe^(+2) ions would result in the formation of :

Adding powdered Cu and Zn to a solution containing 1.0 M is each of Cu^(​2+) and Zn^(​2+) ions would result into the formation of :-

Adding powdered lead and iron to a solution that is 1M in both Pb^(2+) and Fe^(2+) ions, would results to a reaction in which (E^@ Fe^(2+)//Fe = -0.44V, E^@Pb^(2+)//Pb = -0.13V)

An iron rod is immersed in a solution containing 1.0 M NiSO_(4) and 1.0 M ZnSO_(4) Predict giving reasons which of the following reactions is likely to proceed ? (i) Fe reduces Zn^(2+) ions, (ii) Iron reduces Ni^(2+) ions. Given : E_(Zn^(2+)| Zn )^(0) = -0.76 volt and , E_(Fe^(2+)|Fe)^(0) = -0.44 volt and E_(Ni^(2+)|Ni)^(0) = -0.25 V

An iron rod is immersed in a solution containing 1.0 M NiSO_(4) and 1.0 M ZnSO_(4) Predict giving reasons which of the following reactions is likely to proceed ? (i) Fe reduces Zn^(2+) ions, (ii) Iron reduces Ni^(2+) ions. Given : E_(Zn^(2+)| Zn )^(0) = -0.76 volt and , E_(Fe^(2+)|Fe)^(0) = -0.44 volt and E_(Ni^(2+)|Ni)^(0) = -0.25 V

Assertion (A): A solution contains 0.1M each of pB^(2+), Zn^(2+),Ni^(2+) , ions. If H_(2)S is passed into this solution at 25^(@)C . Pb^(2+), Ni^(2+), Zn^(2+) will get precpitated simultanously. Reason (R): Pb^(2+) and Zn^(2+) will get precipitated if the solution contains 0.1M HCI . [K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]

Assertion (A): A solution contains 0.1M each of pB^(2+), Zn^(2+),Ni^(2+) , ions. If H_(2)S is passed into this solution at 25^(@)C . Pb^(2+), Ni^(2+), Zn^(2+) will get precpitated simultanously. Reason (R): Pb^(2+) and Zn^(2+) will get precipitated if the solution contains 0.1M HCI . [K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]

The ratio of Fe^(3+) and Fe^(2+) ions in Fe_(0.9)9S_(1.0) is