Consider the standard electrode potential values (M2+/M) of the elements of the first transition series. {:(Ti,,V,,Cr,,Mn,,Fe,,CO,,Ni,,Cu,,Zn),(-1.63,,-1.18,,-0.90,,-1.18,,-0.44,,-0.28,,-0.25,,+0.34,,-0.76):} Explain : (i) E^(@) value for copper is positive. (ii) E^(@) value of Mn is more negative as expected from the trend. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+).

Consider the standard electrode potential values (M2+/M) of the elemen

1.	Consider the standard electrode potential values (M2+/M) of the elements of the first transition series. {:(Ti,,V,,Cr,,Mn,,Fe,,CO,,Ni,,Cu,,Zn),(-1.63,,-1.18,,-0.90,,-1.18,,-0.44,,-0.28,,-0.25,,+0.34,,-0.76):} Explain : (i) E^(@) value for copper is positive. (ii) E^(@) value of Mn is more negative as expected from the trend. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+).
Answer» Solution :(i) HIGH ionisation enthalpy to convert Cu (s) to `Cu^(2+)` (aq) is not balanced by its hydration enthalpy. (ii) `Mn^(2+)` ion has stable `d^(5)` (half-filled) configuration. (iii) `d^(4)` to `d^(3)` occurs in the case of `Cr^(2+)` to `Cr^(3+)`, it is more stable. `d^(6)` to `d^(5)` occurs in case of `Fe^(2+)` to `Fe^(3+)`. Therefore `Cr^(2+)` is a STRONGER reducing agent than `Fe^(2+)`.