How does Fe change to `Fe^(2+)`?

The standard reduction potentials E^@ for the half reactions are as Zn=Zn^(2+)+2e, E^@=+0.76V Fe=Fe^(2+)+2e, E^@=+0.41V the emf of the cell reaction, Fe^(2+)+Zn=Zn^(2+)+Fe is

Fe forms two chlorides, FeCI_2 and FeCI_3 . Does Fe have the same value of equivalent weight in its compounds?

IF the following half cells have the E^@ values as Fe^(3+)+e to Fe^(2+), E^@=0.77Vand Fe^(2+)+2e to Fe, E^@=-0.44V the E^@ of the cell half cell Fe^(3+)+3e to Fe will be

Complete the following ionic equations : Fe^(2+)rarr Fe^(3+)+………

At equimolar concentration of Fe^(2+) and Fe^(3+) what must [Ag^(+)] be so that the voltage of the galvanic cell made from Ag^(+)//Ag and Fe^(3+)//Fe^(2+) electrodes equals zero? The reaction is Fe^(2+) + Ag^(+) hArr Fe^(3+) + Ag . Determine the equilibrium constant at 25^(@)C for the reaction. Given E_(Ag^(+)//Ag)^(@)= 0.799V and E_(Fe//Fe^(2+))^(@)= 0.771V

The standard potentials, E^@ for the half reaction are as Zn=Zn^(2+)+2e^(-) , E^@=0.76V Fe=Fe^(2+)+2e^(-) , E^@=0.41V The emf for the cell reaction Fe^(3+) +Zn to Zn^(2+)+Fe is

Write the half reaction for the following redox reaction. 2Fe^(3+)(aq) + 2l (aq) rarr I_2 (s) + 2Fe^(2+) (aq)