Question

electrode potential for the following half cell reactions are Zn gives zn2+ ;E°=+0.76V and Fe gives Fe2+ +2e- ; E°=+0.44V. The EMF for the cell reaction will be Fe2+ + Zn gives Zn2+ +Fe will be

Answer 1

Answer : The EMF for the cell reaction will be, -0.32 V

Solution :

The balanced cell reaction will be,  

$Zn(s)+Fe^{2+}(aq)\rightarrow Zn^{2+}(aq)+Fe(s)$

Here, zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Iron (Fe) undergoes reduction by gain of electrons and thus act as cathode.

Now we have to calculate the EMF for the cell reaction.

As we are given the standard oxidation cell potential. Thus, the reduction cell potential will be,

$E^o_{[Fe^{2+}/Fe]}=-0.44V$

$E^o_{[Zn^{2+}/Zn]}=-0.76V$

$E^o=E^o_{[Zn^{2+}/Zn]}-E^o_{[Fe^{2+}/Fe]}$

$E^o=(-0.76V)-(-0.44V)=-0.32V$

Therefore, the EMF for the cell reaction will be, -0.32 V

Answer 2

Answer:

The emf for the cell reaction is +0.32V

Explanation: