Answer :
(a) The anode and cathode will be 
and 
respectively.
(b) The emf of cell potential is 1.022 V
Explanation :
(a) The standard reduction potentials for iron and copper are:
In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
From the standard reduction potentials we conclude that, the substance having highest positive 
potential will always get reduced and will undergo reduction reaction.
So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.
The given cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
Thus, the anode and cathode will be
and respectively.
(b) Now we have to calculate the potential of a cell.
Oxidation half reaction (anode):
Reduction half reaction (cathode):
In order to balance that electrons, we will multiple the reduction reaction by 2, we get:
Oxidation half reaction (anode):
Reduction half reaction (cathode): 
The overall cell reaction will be,
![E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BFe%5E%7B3%2B%7D%2FFe%5E%7B2%2B%7D%5D%7D%3D2%5Ctimes%200.77V%3D1.54V)
![E^o_{[Cu^{2+}/Cu]}=0.34V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D0.34V)
To calculate the 
of the reaction, we use the equation:
![E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5Eo%3DE%5Eo_%7B%5BFe%5E%7B3%2B%7D%2FFe%5E%7B2%2B%7D%5D%7D-E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the cell potential.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BFe%5E%7B2%2B%7D%5D%5E2%5BCu%5E%7B2%2B%7D%5D%7D%7B%5BFe%5E%7B3%2B%7D%5D%5E2%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= emf of the cell = ?
Now put all the given values in the above equation, we get:
Therefore, the emf of cell potential is 1.022 V
