Hey there!:
From the given information , we conculde that the in the given cell .
* Fe (s) is oxidized and M⁺² ion is reduced, hence
* The cell reaction would be:
Fe(s) + M⁺² (aq) ---------> M (s) + Fe⁺²(aq)
* all are in 1.0 M concentrations
*Hence we conclude this cell is at the standard conditions.
Thus :
Eo(cell) = Eo(cathode) + Eo (anode)
from the standard data : Fe⁺² / Fe = -0.41 V
Therefore :
1.24 V = Eo(cathode) + 0.41 V
Eo(cathode) = 1.24 -0.41
Eo(cathode) = + 0.83 V
Hope that helps!