$Cd(Hg).CdSO_4.\large\frac{8}{3}$$H_2O(s)\parallel CdSO_4(sat)Hg_2SO_4(s)Hg$ and cell reaction is $Cd(Hg)+Hg_2SO_4(s)+\large\frac{8}{3}$$H_2O(l)\rightarrow CdSO_4.\large\frac{8}{3}$$H_2O(s)+2Hg(l)$.EMF of cell is 1.1V at $25^oC$.Then $\Delta H^o$ will be
