Given ELi+/Lio = −3.05V and EAl+3/Alo = −1.66V then the emf of the cell Lis /Li+1M//Al+31M/Als will be
LHE … Anode
RHE … Cathode
∴ Ecell0 = ERHE0−ELHE0
Ecello = −1.66V−−3.05V
Ecello = −1.66V+3.05V
Ecello = +1.39V