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

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Detailed Solution

LHE … AnodeRHE … Cathode∴ Ecell0 = ERHE0−ELHE0Ecello = −1.66V−−3.05VEcello = −1.66V+3.05VEcello = +1.39V

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