The answer is:
E per gram = 0.45 V
The explanation:
when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.
and when the molar mass of Li is = 7 g/mol
and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy
that means that :
7 g of Li gives → 3.15 V
So 1 g of Li will give→ ???
∴ The E per gram = 3.15 V / 7 g of Li
= 0.45 V
