In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging, approximately \(50 \%\) of the \(\mathrm{Li}^{+}\) ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO}_{2}\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

To find the composition of the cathode when the battery is fully charged, we need to find out the formula of the cathode material after 50% of the Li+ ions have been extracted. Since the cathode is represented by the formula \(\mathrm{LiCoO}_{2}\), and upon charging, approximately 50% of the Li+ ions are extracted. Thus, the number of Li+ ions in the cathode composition when fully charged is 50% less than the initial number of ions, or 0.5 moles of Li+ ions per mole of \(\mathrm{LiCoO}_{2}\). So, the composition of the cathode when the battery is fully charged is \(\mathrm{Li_{0.5}CoO_{2}}\).

First, determine the moles of \(\mathrm{LiCoO}_{2}\) in the 10 g mass of the cathode. To do this, we need to find the molar mass of \(\mathrm{LiCoO}_{2}\). Molar mass of \(\mathrm{LiCoO}_{2} = \mathrm{Li} + \mathrm{Co} + 2\mathrm{O}\) Molar mass of \(\mathrm{LiCoO}_{2} = 6.94 \mathrm{~g/mol} + 58.93 \mathrm{~g/mol} + 2 \times 16.00 \mathrm{~g/mol} = 97.87 \mathrm{~g/mol}\) Next, find the moles of \(\mathrm{LiCoO}_{2}\): \(moles\ of\ \mathrm{LiCoO}_{2} = \dfrac{10\ \mathrm{g}}{97.87\ \mathrm{g/mol}} = 0.1022\ \mathrm{mol}\) Since each \(\mathrm{LiCoO}_{2}\) contains one Li+ ion, when the battery is fully charged and 50% of Li+ ions are extracted, there will be 0.5 times the original number of Li+ ions remaining in the cathode. Therefore, the number of Li+ ions extracted is equal to: \(0.5 \times 0.1022\ \mathrm{mol} = 0.0511\ \mathrm{mol}\) The charge produced by each Li+ ion is equal to its charge (1+), multiplied by the elementary charge (e): Charge produced by one Li+ ion = \(1 \times 1.602 \times 10^{-19}\ \mathrm{C}\) Total charge (coulombs) produced by 0.0511 mol of Li+ ions is: Total charge = \((0.0511\ \mathrm{mol})(6.022 \times 10^{23}\ \mathrm{ions/mol})(1 \times 1.602 \times 10^{-19}\ \mathrm{C}) = 4945\ \mathrm{C}\) So, a fully charged battery with a 10 g cathode can deliver approximately 4945 coulombs of electricity upon complete discharge.