IChO 2010 · Theory problem 4

The lithium-ion battery

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The rechargeable lithium ion battery has been developed in Japan.

The standard electromotive force of the battery is 3.70 V. Assume that the half-reaction at the cathode is

Figure 1

and the half-reaction at the anode is

Figure 2

4.1 Write the total reaction equation of the battery and calculate the value of the standard Gibbs energy of the reaction [\(\mathrm{kJ\,mol^{-1}}\)].

The battery cell is constructed using \(\ce{LiCoO2}\) and graphite (\(\ce{C}\)) as the electrode materials.

4.2 Calculate the mass of the anode in the completely charged state and that in completely discharged state if 10.00 g of \(\ce{LiCoO2}\) and 10.00 g of graphite (\(\ce{C}\)) are present initially.

4.3 Calculate the maximum energy generated per mass of the lithium ion battery cell [\(\mathrm{kJ\,kg^{-1}}\)]. Assume that the correct ratio for complete reaction between the cathode and anode materials is used and the sum of the mass of electrodes is 50.0 % of the total mass of the battery cell. In comparison, the energy density of lead-acid batteries used for vehicles is about 200 \(\mathrm{kJ\,kg^{-1}}\).

Because an aqueous solution cannot be used as an electrolyte, an organic solution is used in the lithium ion battery cell.

4.4 Give the chemical formula of the gas generated if water is present in the electrolyte.

Official English text of the problem.

Answers

  • 4.1 \[ \boxed{\ce{CoO2 + LiC6 -> LiCoO2 + 6C}} \] \[ \boxed{\Delta_\mathrm rG^\circ=-357\ \mathrm{kJ\,mol^{-1}}} \]

  • 4.2 Completely charged anode: \(\boxed{10.71\ \mathrm g}\). Completely discharged anode: \(\boxed{10.00\ \mathrm g}\).

  • 4.3 \[ \boxed{1.05\times10^3\ \mathrm{kJ\,kg^{-1}}} \]

  • 4.4 Hydrogen gas (dihydrogen), \(\boxed{\ce{H2}}\); SMILES: [H][H].

Solution

For numerical evaluation, use \[ F=96\,485\ \mathrm{C\,mol^{-1}}, \] and atomic molar masses \[ M_{\mathrm{Li}}=6.94,\quad M_{\mathrm{Co}}=58.93,\quad M_{\mathrm O}=16.00,\quad M_{\mathrm C}=12.01 \qquad(\mathrm{g\,mol^{-1}}). \]

The electrode materials—cobalt(IV) oxide, \(\ce{CoO2}\); lithium cobalt(III) oxide, \(\ce{LiCoO2}\); lithiated graphite, \(\ce{LiC6}\); and graphite, \(\ce C\)—are extended solids. They have no discrete molecular structures to represent by molecular SMILES.

4.1

The given discharge half-reactions are \[ \begin{aligned} \text{Cathode:}\quad&\ce{CoO2 + Li+ + e- -> LiCoO2},\\ \text{Anode:}\quad&\ce{LiC6 -> 6C + Li+ + e-}. \end{aligned} \]

Adding them cancels the lithium ions, \(\ce{Li+}\) (SMILES: [Li+]), and electrons: \[ \boxed{\ce{CoO2 + LiC6 -> LiCoO2 + 6C}}. \]

One mole of electrons is transferred per mole of reaction as written, so \(n=1\). Hence \[ \begin{aligned} \Delta_\mathrm rG^\circ &=-nFE^\circ\\ &=-(1)(96\,485\ \mathrm{C\,mol^{-1}})(3.70\ \mathrm V)\\ &=-3.569945\times10^5\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-357\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The negative sign corresponds to spontaneous discharge under standard conditions.

4.2

Charging reverses the overall discharge reaction: \[ \ce{LiCoO2 + 6C -> CoO2 + LiC6}. \]

Thus lithium moves from the positive electrode into graphite. The amount available from the initial lithium cobalt oxide is \[ M(\ce{LiCoO2})=6.94+58.93+2(16.00) =97.87\ \mathrm{g\,mol^{-1}}, \] \[ n_{\mathrm{Li,available}} =\frac{10.00\ \mathrm g}{97.87\ \mathrm{g\,mol^{-1}}} =0.102176\ \mathrm{mol}. \]

The graphite can accommodate one lithium atom per six carbon atoms: \[ n_{\mathrm{Li,capacity}} =\frac{10.00\ \mathrm g}{6(12.01\ \mathrm{g\,mol^{-1}})} =0.138773\ \mathrm{mol}. \]

Since the graphite capacity exceeds the available lithium, all the lithium can transfer, with some graphite remaining unintercalated. The transferred mass is \[ m_{\mathrm{Li}} =(0.102176\ \mathrm{mol})(6.94\ \mathrm{g\,mol^{-1}}) =0.709104\ \mathrm g. \]

Therefore, the completely charged anode has mass \[ m_{\mathrm{charged}} =10.00+0.709104 =\boxed{10.71\ \mathrm g}. \]

Complete discharge transfers this lithium back to the cobalt oxide electrode, leaving the original graphite: \[ \boxed{m_{\mathrm{discharged}}=10.00\ \mathrm g}. \]

4.3

For maximum energy per mass, use the stoichiometric electrode ratio. One mole of reaction requires one mole of \(\ce{LiCoO2}\) and six moles of carbon in the discharged state. Their combined mass is \[ \begin{aligned} m_{\mathrm{electrodes}} &=97.87+6(12.01)\\ &=169.93\ \mathrm g. \end{aligned} \]

Charging redistributes lithium between the electrodes, so their combined mass is unchanged.

Because the electrodes constitute \(50.0\%\) of the cell mass, \[ m_{\mathrm{cell}} =\frac{169.93\ \mathrm g}{0.500} =339.86\ \mathrm g =0.33986\ \mathrm{kg}. \]

The maximum electrical energy delivered per mole of reaction is \[ W_{\max}=-\Delta_\mathrm rG^\circ =356.9945\ \mathrm{kJ}. \]

Consequently, \[ \begin{aligned} \frac{W_{\max}}{m_{\mathrm{cell}}} &=\frac{356.9945\ \mathrm{kJ}}{0.33986\ \mathrm{kg}}\\ &=\boxed{1.05\times10^3\ \mathrm{kJ\,kg^{-1}}}. \end{aligned} \]

This is approximately \(5.25\) times the stated lead-acid battery energy density.

4.4

The lithiated graphite electrode is strongly reducing. Water, \(\ce{H2O}\) (SMILES: O), can accept electrons and form hydrogen gas and hydroxide ions, \(\ce{OH-}\) (SMILES: [OH-]): \[ \ce{2H2O + 2e- -> H2 + 2OH-}. \]

Combining this with oxidation of lithiated graphite gives \[ \ce{2LiC6 + 2H2O -> 12C + 2Li+ + 2OH- + H2}. \]

Thus the gas is hydrogen (dihydrogen): \[ \boxed{\ce{H2}}, \] with SMILES [H][H].

The solution was checked against the official solution.