What is the balanced chemical equation for a redox reaction between a lithium-cobalt battery and a graphite electrode?

Question

I'm doing a project for my last year of secondary school on lithium cobalt batteries and got confused about the equation.

This is the only equation I can find (source: www.science.org.au):

\begin{align} \ce{LiC6 &→ $x$ Li+ + $x$ e- + C6}\tag{anode}\\ \ce{Li_{(1−x)}CoO2 + $x$ Li+ + $x$ e- &→ LiCoO2}\tag{cathode}\\ \hline \ce{C6 + LiCoO2 &→ Li_xC6 + Li_{(1−x)}CoO2}\\ \end{align}

But I don't understand it at all! I figured it must be balanced, but the '$x$' part of the equation is beyond my level of understanding. Is there a simpler way of writing this formula for someone at my level? If not, can someone please explain what this equation means?

Answer 1

Below we have a simpler way of the $\ce{Li-Co}$ cell reactions.

Cathode half-reaction:

$$\ce{CoO2 + Li+ + e- <=> LiCoO2}$$

Anode half-reaction:

$$\ce{LiC6 <=> C6 + Li+ + e-}$$

Net-cell reaction:

$$\ce{LiC6 + CoO2 <=> C6 + LiCoO2}$$

Over-discharging supersaturates lithium cobalt oxide, leading to the production of lithium oxide.

$$\ce{Li+ + e- + LiCoO2 -> Li2O + CoO}$$

Similarly overcharging (up to $\pu{5.2 V}$) leads to following reaction:

$$\ce{LiCoO2 -> Li+ + CoO2 + e-}$$