Observe the diagram of the electrolytic cell (A aqueous solution of molten salt with a power supply). The molten salt is lithium flouride.

Write the half reaction occuring at the cathode, following the usual style guide and making sure to balance electrons.

At anode(+)electrode.

2F^- ==> F2
+ 2e
At the cathode (-) electrode
2Li^+ + 2e ==> 2Li

I don't have any idea what the "usual" style guide is.

To determine the half-reaction occurring at the cathode in the electrolytic cell with a molten salt of lithium fluoride (LiF), we need to consider the reduction of species present at the cathode.

In the electrolytic cell, the cathode is the electrode where reduction takes place. Reduction involves a gain of electrons, so we need to balance the electron transfer in the half-reaction.

Since the molten salt is LiF, we can identify the cation (Li+) and the anion (F-) present in the molten salt. At the cathode, the cation will be reduced to its neutral form, while the anion will not be involved in the reduction process.

The balanced half-reaction at the cathode for the reduction of Li+ can be determined as follows:

1. Write the ionic form of the species undergoing reduction:
Li+ + x e- → Li (neutral lithium)

2. Determine the value of x, the number of electrons required to balance the charge on the cation. Since Li+ has a charge of +1, it requires one electron for neutralization.

3. Multiply the half-reaction by the appropriate factor to balance the electrons. In this case, multiplying the equation by 1 will suffice, since we only need one electron to balance the charge.

The balanced half-reaction is:
Li+ + 1 e- → Li

Therefore, the half-reaction occurring at the cathode in the electrolytic cell with a molten salt of lithium fluoride (LiF) is:
Li+ + 1 e- → Li