Question

A common primary battery for pacemakers is the lithium–iodine cell. The negative electrode is lithium metal, the positive electrode is a paste made with I2 and a small amount of polyvinylpyridine (PVP), and the separator is the ionic salt LiI. The overall reaction is

2Li + I2 « 2LiI(s)

1. Write out the half-cell reactions and using the DG°f,LiI = -270.3 kJ mol-1 calculate the equilibrium potential and the theoretical capacity in A·h·g−1. You may treat the positive electrode paste as pure iodine.

2. Note that LiI is produced during discharge, and this salt adds to the thickness of the separator. The nominal current is 28 μA and is assumed constant over the life of the cell. How much active material is needed for a 5-year life, hint calculate capacity in A h? At 37 °C, the LiI electrolyte has an ionic conductivity of 4 × 10-5 S·m-1. If the separator is formed in place from the overall reaction, and LiI has a density of 3494 kg·m-3, what is the voltage drop across the separator due to ohmic losses in the separator after 2.5 years? The cell area is 13 cm2.

Answer 1