This question is about the development of lithium cells - AQA - A-Level Chemistry - Question 9 - 2021 - Paper 1

To calculate the cell EMF, we find the reaction of lithium and iodine. Considering the standard potentials, we can add lithium’s E° value and the E° for reduction of iodine. The EMF of the cell can be calculated as follows:

$E_{cell} = E°(Li^+/Li) + E°(I_2/I^-)$

Thus:

$E_{cell} = (-3.04 ext{ V}) + (0.54 ext{ V}) = -2.50 ext{ V}$

The discrepancy between the calculated EMF value of -2.50 V and the commercial EMF of 2.80 V could be attributed to non-standard conditions such as concentration, temperature, or the presence of stabilizing agents in the commercial cell that enhance the overall voltage produced.

In lithium perchlorate (LiClO4), lithium has a +1 oxidation state. The overall charge of the molecule is neutral, thus the oxidation state of chlorine can be calculated by considering that the four oxygen atoms each have a -2 oxidation state. Therefore, the oxidation state of chlorine must be +7 to balance the overall charge:

$+1 + x + (4 imes -2) = 0 \implies x = +7$

The reaction at the positive lithium cobalt oxide electrode can be represented as:

$LiCoO_2 + Li^+ + e^- \rightarrow Li_2CoO_2$

At the negative lithium electrode, the reduction reaction can be expressed as follows:

$Li^+ + e^- \rightarrow Li(s)$