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

To calculate the cell EMF, we observe the half-reactions. Assuming that lithium iodide is formed from lithium and iodine, we use the electrode potential for lithium:

• From Li+(aq) + e- → Li(s) : E° = -3.04 V
• From 1/2 I2(g) + e- → I-(aq) : (not given but assumed from standard values) let's assign E° = +0.54 V for the iodine half-reaction.

The total cell EMF (E°cell) can be calculated as follows:

$E°_{cell} = E°_{cathode} - E°_{anode} = 0.54 - (-3.04) = 3.58 ext{ V}$

The EMF value of a commercial lithium-iodine cell is 2.80 V, which is lower than the theoretical value calculated. This discrepancy could be due to non-standard conditions such as concentration or temperature differences, or internal resistance within the cell that reduces its effective EMF.

To deduce the oxidation state of chlorine in LiClO4, we consider that lithium (Li) has an oxidation state of +1, and each oxygen (O) has an oxidation state of -2. The oxidation state of chlorine (Cl) can be determined by evaluating the overall charge balance:

Let the oxidation state of Cl be x.

$+1 + x + 4(-2) = 0$
$x - 8 + 1 = 0$
$x = +7$

Thus, the oxidation state of chlorine in LiClO4 is +7.

At the positive lithium cobalt oxide electrode, lithium ions are released into the electrolyte as cobalt oxide gets reduced. The half-reaction can be written as:

At the negative lithium electrode, lithium ions gain electrons to form solid lithium. The reaction can be expressed as: