To determine the cathode and anode, we need to compare the reduction potential (E˚red) for each half-reaction. These can be found in textbooks or online.
Li+(aq) + e– → Li(s) E˚red = –3.040 V
I2(s) + 2 e– → 2 I-(s) E˚red = 0.535 V
In an electrolytic cell, energy is being provided to the cell.
Therefore, the half-reaction with the more negative E˚red is designated as cathode/reduction.
The half-reaction with the less negative E˚red is designated as the anode/oxidation.
I2(s) + 2 e– → 2 I-(s) E˚red = 0.535 V → Anode/Oxidation
Li+(aq) + e– → Li(s) E˚red = –3.040 V → Cathode/Reduction
Current is applied to an aqueous solution of lithium iodide.
What is produced at the cathode?
What is produced at the anode?
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