Problem: Current is applied to an aqueous solution of lithium iodide. What is produced at the cathode? a) H2(g) b) Li(s) c) O2(g) d) I2(s) What is produced at the anode? a) I2(s) b) Li(s) c) H2(g) d) O2(g)
FREE Expert Solution
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
Problem Details
Current is applied to an aqueous solution of lithium iodide.
What is produced at the cathode?
a) H2(g)
b) Li(s)
c) O2(g)
d) I2(s)
What is produced at the anode?
a) I2(s)
b) Li(s)
c) H2(g)
d) O2(g)