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)
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Galvanic Cell concept. You can view video lessons to learn Galvanic Cell. Or if you need more Galvanic Cell practice, you can also practice Galvanic Cell practice problems.