Question

the electrolysis of aqueous potassium iodide
anode
a power supply is connected to two graphite electrodes which are submerged into a beaker holding 1.0 m ki (aq) at 25 °c. hydrogen gas is produced at one of the electrodes.
review the potential standard reduction potentials for the electrolysis reaction to help answer the question.
what is produced at the anode?
water
oxygen
iodine
standard reduction potentials
half - reaction | e° (v)
k⁺ + e⁻ → k | - 2.92
2h₂o + 2e⁻ → h₂ + 2oh⁻ | - 0.83
i₂ + 2e⁻ → 2i⁻ | + 0.34
o₂ + 4h⁺ + 4e⁻ → 2h₂o | + 1.23

Explanation:

At the anode, oxidation occurs (loss of electrons). In aqueous KI, possible species to oxidize are \( \text{I}^- \) and \( \text{H}_2\text{O} \). The oxidation half - reactions (reverse of reduction) are: For \( \text{I}^- \): \( 2\text{I}^-
ightarrow \text{I}_2 + 2e^- \) (reverse of \( \text{I}_2+2e^-
ightarrow 2\text{I}^- \), \( E^{\circ}_{ox}= - 0.34\space V \)). For \( \text{H}_2\text{O} \): \( 2\text{H}_2\text{O}
ightarrow \text{O}_2 + 4\text{H}^++4e^- \) (reverse of \( \text{O}_2 + 4\text{H}^++4e^-
ightarrow 2\text{H}_2\text{O} \), \( E^{\circ}_{ox}=- 1.23\space V \)). The less negative (more positive) oxidation potential means the reaction is more favorable. Since \( - 0.34>-1.23 \), oxidation of \( \text{I}^- \) to \( \text{I}_2 \) (iodine) is more favorable at the anode.

Answer:

Iodine