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Iodide ions can be oxidized to form iodine. Write the balanced oxidation half-reaction for the oxidation of iodide to iodine.

User WhirlWind
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Final answer:

The oxidation half-reaction for iodide to iodine is 2 I- (aq) → I2 (s) + 2 e-. This reaction demonstrates the loss of electrons by iodide ions, indicating their oxidation to molecular iodine.

Step-by-step explanation:

The balanced oxidation half-reaction for the oxidation of iodide (I⁻) to iodine (I2) can be written as follows:

2 I⁻ (aq) → I2 (s) + 2 e−

This half-reaction shows that two iodide ions (I⁻) lose a total of two electrons to form molecular iodine (I2), which is the oxidation process. During this process, the oxidation state of iodine changes from -1 in I⁻ to 0 in I2. Since iodine is a relatively weak oxidizing agent, it allows for the selective titration of strong reducing agents.

User Obcure
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Answer:

Step-by-step explanation:

The oxidation of iodide ions (I⁻) to form iodine (I₂) involves the loss of electrons. The balanced oxidation half-reaction is as follows:

2I⁻ ->I₂ + 2e⁻

This reaction shows that two iodide ions lose electrons to form one molecule of elemental iodine (I₂) and two electrons.₋⁻

User Marco Magrini
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