potassium iodide solution using carbon electrodes

The answer is D.

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Reason :
In the solution, potassium iodide give K+(aq) ion and I-(aq) ion, and water is slightly ionized to give H+(aq) ion and OH-(aq) ion.

Anode : I-(aq) and OH-(aq) ions migrate to the anode. Since I-(aq) is more concentrated than OH-(aq), I-(aq) ion is preferentially discharged to give iodine solution.
2I-(aq) → I2(aq) + 2e-

Cathode : K+(aq) and H+(aq) ions migrate to the cathode. Since K+(aq) ion cannot be discharge at the cathode, H+(aq) ion is preferentially discharged to give hydrogen gas.
2H+(aq) + 2e- → H2(g)

你呢條問題 問得好唔清楚
concentrated of ions will preventially discharge ge
咁我當佢係dilute ge話 答案就 C
當佢係concentrated 就 D
A, B一定唔會... 除非你用以mercury 做cathod ge electrode

c.
因為當進行個process 既時候,, 會出左potassium ions,, iodide ions,, hydrogen ions 同hydroxide ions... 起chemical cell 入面,, 黐住負極果邊會lose electrons,, 會lose electrons 既只會係iodide ions 同hydroxide ions... 根據electrochemical series,, hydroxide ions 既reducing power 係高過iodide ions,, 所以就會負極出oxygen...
( 4OH¯ →O2 + 2H2O+ 4e¯ )
而另一邊出hydrogen 係因為hydrogen ions 既oxidizing power 高過potassium ions..
( 2H+ + 2e¯ →H2 )