Dealing with chemistry question?

1.
At electrode A (cathode) : 2H⁺(aq) + 2e⁻ → H₂(g)
At electrode B (anode) : 2I⁻(aq) → I₂(aq) + 2e⁻

2.
At electrode B, the colorless solution turns brown.

3.
The colorless solution at electrode A turns blue.
(This is because H⁺ ions are consumed and OH⁻ ions are left which turn litmus blue.)

4.
When dilute KI solution is used, colorless gas are involved at both the electrodes and the solution has no color change.
(This is because H⁺ and OH⁻ are preferentially discharged at electrodes A and B to give hydrogen and oxygen gases respectively.)

Not so easy