Answer to Question #153956 in Chemistry for sd

Question

2H+ (aq) + ClO- (aq) + 2I- (aq) → Cl- (aq) + I2 (aq) + H2O (l)          Equation 2

In equation 2 on the lab information sheet, what element is oxidized? What element is reduced? Justify using oxidation numbers. 

Answer 1

Solution:

2H⁺(aq) + ClO^-(aq) + 2I^-(aq) → Cl^-(aq) + I₂(aq) + H₂O(l) 

Oxidation is represented by an increase in oxidation number.

Reduction is represented by a decrease in oxidation number.

ClO^- + 2H⁺ + 2e^- → Cl^- + H₂O (oxidation)

2I^- - 2e^- → 2I⁰ (reduction)

- I^- was oxidized (Oxidation number of I: -1 → 0); I^- is the reducing agent.

- ClO^- was reduced (Oxidation number of Cl: +1 → -1); ClO^- is the oxidizing agent.

Answer:

I^- was oxidized (element - iodine, I);

ClO^- was reduced (element - chlorine, Cl).